FIITJEE Question Papers from Rankers Study Material for IIT JEE

FIITJCC RANKERS STUDY MATERIAL IIT-JEE, 2002 PHASE -1

CHEMISTRY Time: Two Hours

Maximum Marks : 100

Note: i)

There are NINE questions in this paper. Attempt ALL questions.

ii)

Answer each question starting on a new page. The corresponding question number must be written in the left margin. Answer all parts of a question at one place only.

iii)

Use of Logarithmic tables is permitted.

iv)

Use of calculator is NOT PERMITTED

Useful Data: Gas Constant

R

=

8.314 J mol"1K"1 0.0821 lit atm mol"1 K"1 2 Cal mol"1

=

=

Avogadro's Number

N

=

6.023 x 1023

Planck's constant

h

=

6.625 x 10"34 J sec.

Velocity of light

c

=

3 x 108 m sec"1

1 electron volt

ev F

=

1.6 x 10"19 J

=

96500 C

Atomic Masses

40, C = 12, O = 16, K = 39, Ag = 108, Mn = 55, Cr = 52, Ca CI = 35.5, N = 14, 4, S = 32, Na = 23, 2 H = 1, P = 31, I = 127, As = 75, Fe = 56, Ag = 108

Name of the Candidate Enrollment Number

(F!ITJ€€ ICES House, Sarvapriya

Vihur (Near Hauz Khas Term.), New Delhi - 16, Ph : 686 5182, 685 4102, Fax :

6513942

RSM12-P2-T(M)-CH-2

1.

a) A 2.36 g sample of dolomite containing CaC0 3 and MgC0 3 were dissolved in 700 ml_ of decinormal solution of HCI. The solution was diluted to 2.5L and 25mL of this solution required 20 mL 0.01 N NaOH solution for complete neutralisation. Compute the composition of the ore. [7] b) A sample of hydrazine sulphate (N 2 H 6 S0 4 ) was dissolved in 100 mL water 10mL of this solution was reacted with excess FeCI3 solution and warmed to complete the reaction. Ferrous ion formed was estimated and it required 20 ml of M/50 KMn0 4 solution. Estimate the amount of sulfate in one litre of solution. [7]

2.

3

A mixture contains NaCI and an unknown chloride MCI. i)

1 g of this is dissolved in water. Excess of acidified AgN0 3 solution is added to it. 2.567 g of a white precipitate is formed.

ii)

1 g of the original mixture is heated to 300°C. Some vapours come out which are absorbed in acidified AgN0 3 solution. 1.341 g of a white precipitate is obtained. Find the molecular weight of the unknown chloride. [10]

a) 1.6 g of Pyrolusite ore was treated with 50cm3 of 1.0 N, oxalic acid and some sulfuric acid. The oxalic acid left undecomposed was raised to 250 cm3 in a flask. 25cm3 of this solution when titrated with 0.1N KMn0 4 required 32cm3 of the solution. Find out the percentage of pure Mn0 2 in the sample and also the percentage of available oxygen. [6] b) A mixture of pure K 2 Cr 2 0 7 and pure KMn0 4 weighing 0.561 g was treated with excess of Kl in acid medium. Iodine liberated required 100 mL of 0.15 N of hypo solution for exact oxidation. What is the percentage of each in the mixture? [8]

4.

Consider the following two possibilities for electron transition in a hydrogen atom,pictured below. i)

The electron drops from the Bohr orbit n = 3 to the orbit n = 2, followed by the transition from n = 2 to n = 1

ii) The electron drops from the Bohr orbit n = 3 directly to the orbit n = 1 a) show that the sum of the energies for the transitions n =3 n = 2 and n = 2 n = 1 is equal to the energy of transition n = 3 -> n = 1

\

n=1

n=2

n=3

b) Are either wavelength (or) the frequencies of the emitted photons additive in the same way as their energies? Explain? [5] 5.

a) If uncertainties in the measurement of position and momentum are equal, calculate uncertainty in the measurement of velocity. [4] b) What accelerating potential is needed to produce an electron beam with an effective wavelength of 0.090A? [4]

6.

a) Calculate the percentage of 'free volume' available in 1 mol gaseous water at 1.00 atm and 100°C. Density of liquid water at 100°C is 0.958 g/mL [4] b) At room temperature the following reactions proceed nearly to completion: 2NO + 0 2 » 2N0 2 > N204 the dimer N 2 0 4 sdolidifies at 262K. a 250 mL flask and a 100 mL flask are separated by a stop cock. At 300K, the nitric oxide in the larger flask exerts a pressure of 1.053 atm FIITJCC ICES House, Saruapriya

Vihar (Near Hauz Khas Term.), New Delhi - 16, Ph : 686 5182, 685 4102, Fax : 6513942

RSM12-P1 - T ( M )-C H - 3

and the smaller one contains oxygen at 0.789 atm. The gases are mixed by opening the stop cock and after the end of the reaction the flasks are cooled to 220K. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at 220K. (Assume that the gases behave ideally). [6] 7.

a) 5.0L of water is placed in a closed room of volume 2.5 x 104L and having temperature 300K. If vapour pressure of H 2 0 is 27.0 mm and density is 0.990 gem -3 at this temperature, how much water is left in liquid state? [4] b) At 1200°C, molecular chlorine dissociates as Cl2(g) ^ 2CI(g). At 1200°C and constant pressure, the mixture effuses1.16 times as fast as S0 3 effuses. Calculate the degree of dissociation of Cl2. [6]

8.

a) At 21,5°C and a total pressure of 0.0787 atm, N 2 0 4 is 48.3% dissociated into N0 2 . Calculate Kc for the reaction N 2 0 4 (g) 2N0 2 (g). At what total pressure will the percent dissociation be 10%. [6] b) A container whose volume is V contains an equilibrium mixture that consists of 2 mol each of PCI5, PCI3 and Cl2 (all as gases). The pressure is 30.3975 kPa and temperature is T. A certain amount of Cl2 (g) is now introduced keeping the pressure and temperature constant until the equilibrium volume is 2V. Calculate the amount of Cl2 that was added and the value of Kp. [8]

9.

a) A 0.20 ml sample of solution containing 1.0 x 10"7Ci ( 1Ci = 3.7 x 1010 dis/sec) of 3 H is injected into blood stream of an animal. After sufficient time, 0.10 ml of animal blood is found to have activity of 20 dis/minute. Calculate the blood volume of the animal. [4] b) In a particular experiment the specific rate constant for the decomposition of gaseous N 2 0 5 is 1.68 x 10"2s~1 at a particular temperature. If 2.5 moles of N 2 0 5 gas are placed into 5 L container at the temperature, how many moles of N 2 0 5 would remain after 1 minute? How much oxygen would have been produced? [5] c) An optically active drug has one chiral center and only dextrorotatory isomer is effective. Moreover, it becomes ineffective when original activity is reduced to 35% of original. The rate constant is 1 x 10-8 sec -1 . Find the expiration time of drug in years. [6]

FIITJCC ICES House, Sarvapriya

Vihar (Near Hauz Khas Term.), New Delhi - 16, Ph : 686 5182, 685 4102, Fax :

6513942

FIITJCC Rankers Study Materiel NT - JEE 2002 PHASE TEST - 1

MATHEMATICS Time : Two hours

Maximum Marks:100

NOTE : 1.

This paper consists of twelve questions only

2.

Attempt All questions.

3.

Marks for question or its sub-questions are shown in the right hand margin.

4.

Use of Calculator is NOT

Name of the Enrollment

PERMITTED

student Number

FIITJCC. ICES HOUS£'(Opp. VIJAY MANDAL ENCLAVE), SARVAPRIYA VIHAR, NEW DELHI - 16. Ph:6854102, 6865182, FAX: 6513942

R S M 1 2 - P 1 -T(M)-M A-2

1

Two of the vertices of the triangle formed by the tangents to the parabola y2 = 4ax lie on the rectangular hyperbola xy = c2. Prove that the remaining vertex also lies on the hyperbola.[8]

/2.

Find the equation of the circum-circle of the triangle formed by the lines y2- 4xy + 3x2 = 0 and x + 2y = 1. [8] a, b, c are three positive and distinct numbers and a is harmonic mean of b and c. Prove that one of the solutions of ax2 + bxy +cy2=bx2 +cxy +ay2 = d is j X

dc(b + c) 2

_ I 2

^2b(b +bc + c ) '

y

db(b + c) 2

]j2c(b +bc + c 2 )

[ 1

the length of the focal chord of the parabola x2= 4cy which touches the ellipse y2 Z_2 = 1 ( b < c ) . [8] b

/A.

Find x2 _2 + a

f 5.

Given that a, b, c are positive distinct real numbers such that quadratic expressions ax2 + bx + c, bx2 + cx + a and cx2 + ax + b are always non-negative, prove that the a2 + b2 + c2 expression can never lie in (—ao, i ) u [4,a>). [81 ab + bc+'ca

-r6.

A circle is made to pass through the point (1,2), touching the straight lines 7x = y +5 and x + y + 13 = 0. Find the area of the quadrilateral formed by these tangents and the corresponding normals, of the circle with smaller radius. [8]

7.

Tangents are drawn from a point on the line x + 4 = 0 to the parabola x2 = 4y. Find the locus of a point P which divides the length of tangents in the ratio 1:2 internally. [8] 2

8.

X V Tangents are drawn from a point. P to the ellipse —2 + ~ = 1. The line joining the points of a b contact subtends an angle 45° at the centre of the ellipse. Prove that the locus of P is (b4x2 + a 4 y 2 - a2b4 - b2a4 f - 4a4b4 (b2x2 + a2y2 - a2b2) = 0

9.

[8]

If a line is such that its intercepts on the co-ordinates axes are e and e' where e and e' are equal to the eccentricities of a hyperbola and its conjugate hyperbola, then prove that this line always touches a fixed circle. [8] 1

r=

V10 yf50_ l(n2 + l)(n2 + 2 - 2 n ) 1 1 + sec' 3 + sec" 7 +, . . + sec" J ( n 2 _ n + -|J

10.

Sum the series sec" ^

11.

If x and y are real variables satisfying x2 + y2 +8x - 1 Oy +40 = 0 , and a = max.[(x+2)2 +(y —3)2], b = min.[(x +2)2 +(y -3) 2 ] then prove that a +b = 18.

12.

Find the values of 'a' for which the equation (x2 +4x +7)2 - ( a -2)(x 2 +4x +7) (x2 +4x +6)+(a -3) (x2 +4x +6)2 = 0 has at least one real root.

FIITJCC.

ICES HOUS£'(Opp. VIJAY MANDAL ENCLAVE), SARVAPRIYA VIHAR, NEW DELHI - 16. Ph:6854102, 6865182, FAX: 6513942

FIITJCC Rankers Study Material IIT - JEE 2002 P H A S E - I

TEST

PHYSICS Time : Two Hours

Maximum Marks : 100

NOTE : 1.

Attempt ALL questions.

2. There is NO NEGATIVE

MARKING.

3. Use of logarithmic tables is

PERMITTED.

4. Use of Calculator is NOT PERMITTED.

5

Useful Data : Acceleration due to gravity Density of water

g p

= 1 0 m/s 2 = 1000 kg/m 3

Name of the student Enrollment Number


RSM12-P1 -T-PH-2

A particle is projected with a velocity of 20m/s at an angle of 30° to an inclined plane of inclination 30° to the horizontal. The particle hits the inclined plane at an angle of 30°. During its downward journey. Find the (a) time of hitting (b) the height of the point of hitting from the horizontal plane passing through the point of projection. [5+5] 2.

A car accelerating at the rate of 2m/s2 from rest is carrying a man at the rear end who has a gun in his hand. The car is always moving along +Ve x-axis. At t = 4s, the man fires from the fun and the bullet hits a bird at t = 8 s. The bird has a position vector 40i + 80j+40k find velocity of projection of the bullet with respect Take y-axis in the horizontal plane.

[10]

A particle in moving with a velocity of = (3 + 6t + 9t2) cm/s. (a) find out the acceleration of the particle at t = 3second. (b) find out the displacement of the particle in the interval t = 4 sec to t = 6 sec. (c) Find out ht average velocity of the particle in the interval t = 5 second to t = 8 second. [3+3+4] 4 / (I) Two masses mi = 10 kg and m2 = 5 kg are connected by an ideal string as shown in the figure. The coefficient of friction between mi and the surface is jlx = 0.2. Assuming that the system is released from rest, Calculate the velocity when m2 has descended by 4m . (ii/A block of mass m is attached to two unstretched springs of spring constant ^ and k2 as shown in figure. The block is displaced towards right through a distance 'x' and is released. Find the speed of the block as it passes through a distance x/4 from its mean position.

5.

Three blocks A, B and C have masses 1kg , 2kg and 3 kg respectively are arranged as shown in figure. The pulleys P and Q are light and frictionless. All the blocks are resting on a horizontal floor and the pulleys are held such that strings remain just taut. At moment t = 0 a force F = 401 Newton starts acting on pulley P along vertically upward direction as shown in figure. Calculate (i) the times when the blocks lose contact with ground. (ii) the velocities of A when the blocks B and C loses contact with ground. (iii) the height by which C is raised when B loses contact with ground

[5]

JUULMISL^ m JUUULUM/

[5]

A B C [ 3 + 3 + 4 = 10]


1

RSM12-P1 -T-PH-3

3-

In a car race, car A takes a time t sec less than car B at the finish and passes the finishing point with speed v m/s more than the car B. Assuming that both the cars starts from rest and travel with constant acceleration ai and a2 respectively. Show that v=( A /aja^) t [10] On a cricket field the batsman is at the origin of co-ordinates and a fielder stands in position (46 i + 28 j ) m. The batsman hits the ball so that it rolls along the ground with constant velocity (7.5 i + 1 0 j ) m/s. The fielder can run with a speed of 5 m/s. If he starts to run immediately the ball is hit, what is the shortest time in which he could intercept the ball ? [10] A smooth fixed wedge has one face inclined at 30 to the horizontal and a second face at 45° to th horizontal. The faces are adjacent to each other a the top of the wedge. Particles of masses 2m an 5m are held on these respective faces connected b a taut inelastic string passing over a smooth pulle at the top of the wedge as shown in the figure. Fin the acceleration of the system if the particles ar simultaneously released and show that the fore

10

1

acting on the pulley is — mg (1 + V2) cos ( 5 2 - ) ° . 9.

A small ball is suspended from point O by a thread of length /. A nail is driven into the wall at a distance of 1/2 below O, at A. The ball is drawn aside so that the thread takes up a horizontal position at the level of point O and then released. Find (i) At what point to the ball trajectory, will the tension in the thread disappear. (ii) What will be the highest point to which it will rise? 5+5=10]

10.

There are two parallel planes each inclined to the horizontal at an angle a. A particle is projected from a point mid-way between the two planes so that it grazes one of the planes and strikes the other at right angle. Find the angle of projection. [10]

* * *

FIITJCC. ICES HOUS£'(Opp. VIJAY MANDAL ENCLAVE), SARVAPRIYA VIHAR, NEW DELHI - 16. Ph:6854102, 6865182,FAX:6513942

FIITJCC Rankers Study Material IIT - JEE 2002 PHASE-II TEST CHEMISTRY Time: Two Hours

Maximum Marks: 100

Note: 1.

Each questions carries EQUAL MARKS

2.


3.

There is NO NEGATIVE MARKING.

4.

Use of logarithmic tables is PERMITTED.

5.

Use of Calculator is NOT PERMITTED.

6.

Useful Data : Gas Constant

R =

8.314 J mor1K~1 0.0821 lit atm mol -1 K~1 2 Cal mol"1

Avogadro's Number

N

=

6.023 x 1023

Planck's constant

h

=

6.625 x 10~34 J sec.

Velocity of light

c

=

3 x 108 m sec -1

1 electron volt

ev = F

Atomic Masses

1.6 x 10"19 J

=

96500 C Ag = 108, Mn = 55, Cr = 52, Ca = 40, C = 1 2 , 0 = 16, K = 39, CI = 35.5, N = 14, S = 32, Na = 23, H = 1, P = 31, I = 127, As = 75

/

Name of the Candidate Enroll



6513942

RSM12-P2-T(M)-CH-2

1.

The orthorhombic lattice of sulphur is shown here, a = 10.46 A, b = 12.87 A, c = 24.49A. This unit cell contains 128 atoms of sulphur (atomic weight = 32). Calculate the theoretical density [10]

The enthalpy change for the reaction C2H4(g) + H 2 0(g) ->C2H5OH(g) is—11.21 kcal/mol. The difference in the enthalpies of formation of C2H5OH(g) and H 2 0 (g) is 1.288 kcal/mol. Calculate the enthalpy of formation of ethylene. The enthalpy of formation of C2H5OH bears the ratio 0.9777 to the enthalpy of formation of H 2 0 (g) . Both these enthalpies have negative values. Calculate them separately. [10] The solution of 1 mole of CuS0 4 , 1 mole of CuS0 4 .H 2 0 and 1 mole of CuS0 4 .5H 2 0 in 800 moles of water is accompanied by the liberation of 15.90, 9.33 and the absorption of 2.80 kcal respectively. Calculate the enthlpy change for the following processes.

CuS0 4

H,0

-> CuS0 4 .H 2 0; CUS0 4 .H 2 0

> CuS0 4 .5H 2 0

[10]

Calculate the enthalpy change for the reaction CH4(g) + Cl2(g) = CH3CI(g) + HCI(g), given the following CH4(g) + 20 2 (g) = C0 2 (g) + 2H 2 0(I); AH, = -212.79 kcal/mol CH3CI(g) + | o 2 ( g ) = C0 2 (g) + H 2 0(l) + HCI(g); AH2 = - 164 kcal/mol H2(g) +

(g) = H 2 0(l); AH3 = -68.317 kcal/mol [10]

(g) + l c i 2 (g) = HCI(g); AH4 = -22.06 kcal/mol 5.

The saturated vapour pressure of a certain liquid is expressed by the equation log p(mm) = 798 5 ^ 6.857 where T is in degree Kelvin. Calculate its molar latent heat of T vazpourization. Calculate also the normal boiling point. [10] When Wi g of a non-volatile solute are dissolved in w2g of a solvent, the relative lowering of w Mt vapour pressure is R. If the ration — L is r and the ratio of the molar masses — = m, show Wo M, that -1 = 1 + — . Calculate the number of grams of nonvolatile solute dissolved in a kilogram R r of ethanol or R = 0.022. Molar mass of the solute = 61,33g. [10] Two solutions, one of a compound S in water and the other an aqueous solution of urea were allowed to attain isopiestic equilirbium in a closed system see Figure. It was observed that when both solutions were under equilibrium vapour pressure, solution 1 contained 2% of S by weight and solution 2 contained 5% of urea by weight. Calculate the molar mass of S.

1. Solution of s 2. Solution of urea

[10] Faust and Montillion reported that in the electrolysis of a solution containing K3[Cu(CN)4], K2[Ni(CN)4] and K2[Zn(CN)4] a cathodic deposit was obtained containing 72.8% by weight of copper, 4.3% by weight of nickel and 22.9% of zinc by weight. If the deposit had a weight of 0.175g, how many coulombs were passed through the solution? Assume that no hydrogen was evolved. Atomic weights: Cu = 63.6, Ni = 58.7 and Zn = 65.4. [10] FIITJCC ICES House, Saruapriya

Vihar (Near Hauz Khas Term.), New Delhi - 16, Ph : 686 5182, 685 4102, Fax :

6513942

RSM12-P2-T(M)-CH-3

9.

i)

The cell Pt

Fe3+ Cu+ Fe

2+

Cu

2+

Pt in which [Fe3+] = [Cu+] and [Fe2+] = [Cu2+] exhibits an emf.

Assuming that [Fe3+] » [Fe2+] i.e., [Cu+] is » [Cu2+], what chemical changes may be expected, if the two platinum electrodes are externally connected under equilibrium the cell emf would be zero? ii) What then would be the electrode potential for the Pt electrode in each half-cell? Given l 2+, Cu+) +0.17V [6 + 4 = 1 0 ] FeJ\ Fe = +0.77V, E(cu 10.

Given that 2 x 10~4 mol each of Mn2+ and Cu2+ was contained in 1L of a 0.003M HCI0 4 solution and this solution was saturated with H2S. Determine (i) whether or not each of these ions, Mn2+ and Cu2+, will precipitate as sulphide (ii) how much Cu2+ escapes precipitation (iii) If the solution is made neutral by lowering the [H+] to 10-7, will MnS precipitate. If it precipitates calculate the percentage precipitation. Given solubility of H2S is 0.1 mol I -1 (assumed to be independent of the presence of other materials in the solution). Ksp (MnS) = 3 x 10~14, Ksp (CuS) = 8 x 10"37 K, and K2 for H2S are 1 x 10~7 and 1.3 x 10~14 respectively. [10]

FIITJCC ICES House, Saruapriya


FIIIJCC Rankers Study Materiel I IT - J EE 2002 PHASE T E S T - I I


Maximum Marks: 100

NOTE : 1.

This paper consists of twelve questions only.

2.


3.


4.


PERMITTED.

Name of the student Enrollment

Number

FIITJCC, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya

Vihar, New Delhi - 16. Ph: 6865182, 6854102. Fax: 6513942

RSM12-P2-T(M)-MA-2

1.

Show that -a < cos6(sin0 + Vsin2 0 + sin2 a ) < a, where a = Vl + sin2 a .

2.

Functions f(x) and g(x) are defined in [a, b] such that f(x) is monotonically increasing while g(x) is monotonically decreasing. It is given that the range of f(x) as well as that of g(x) are subsets of [a, b]. Find the range and domain of h(x) = fog(x) + gof(x). [8] J_

_1_ +

3X

1 +

[10]

l/x

3.

X Evaluate Lt 2 x->0

"'10X

4.

Evaluate lim

5.

Prove that there exist exactly three non-similar isosceles triangles ABC such that tanA + tanB + tanC = 100. [8]

6.

j e t a n 0 (sec 0 - sin 0) d9.

7.

Let f be a real valued function satisfying f(x) + f(x + 4) = f(x + 2) + f(x + 6). Prove that x+8 Jf(t)dt is a constant function. [8]

[8]

(1 + x ) 1 / x + e ( x - 1 )

x->0

[8]

sin - 1 x

[8]

8.

1 J| 1/x Prove the identity f — 2 f -HL. (x>0) J 1+t J 1 + t2

9.

Find the ratio in which y = V 5 - x 2 divides the area of the region bounded by

[8]

x2 - 4y = 0 and 2x - y = 0.

[8]

10.

Solve the differential equation y4dx + 2xy3dy =

11.

Find the range of the function f(x) =

12.

Let f be the inverse function of f, where f'(x) = 1 +x3 and f(0) = 1. Prove that

~^ ^ . x y

x4 -V2x + 2 x 4 - V 2 x +1

[8]

[8]

[10]

g(x) = J — ! -dt i 1 +g(t Q )r

*

"k

FIITJCC, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya

ic

Vihar, New Delhi - 16. Ph: 6865182, 6854102. Fax:

6513942

FIITJCC Rankers Study Material i i t - JEE 2002 P H A S E - I I

TEST


Maximum Marks : 100

NOTE : 1. Attempt ALL questions.

2. There is NO NEGATIVE

MARKING.

3. Use of logarithmic tables is

4. Use of Calculator is NOT

PERMITTED.

PERMITTED.

5. Useful Data :

Acceleration due to gravity Density of water

g p

= 10 m/s 2 ,3 = 1000 kg/m :



RSM12-P2-T-PH-2

A small sphere of radius r is released from point 'A' inside the fixed large hemispherical bowl of radius R as shown in figure. If the friction R between the sphere and the bowl is sufficient enough to prevent any slipping, then find (a) what fractions of the total energy are translational and rotational , when the small sphere reaches the bottom of the hemisphere. (b) and the normal force exerted by the small, sphere on the hemisphere when it .is at the bottom of the hemisphere. [10] A wire of uniform cross-section is stretched between two points 1m apart. The wire is fixed at one end and a weight of 9 kg is hung over a pulley at the other end, produces fundamental frequency of 750 Hz. (a) What is the velocity of transverse waves propagating in the wire. (b) If now the suspended weight is submerged in a liquid, of density (5/9) that of the weight, what will be the velocity and frequency of the waves propagating along the wire? [10] A pipe of length 1.5m closed at one end is filled with a gas and it resonates in its fundamental with a tuning fork. Another pipe of the same length but open at both ends is filled with air and it also resonates in its fundamental with the same tuning fork. Calculate the velocity of sound at Q°C in the gas. Given that the velocity of sound in air is 360 m/s at 30°C where the experiment is performed.

[10] A ball is suspended by a thread of length L at the point O on the wall PQ which is inclined to the.yertical by an angle a. The thread with the ball is now displaced through a small angle p away from the vertical and also from the wall. If the ball is released, find the period of oscillation of the pendulum when (a) P < a (b) a > p. Assume the collision on the wall to be perfectly elastic.

[10]

Find the time period of oscillation for the arrangement shown in the figure. The pulleys are smooth and massless.

FIITJCC, ICES HOUSE (Opp. V1JAY MANDAL ENCLAVE), SARVAPRIYA VIHAR, NEW DELHI -16. Ph:6854102, 6865182, FAX: 6513942

RSM12-P2-T-PH-3

Two identical spherical bodies A & B rolling without slipping in opposite directions with speeds v & 2v on a rough horizontal plane collide elastically. The coefficient of static friction between the sphere and the horizontal surface is ^ . Obtain the velocities of the spheres when they have started pure rolling again .

[10]

7.

Two small dense stars rotate about their common centre of mass as a binary system with the period of 1 year for each. One star is of double the mass of the other and the mass of the lighter one is of (1/3) the mass of the sun. Given the distance between the earth and the sun is R. If the distance between the two stars is r, then obtain the relation between r and R. [10]

8.

Using 4 moles of air and assuming this is an ideal diatomic gas, an engine goes through the following reversible changes in one cycle. (i) Isothermal expansion from a volume of 0.02 m3 to 0.05 m 3 at100°C. (ii) At constant volume , cooling to 27°C. (iii) Isothermal compression at 27°C to 0.02 m 3 volume . (iv) At constant volume, compression to original pressure , volume and temperature. (a) Show this cycle in a PV diagram. (b) Calculate efficiency of the cycle. [10]

9.

One mole of an ideal gas whose pressure changes with volume P = aV, where a is a constant, is expanded so that its volume increases r\ times. Find the change in internal energy and heat capacity of the gas. [10]

10.

A thin metal pipe of 1 m length and 1 cm radius carries steam at 100°C. This is covered by two layers of lagging. The thermal conductivity of outer layer which is 2 cm thick is 3.6 x 10"4 cal.cm"1 deg"1s"1 and that of inner layer is 2.4 x 10'4 cal.cm"1 deg"1s"1 . If the outer surface of the lagging is at 30°C, find the temperature of the cylindrical interface of the two lagging materials.

*

*

2 cm

IT,

Z 1 cm

30°C

[10]

*

FIITJCC, ICES HOUSE (Opp. V1JAY MANDAL ENCLAVE), SARVAPRIYA VIHAR, NEW DELHI -16. Ph:6854102, 6865182, FAX: 6513942

FIITJ€€ RANKERS STUDY MATERIAL IIT-JEE, 2002

PHASE I II CHEMISTRY Time: Two Hours

Maximum Marks: 100

Note: i)

There are TWELVE questions in this paper. Attempt ALL questions.

ii)


iii)


iv)



R =

8.314 J mor1K~1 0.0821 lit atm mol"1 K"1 2 Cal mor 1

Avogadro's Number

N =

6.023 x 1023

Planck's constant

h

=

6.625 x 10~34 J sec.

Velocity of light

c

=

3 x 10s m sec -1

1 electron volt

ev =

1.6 x 10-19 J

F

96500 C

Atomic No:

=

Ca = 20, C = 6, O = 8, K = 19, CI = 17, F = 9, N = 7, S = 16, Na = 11. Cu = 29, Co =27, Mn = 25, Y = 39, Zr = 40, Nb = 41, La = 57, Hf = 72, Ta = 73.

Atomic Masses: Ag = 108, Mn = 55, Cr = 52, Ca = 40, C = 12, O = 16, K = 39, CI = 35.5, N = 14, S = 32, Na = 23, H = 1, P = 31, I = 127, As = 75, Fe = 56, Ag = 108

Name of the Candidate : Enrollment Number

FIITJCC, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Ph: 6865182, 6854102. Fax: 6513942

RSM12-P1-2-T(M)-CH-2

1.

a) A certain liquid has a viscosity of 104 poise and a density of 3.2g/mL. How long will it take for a platinum ball with a 2.5 mm radius to fall 1 cm through the liquid? The density of platinum is 21.4 g/cc [5] b) At 18°C, the E.M.F. of the following concentration cell is 0.51V A g | ^ A g N 0 3 1 | saturated AgCI in 1N KCI|Ag N If the degree of ionisation of -ÂgNOs and 1N KC! are 0.813 and 0.76 respectively, calculate the solubility product and solubility of silver chloride in pure water.

2.

[6]

a) 0.1 mole of H2 and 0.2 mole of C0 2 are introduced in an evacuated flask at 723 K and the following reaction occurs to an equilibrium pressure of 50.60 kPa. H2(g) + C0 2 (g) ^ H 2 0(g) + CO(g) ...(1) Analysis of the mixture shows that it contains 10 mole % of H 2 0. A mixture of CoO (s) & Co(s) is then introduced such that the additional equilbria (2) & (3) are established. CoO(s) + H2(g) Co(s) + H 2 0(g) ... (2) CoO(s) + CO(g) Co(s) + C0 2 (g) ... (3) Analysis of the new mixture thus obtained is found to contain 30% mole H 2 0. Calculate the standard equilibrium constants Ki, K2, K3 for the reactions (1), (2) & (3) respectively.

[8] b) A large cylinder of helium filled at 2000 Ibf/in. had a small thin orifice through which helium escaped into an evacuated space at the rate of 6.4 mmol/hr. How long would it take for 10 mmol of CO to leak through a similar orifice if the CO were confined at the same pressure? [5] 3.

For CdS, the solubility product is 7 x 10~25 Will CdS be precipitated from a solution of cadmium salt containing 1.0 mol/m3 Cd2+ saturated with H 2 S gas (equilibrium concentration 100mol/m3 H2S) at pH = 0? The first and second dissociation constants of H 2 S are 10~7 and 10"14 respectively. What will happen at pH = 2? [5]

4.

1 gm of an impure sample of Iron pyrites (FeS2) is treated with 100 ml barium permanganate solution in acidic medium where FeS2 oxidised into Fe3+ and SO;;". The remaining permanganate solution was treated with excess KI in acidic medium and l2 liberated required 050 ml of 0.12 M solution of barium thiocyanate solution. In another case 1.5 gm of same Iron pyrites sample requires 10 ml of 0.35 M KMn0 4 solution in acidic medium. Calculate the percentage purity of FeS2 and molarity of Ba(Mn0 4 ) 2 solution. [Atomic weight of Fe = 56 and S = 32] [10]

5.

Standard free energy change (AG0) of the esterification reaction of n-propanoic acid and ethanol is 440 cal at 25°C according to the following reaction. CH3CH2COOH(l) + C2H5OH(l) CH3CH2COOC2H5(l) + H 2 0(l) Calculate the degree of esterification if an equimolar mixture of n-propanoic acid and C2H5OH are allowed to attain equilibrium and also calculate the number of moles of the ester produced if 0.5 moles of acid added with 0.5 moles of C2H5OH. [6]

6.

A solution contains a mixture of Ag+ (0.10M) and Hg2+ (0.10M) which are to be separated by selective precipitation. Calculate the maximum concentration of I" ion at which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated ? Ksp. (Agl) = 8.5 x 10~17 and Ksp (Hg2l2) = 2.5 x lO"26 [6]

7.

A salt MiM 2 C 4 H 4 0 6 , 2H 2 0 is heated until to get a constant weight. On constant ignition, it is converted into M 1 M 2 C0 3 . After complete burning the resultant solid was treated with 45 ml 0.2 l\ZhH2S04. The resulting mixture was back titrated by 0.1 M NaOH. If the volume of the NaOH required for back titration is 2.39ml and initial wt. of the sample is 10.732 gm then calculate the percentage purity of the sample. Consider Mi and M2 are two metals of valency one. Atomic wt. of Mi =23, M2 = 39, C = 12, O = 16. [10]

FIITJCC, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi -16. Ph: 6865182, 6854102. Fax: 6513942


8.

A mixture of CO2, N2 and water vapour is introduced into a flask which contains solid dry CaCI2. Immediately after introduction of the gas mixture the pressure of the flask is observed 765mm. Now the gas mixture is cooled to 0°C. After sometime the pressure reached a steady value 645mm. Then through the gas mixture potassium hydroxide is passed and pressure of the gas falls to 345 mm at 25°C. If the initial temperature of the gas mixture was 35°C, then calculate the composition in mole % of original mixture. Neglect the volume change of the flask. [8]

9.

A stationary He+ ion emitted a photon corresponding to the first line (Ha) of the Lyman series. That photon liberated a photo electron from a stationary H atom in ground state. What is the velocity of photo electron? (RH = 109678 cm -1 .) [8]

10.

A small amount of solution containing Na24 radio nucleide with activity A = 2 x 103 dps was administered into blood of a patient in a hospital. After 5 hours, a sample of the blood drawn out from the patient showed an activity of 16 dpm per c.c. t1/2 for Na24 = 15 hr. Find: a) Volume of the blood in patient b) Activity of blood sample drawn after a further time of 5 hr. [3+3]

11.

a) One way of writing the equation of state for real gases is PV = RT

where B is V a constant. Derive an approximate expression for B in terms of Vander Waal's constant 'a' and 'b'. [5]

(b) The half life period of ^ C e is 29.82 days. It is a p - emitter and the average energy of the p-particles emitted is 0.4132 MeV. What is the total energy emitted per second in watts by 10 mg of ^ C e ? [6] 12.

An element A (Atomic wt - 100) having bcc structure has unit cell edge length 400 pm. Calculate the density of A and number of unit cells and number of atoms in 10 gm of A. [6]

FIITJCC, ICES House, (Opp. Vijay Mandal Enclave),

Sarvapriya

Vihar, New Delhi -16.

Ph: 6865182, 6854102. Fax:

6513942

"X

finJ€i RANKERS STUDY MATERIAL

IIT - JEE, 2002 PRACTICE TEST - PHASE-I, II MATHEMATICS Time: Two hours

Maximum Marks: 100

Note: i).

This paper consists of TEN questions only.

ii).


iii).


iv).



Candidate Number

fH?J€€, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi -16. Ph: 6865182, 8854102. Fax: 6513942

RSM12-PT-PH-I, II (M)-MA-2

Let f(x) = Va 2 cos 2 x + b 2 sin 2 x + Va 2 sin 2 x + b 2 cos 2 x, (a * b) Then find the range of f(x).

1.

Discuss the continuity of the function: f(x)

f[x]+Vx-[x]

x>0

sinx

x<0

[10] [10]

X

3.

Find out the least value of the function F (x) = J (4sint + 3cost)dtin

n 3n

[10]

4'T

Evaluate /

71

cosec 2x

(a). J-

V

sin 2x V

\

tan dx

7T

(b). J

2

3

^ 71 v4

X

^ ,

dx.

[5+5]

2

cos xVtan x + tan x + tanx

vj

Find the equation of the circum-circle of the triangle formed by the lines y2- 4xy + 3x2 = 0 and x + 2y = 1. [10] a, b, c are three positive and distinct numbers and a is harmonic mean of b and c. Prove that one of the solutions of ax2 + bxy +cy2=bx2 +cxy +ay2 = d is dc(b + c) I db(b + c) x= [10] 2 2 Y V2b(b +bc + c ) ' y 2c(b2 +bc + c 2 ) ' 7.

Given that a, b, c are positive distinct real numbers such that quadratic expressbns ax2 + bx + c, bx2 + cx + a and cx2 + ax + b are always non-negative, prove that the a2 + b2 + c2 expression can never lie in (-oo,1) u [4,oo). [10] ab + be + ca

8.

A circle is made to pass through the point (1,2), touching the straight lines 7x = y +5 and x + y + 13 = 0. Find the area of the quadrilateral formed by these tangents and the corresponding normals, of the circle with smaller radius. [10]

9.

Tangents are drawn from a point on the line x + 4 = 0 to the parabola x2 = 4y. Find the locus of a point P which divides the length of tangents in the ratio 1:2 internally. [10]

10.

Find the values of 'a' for which the equation (x2 +4x +7)2 - ( a -2)(x 2 +4x +7) (x2 +4x +6)+(a - 3 ) (x2 +4x +6)2 = 0 has at least one real root. [10]

* * * *

FIITJCC, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi -16. Ph: 6865182, 6854102. Fax: 6513942

FIITJCC RANKERS STUDY MATERIAL IIT - JEE, 2002 PRACTICE TEST PHASE -1, II PHYSICS Time: Two hours

Maximum Marks: 100

Note: i).


ii).


iii).


iv).

Use of Calculator is NOT PERMITTED

Name of the Candidate Enrollment

:

Number


RSM12-PT-PH-L-LL ( M ) - P H - 2

1.

Distance between the centers of two stars is 10 a. The mass of these stars are M and 16M and their radii a and 2a respectively. A body of mass m is fired straight from the surface of the larger star towards the smaller star. What should be the minimum initial speed to reach the surface of the smaller star. [10]

2.

In the shown figure mass of the pulley is m and radius 2R. A light concentric spool of radius R is rigidly attached with the pulley. Two blocks' A and B having masses m & 4m -respectively, are attached with the pulley by means of light strings. Lower surface of the block B is attached to a spring Of- ••• stiffness k and block B rests on asmooth inclined plane inclined 30° with horizontal. Other end of the spring is fixed to the-ground , Initially the blocks are held at a height such that spring is in relaxed position . The block A is streatched a small distance and released then find the frequency of the , w oscillation of block B. (The strings do not slip on the pulley}.

3.

4.

[10] -v-V? • ; , - .

v.,7-;

An open tank containing a non-viscous liquid tp a height of 5m is placed over the ground. A heavy spherical ball falls from height 40 m oVer'the ground in the tank. Ignoring air resistance find the height to which ball will go back. Collision between bail and bottom of tank is perfectly elastic. [10] An adiabatic cylinder of length 1m and cross-sectional area 100 cm2 is closed at both ends. A freely moving non conducting thin piston divides the cylinder in to two equal parts. The piston is connected with right end by a spring having force constant k = V2 x 103 N/m and natural length 50 cm. Initially left part of the 28 gms of nitrogen out of which one third of molecules are dissociated into atoms. If the initial pressure of gas in each parts is P0 = A/2 x 10s N/m2. Calculate the 3

(a) heat supplied by the heating coil connected to compress the spring by — I 4 (b) work done by gas in right part.

A sphere of mass m is placed on a rough plank of mass 2m which in turn is placed on an rough inclined plane, inclined at an angle 9, with horizontal. Friction coefficient between plank and sphere is \i2 and that between plank and incline

[5+5=10]

m J

plane i s j ^ . Find the maximum-value of — to ensure pure rolling at plank and sphere surface.

[10]

FIITJCC Ltd. ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi -16. Ph: 6865182, 6854102. Fax: 651-394

RSM12-PT-PH-L-LL (M)-PH- 3

Three identical springs A, B &C each of natural length I are connected to a point mass m as shown in the figure. A & B are horizontal and C is vertically fixed with rigid supports. What is the work done by the external agent in slowly lowering the mass m till it attains equilibrium when the springs A & B make an angle 2sin"13/5 between them. Neglect the masses of the springs.

external agent

B

[10] Two masses of 1 kg and 2 kg are attached with a light string, which passes over a pulley of mass 8 kg. Initial length of string on each side was 1m. Two-wave pulses are generated at time t = 0 and blocks are allowed to move. Find the ratio of time in which each pulse reaches to pulley on either side. Assume no slipping between pulley and string and mass per unit length of string is 0.01 kg/m, which can be neglected with respect to mass 1kg, & 2kg.

Fixed axis

B AC±3 • 1kg 2kg

[10] 8.

A ball of mass 5 kg is projected horizontally with velocity 20 m/s. After time t = 2 sec. it hits a wedge of 20 kg which is attached through a spring with a fixed wall. The surface is smooth and spring constant is 1000 N/m. If the collision is inelastic with coefficient of 1 restitution e = - j = , find the maximum compression of spring.

V2

The equation of a resultant wave received by the detector is 5 1 Y=(cos 7it- —) sin 5007it If the source of lowest frequency is eliminated. Find the ratio of the beat frequencies before and after eliminating the source of lowest frequency. [10]

10.

A man with some passengers in his boat, starts perpendicular to flow of river 200 m wide and flowing with 2 m/s. Boat speed in still water is 4 m/s. when he reaches half the width of river the passengers asked him that they want to _reach the just opposite end from where they have started. (a) Find the direction due which he must row to reach the required end. (b) how many times more time, it would take to that if he would have denied the passengers. [10]

*

*

*

FIITJCCLtd.ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 1 6 . Ph: 6865182, 6854102. Fax: 651-394

FI8TJCC RANKERS STUDY MATERIAL IIT - JEE 2002 P H A S E - I I I TEST CHEMISTRY Time: Two Hours

Maximum Marks:

100

Note: 1.

Each questions carries EQUAL MARKS.

2.

Attempt ALL questions,

3.

There is NO NEGATIVE MARKING.

4.


5.


6.

Useful Data : Gas Constant

R

=

8.314 J mol"1K~1

=

0.0821 lit atm moi"1 K"1

=

2 Cal mor 1

Avogadro's Number

N

=

6.023 x 1023

Planck's constant

h

=

6.625 x 10~34 J sec.

Velocity of light

c

=

3

1 electron volt

ev =

1.6

F

96500 C

Atomic

Masses

=

x

10® m sec -1 x

10~19 J

Ag = 108, Mn = 55, Cr = 52, Ca = 40, C = 12, O = 16, K = 39, CI = 35.5, N = 14, S = 32, Na = 23, H = 1, P = 31, I = 127, As = 75

Name of the Candidate Enrol!


Vihur (Near Hauz Khas Term.), New Delhi - 16, Ph : 686 5182, 685 4102, Fax : 6513942

R S M 1 2 - P 3 - T ( MJ-CH-2

1.

a) Explain the following i) AlCb exists as dimer while BCI3 exists as monomer. ii) Solubility of hydroxides, fluorides or oxalates of the metals of Gr.llA increases down the group. iii) Upon passing HCI(g) into a saturated solution of common salt, the latter begins to precipitate. b) Complete and balance the following equations indicating the colour of the precipitate, if any, formed. i) FeCI3 + K4[Fe(CN)6] — > ? ii) CuS0 4 + KCNS + S0 2 + H 2 0 » ? [2+2+2+2+2]

2.

a) A white inorganic powder poisonous in nature dissolves in dil. HCI with effervescence. The gas so evolved turns lime water milky-. If a burning Mg - ribbon is put into the gas jar filled with this gas, some residue is deposited on the bottom which when treated with dil.acid, leaves behind a black mass on the bottom. Acetic acid solution of the white powder gives a yellowish precipitate with potassium chromate solution. In the flame test the white powder gives yellowish - green flame. Identify the compound and give the chemical equations of the reactions described above. How is its poisonous property taken use of? b) Arrange the following in increasing order of the character as instructed in bracket against each i) BeC0 3 ,MgC03,CaC0 3 ,BaC0 3 , (thermal stability ) ii) HCIO, HCI0 3 , HCI0 4 , HCI0 2 (acidic nature) iii) CCU, MgCI2, AICl3, PCI5, SiCI4 (extent of hydrolysis) c) Arrange the species: 0 2 , 0 2 + , 02~, 0 2 2 - in order of their increasing bond length giving reason for your answer [4+3+3]

3.

a) An orange coloured solid upon heating gives a green coloured oxide with evolution of nitrogen gas. The oxide is amphoteric. The orange solid acts as an oxidising agent in acid medium and reacts with NaOH solution to give smell of NH3. Identify the orange solid and write the chemical equation of its thermal decomposition. Calculate spin magnetic moment of [Fe(CN)6]4~ which is an inner orbital complex. What is the state of hybridisation and EAN of the central metal in the complex? ii) Write the formula of the complex "penta ammine carbonato cobalt (III) tetrachloro cuprate (II). [5+ 2 + 3 ]

b) i)

4

a) Give reasons for the following: i) In NH3, PH3, ASH3 and SbH3, the bond angle decreases in order of their mention? ii) CCI4 does not hydrolyse with water whereas SiCI4 does hydrolyse? iii) Amongst BF3, BCI3 and BBr3, the lewis acid strength increases in order of their mention? iv) During electrolysis of alumina for getting AI metal, some cryolite and a little fluorspar are added? b) Arrange the following in increasing order of the character as instructed in bracket against each. i) LiCI, LiBr and Lil (solubility in water) ii) HF. HCI, HBr and HI (acidic nature) [ 4 x 2 + 2]

fllTJ€€ ICES House. Sarvapriya


RS M12-P3-T{ M )~C H - 3

a) i)

Write the IUPAC name of the following organic compound.

x

CH 3 ii)

CH3

[C0(en)(NH 3 ) 2 CIBr]N0 3

b) Write balanced chemical equations indicating the action of heat upon the following. i) FeS0 4 , (ii) AgN0 3 , (iii) NH 4 N0 3 and (Iv) K 2 Cr 2 0 7 c) i)

What is the use of hypo in the photography? Give chemical equation(s) of the reaction(s) involved. ii) Give the structure of thiosulphate ion. [3 + 4 + 2 + 1]

a) Arrange the following in increasing order of the property as indicated in the bracket against each. i) CH3COOC3H7, CH3COOCH(CH3)2, CH3COOC(CH3)3 (relative rates of alkaline hydrolysis). ii) HCOOCH3, CH3COOCH3, (CH3)2CHCOOCH3 (relative rates of alkaline hydrolysis) iii) N H 3 , C H 3 N H 2 I ( C H 3 ) 2 N H , ( C H 3 ) 3 N (basic character in aqueous medium) iv) N H 3 , B U N H 2 , B U 2 N H , B U 3 N (basic character in chlorobenzene) v)

OH

OH

OH

OH

OH

-CH3

o.

(acid strength)

NO2

CH 3

(b) The two carbon - oxygen bond lengths in formate ion are equal while they are different in formic acid. Explain. [5 + 3] a) Complete and balance the following chemical equations. Also mention the colour of the precipitate, if any formed, or characteristic colouration of solution, if any, developed. i) Na2[Fe(CN)5NO] + Na2S >? ii) KCI + Na 3 [Co(N0 2 ) 6 ] »? iii) MgCI2 + Na 2 HP0 4 + NH4OH + H 2 0 -> ? b) A white solid (X) upon heating gives a yellow solid with evolution of brown fumes. The yellow solid turns white after sometime. The residue is amphoteric. HCI solution of this residue gives bluish wh'te ppt. With potassium ferrocyanide solution solution while its solution in sodium hydroxide gives a white ppt. with H2S gas. Identify the solid A and give chemical equations of the reactions described in this question. {VA x 3 + SVi] a) Calculate percentage ionic character of AB molecule assuming its dipole moment to be 1.03 D and bond length 127 pm. (Debye, D = 3.3356 x 10-30 C.m, pm. = 10~12 m and unit charge = 1.6 x 10~19C) b) Arrange the following in increasing order of the character as instructed in bracket to each i) NH3,PH3tAsH3 and SbH3 (bond angle) ii)

LiF, LiCI, LiBr and Lil

(covalent character)

(iii) H 2 0, H2S and H2Te

( boiling point)

(iv) F, CI, Brand I

(electron affinity)



RSM12-PLIVtoIV- T ( M ) - C H - 4

(v) Al3+, Mg2+, F", 02~ and Na+

(size)

(vi) NaF, NaCI, NaBrand Nal

(melting point)

(vii)Na, Mg, At and Si

(2nd LP.)

[3+7]

a) Pyrrole is less basic as compare to pyridine. Explain b) Which is more stable carbocation i) CH3 and ii) iii)

CH 3 —CH=CH and C H 2 - C H = CH2 C 6 H 5 - C H 2 and (C6H5)3C+

c) Convert the following structures is to Fischer projection

[2 + 6 + 6] 10.

a) Arrange the following in the increasing order of enol-character. O O O

II

CH3-C-CH3,

i)

O iii)

FI17J€€

CH3 - C - C H

(ii)

CH

II

3

-C-CH

II

2

-C-CH

3

O 2

- C -

OC2HS

I C E S House, Sarvapriya

Vihar (Near Hauz Khas Term.), New Delhi • 16, Ph : 686 5182, 685 4102, Fax : 6513942

RS M12-P3-T( M )-C H-5

O

iv)

o=c

/

\NH

C

\ /

•CH2

O b) Acetophenone on reaction with hydroxylamine hydrochloride can produce two isomeric oximes. Write structures of the oximes c) Which of the following species behaves as (i) nucleophile,, (ii) an electrophile, (iii) both or (iv) neither. CP, H 2 0, H+, AIBr3, CH3OH, BeCI2, Br+, Cr+3, SnCI4, NO£, H 2 C=0, H 3 CC^N, H2, CH4 and H 2 C = CH.CH 3 . [4+2+4]



6513942

nifJCC Rankers Study Materiel N T - J E E 2002 PHASE - III TEST


Maximum Marks:100

NOTE : 1.

This paper consists of ten questions only

2.


3.


4.


PERMITTED.


Number


RSM12-P3-T(M)-MA-2

Consider the equations 2 cos"' x + 4(sin"1 y)2 = pit* tan"1 x + tan"1 y = [p] 4 show that y =

2.

1-x 1+x

V x, y e R+

& xy < 1

where [x] denotes the greatest integer less than or equal to x.

In a right angled triangle ABC right angled at A, a point E is so taken on the side AC such that AE : EC = 1: 2. The median AD on BC intersects BE at F. If BE : AD = 2 : S sin3 9

find the measure of Z ABC.

[10]

cos 3 9 , prove that tan 20 = 2tan (39 + a) cos(20 + a)

3.

If

4.

Find the general value of 9 satisfying the equation

sin(29 + a)

Vl 6 - 4 sin3 9 - 4 sin2 9 + sin 9 + 5.

[10]

[10]

- sin 9 + 4cos 2 (1 + sin 9) = 7

[10]

A, B, C, D are four points situated on the circumference of a circular swimming pool, with D diametrically opposite to B. The angular elevation of a vertical diving stand situated at C are 45°, 60° & 30° form A, B and D respectively. If A is located midway between B an D and at a distance of 10V3 m form either point find the radius of the pool. [10] If (1 + x)n = C0 + CtX + CjX2 + \ ± ( - r C

2

J

+ Cnxn where x is any variable then prove;

+{XHrtC,r_1}

=£Cr'WhereC°>C1'C2

Cn

r=0

r=1

[10]

have their usual meanings. 7.

8.

1

1

7

Find the sum of the series given by:- S = 1 + — + — + + s 6 18 324

Evaluate the determinant:

1

sin 39

sin3 9

2cos 9

sin69

sin3 29

2

QO

[10]

[10]

3

4 c o s - 1 sin 99 sin 39 Find the range of values of p if the system of equation given by> x + 2y + [p + 3] z = 0 2x + [p + 1] y + 3z = 0 [p + 2]x + 3y + z = 0 have a non-trivial solution, where [.] denotes the greatest integer less than or equal to. 10.

[10]

Using the principal of mathematical induction to prove that the last digit of the number represented by (2n - 1 ) ,2 n_1 V n is a prime is either 6 or 8. [10]

FI!TJC€, ICES HOUSE (Opp. VIJAY MANDAL ENCLAVE), SARVAPRIYA VIHAR, NEW DELHI -16. Ph:6854102, 6865182, FAX: 6513942

FllTJiC Rankers Study Material I IT - JEE 2002 PHASE - III


Maximum Marks : 100

NOTE : 1. Attempt ALL questions. 2. There is NO NEGATIVE

MARKING.

3. Use of logarithmic tables is 4. Use of Calculator is NOT

PERMITTED. PERMITTED.

5. Useful Data : 10 m/s2

Acceleration due to gravity

g

Coefficient of permeability of free space

M-o

Coefficient of permittivity of free space

So

Mass of electron

me

=

9.1 x 10~31 kg

Charge of electron

e

=

1.6 x i c r 1 9 c

=

4n x 1(T7 T-m/A 8.8 x 1(T 12



RSM12-P-3-T(M)-PH-2

In the shown figure connector PQ can slide on the two frictionless conducting rails. Resistance of the rails as well as fixed connector AB is negligible. Separation between the rails is I and resistance of the connector is R. A constant force F parallel to the rails is applied to the connector. There exists a uniform magnetic field of induction B perpendicular to the plane of loop. Find the current in the connector when it achieves the terminal velocity . Calculate the charge on each capacitor and the potential difference across it in steady state in the circuit shown for the cases (i) switch s closed, & (ii) switch s open.

A x

[6] 6nF

luF

2nt AMAAA—I gion

—AAA/W200

100 V

[6] A uniform line charge density X C/m over a straight line of length 2a is rotating about an axis passing through its centre and perpendicular to its length. Calculate the equivalent magnetic moment if the angular velocity is co. [6] A capacitor loses a certain fraction of its charge in 30sec because of humidity in the air giving rise to leakage between its terminals. When a 4MC2 resistance is connected between its terminals, in the absence of humidity, the same fraction of charge is lost in 7.5s. Calculate the leakage resistance due to humidity. [7] Three identical metallic plates are kept parallel to one another at seperations a & b as shown in figure. The outer plates are connected by a thin conducting wire and a charge Q is placed on the central plate. Find the charge on all the six surfaces. [8] In an oscillating LC circuit, at t = 0 charge on the capacitor is zero. If maximum charge on the capacitor can have is Q then after what minimum time energy stored in the capacitor and that in the inductor will become equal. L and C are the inductance and capacitance of the circuit respectively. [7] A metal rod OA of mass m and length I is kept rotating with a constant angular speed co in a vertical plane about a horizontal axis at the end O. The free end A is arranged to slide without friction along a fixed conducting circular ring in the same plane as that of rotation. A uniform and constant magnetic field induction B is applied .perpendicular and into the plane of the rotation as shown in the figure. An inductor and an external resistance R are connected through a switch S between the point O and a point C on the ring to form an electrical circuit. Neglect the fISTJii, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Ph: 6865182, 6854102. Fax: 6513942

RSM12-P-3-T(M)-PH-3

resistance of the ring and the rod. Initially, the switch is open. (a) What is the induced emf across the terminals of the switch ? (b) The switch S is closed at time t = 0 (i) obtain an expression for the current as a function of time. (ii) in the steady state, obtain the time dependence of the torque required to maintain the constant angular speed, given that the rod OA was along the positive X - axis at t = 0. [10] 8.

X XLX X ^ •N.X X Ix X A conducting rod is bent as a parabola x xlx x n X IX X y = Kx2 and it is placed in a uniform magnetic field of X x x \x x > X x l x induction B. At t = 0 a conductor of resistance R x x xR> X x j X X starts sliding up on the parabola with a constant X T-STTY X acceleration a and the parabolic frame starts X X X ^V^ _ X X X X rotating with constant angular frequency co about the X X X X > X X X X X X X X > X X X X axis of symmetry, as shown in the figure . Ignore Y Y Y Y } Y Y Y Y resistance of the remaining parts. If at t = 0, the parabola lies in the plane of the paper then find the current in the connector when the angle between the magnetic field and the plane of the parabola becomes 45° for the first time. [10]

Two concentric shells of radii R and 2R are shown in figure . Initially a charge q is imparted to the inner shell. After the keys KT & K2 are alternately closed n times each , find the potential difference between the shells. K r x K, 10.

11.

Consider the circuit shown in figure. Prior to t = 0, the switch is in position A and the capacitor is uncharged. At t = 0, the switch is instantaneously moved to position B. (a) Determine the current in the L-C circuit for t > 0. (b) Find the charge q on the lower capacitor plate for t >0.

[10]

[ 6 + 4 = 10]

An electron flies into a plane horizontal capacitor parallel to its plates with a velocity of v0 = 107m/s. The length of the capacitor I = 5 cm and the intensity of its electric field E = 100V/cm. When the electron leaves the capacitor, it gets into a magnetic field of induction B = 10"2T, whose force lines are parallel to the initial direction of motion of electron. Find (i) the radius of the helical trajectory of the electron in the magnetic field. (ii) the pitch of the helical trajectory of the electron . [10]

FIITJCC. ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Ph: 6865182, 6854102. Fax: 6513942

RSM12-P-3-T(M)-PH-4

12.

Figure shows a rod PQ of length / and an infinitely long current carrying wire in the same plane of the paper. If the rod is rotating with a constant angular velocity co about an axis normal to the plane of the paper and passing through C, find the emf induced across its ends P and Q at the moment it is normal to the wire and also determine which terminal is at higher potential. Perpendicular distance of the centre of the rod from the wire is /. Current in the infinitely long wire is equal to i0. *

*

CO

I

Q

*

[10]

*

fISTJii, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Ph: 6865182, 6854102. Fax: 6513942

FIITJCC RANKERS STUDY MATERIAL

IIT-JEE, 2002 PHASE - IV


Maximum Marks : 100

Note: i)

There are NINE questions in this paper. Attempt ALL questions.

ii)


iii)


iv)



=

8.314 J mor 1 K" 1 0.0821 lit atm moi"1 K~1

=

2 Cal mor 1

R =

6.023 x 1023

Planck's constant

N = h =

Velocity of light

c

3

1 electron volt

ev = F =

Atomic Masses

Ag = 108, Mn = 55, Cr = 52, Ca CI = 35.5, N = 14, S = 32, Na = 2 Fe = 56, Ag = 108

Avogadro's Number

=

6.625 x 10~34 J sec. x

10® m sec"1

1.6 x 10"19 J 96500 C O = 16, K = 39, I, I = 127, As = 75,


(F!ITJ€€ ICES House,

Sarvapriya

Vihur (Near Hauz Khas Term.), New Delhi - 16, Ph : 686 5182,

685 4102, Fax :

6513942

RSM12-P-IV - T ( M ) - C H - 2

1.

Treatment of compound (A) C8H10O with chromic acid pyridine gives (B), C 8 H 8 0. Treatment of (B) with two equivalents of Br2 yields (C), C 8 H 6 OBr 2l which on treatment with caustic soda followed by acidification gives a compound (D) C 8 H 8 0 3 . The latter liberates C0 2 on treatment with NaHC0 3 and is resolvable. Write structures for (A), (B), (C) and (D). Give mechanism of formation of (D) from (C). [8]

2.

An ester (A) is condensed in the presence of sodium methoxide to give a p-keto ester (B) and Methanol. On mild hydrolysis with cold conc. HCI, (B) gives methanol and 3-oxo-acid(C). (C) underwent readily decarboxylation to give cyclopentanone. a) Identify (A), (B) and (C) b) Name the reaction involved in conversion of (A) to (B) c) Give the mechanism of decarboxylation [12]

3.

Identify the products (A) to (I) LIN

i) B U - C S C - H

"2

PHCH0

) A

MN

>B

°2

) C — ^

^

>

D

A H)

Ozonoiysis

E

) p

i) Aq. oh-

> 1

.

ace ty|-2-methyl-cyclopentene

ii) - H 2 0

iii) Ph - C = CH + CH3MgX 4.

>G

ArCH2Br

> H—u

N 3

"

>l

[6 + 4 + 3]

Deduce structures (C) , (D) and (E) where (C) and (D) are positional isomers having eight carbon atoms each, with one of them being major product and other one is the minor product of (A) and (E) reactants. (E) similar to (A) has 4 carbon atoms but they are not isomers of each others. CH3 CH3 - C = CH2 (A)

(C) HzS 4

°

)C

>< \

H SO

* *

(E)

(D)

[10]

5.

An acidic compound A, C 4 H 8 0 3 loses its optical activity on strong heating yielding B, C 4 H 6 0 2 which reacts readily with KMn0 4 . B forms a derivative C with SOCI2, which on reaction with (CH3)2NH gives D. The compound A on oxidation with dilute chromic acid gives an unstable compound, which decarboxylates readily to E, C 3 H 6 0. Give structures of A to E. [8]

6.

How would you bring about following conversions a) CHO b)

c)

d) 7.

>N02 PhCOCH3

^ ^ B r » PhC(CH3) = CHCOPh

[4 + 3 + 4 + 4]

Give reasons for the following: a) When 1-pentene-4-yne is treated with HBr in equimolecular proportion, the addition takes place on double bond and not on triple bond yielding there by the product CH3CH(Br)CH2C = CH. b) Product formed by the reaction between a primary amine and ethylene oxide is less nucleophilic in character than the starting primary amine itself.

fliues ICES House, Sarvapriya

Vihar (Near Hauz Khas Term.), New Delhi - 16, Ph: 686 5182,

685 4102, Fax:

6513942

RSM12-P1

- T ( M ) - C H-38

c) When hydrolysis of optically active 2-bromopropionate is carried out with low concentration of OH" ion in presence of Ag 2 0, it takes place with overall retention of configuration although reaction proceeds by SN2 path way. [4x3] Suggest probable mechanism for each of the following reactions a) ^ /CH2-CH2NHMe

9

C6H5Li

N I

CI

b)

o

Me

CI CHCI.

S—>

(-)

(Na - salt)

[5x2]

Identify the major organic products A to H of the following reactions a) Ph

I

C H 3 - C - CH2I + Ag+

>A

AH b

)

Me2C = CH2

C

^

C

°3

H

>

0 EtMgBr/Et2Q

) R

io H 3 o

c)

i)CH2

CHD2Mgl

° >

ii) H 3 0 +

d)

—

C

Dehydration

COPh PhMgBr/CuCI

H3Q

e) -> E f)

g)

+H-C=CNa

NH (I)

°

NH cl(aq)

>

*

>G

h) i) SnCI2 /HCI ii) H 2 0

)

CH3NQ2

^

NaOH

[8 x

FIITJCC

ICES House, Sarvapriya

Vihar (Near Hauz Khas Term.), New Delhi - 16, Ph : 686 5182,

685 4102, Fax :

V/2]

6513942

FIITJ€€ Rankers Study Material I IT - JEE 2002 PHASE - IV MATHEMATICS Time : Two hours

Maximum Marks:100

NOTE : 1.

This paper consists of ten questions only

2.


3.


4.


PERMITTED.


Number

FIITJ€€ Ltd. ICES House (Opp. Vijoy Mondol Enclave), Sam*-w

Vihar, New Delhi • 110016. Ph:6854102, 6865182, Fax: 6513942

RSM12-P-IV-T(M)-MA-2

1.

Find the complexl' numbers satisfying the equation 2|z|2 +z2 - 5 + i V3 =0.

2

Using the equation z8 +1 = 0 prove that

f

TUY

771 371Y 5 TTv cos 0 - cos — cos 9 - c o s — | . 8 A 8

cos48=8 cos 9 - cos - cos 9 - cos — 8A 8

[10]

[10]

3

If Zi ^ z2 = 0 and Zi z2 +z3 z4 = 0 then prove that the points representing z1t z2, z3, z4 are concyclic. [10]

4.

Show that the number of ways of selecting n objects out of 3n objects, of which n are alike and rest are different is 22n_1 + ^ — ^ j - . n! (n - 1Jf

[10]

5.

There are 6n seats in a row. In how many ways n persons can sit such that between any two person there are at least three seats vacant. If n is even then in how many ways they can sit such that each person has exactly one neighbour. [10]

6.

Find the distance of the point p(i + j + k) in the plane n which passes through the points A(2i + j + k), B(I + 2j + k) and c(i + j + 2k). Also find the vector of the foot of the perpendicular from P on the plane.

position [10]

7.

Let ABC and PQR be any two perpendiculars from the points A, concurrent. Using vector methods P, Q, R to BC, CA, AB respectively

triangles in the same plane. Assume that the B, C to the sides QR, RP, PQ respectively are or otherwise, prove that the perpendiculars from are also concurrent. [10]

8.

Four students after selection in IIT-JEE decide to join anyone of the four NT Kanpur, Delhi, Bombay or Kharagpur. They discuss the placement position after passing and find the degree of attractiveness for joining in proportional to numbers 8, 9, 10, 12. Find the probability that each will join a different NT. [10]

9.

Watson received a telegram from Sherlock Holmes to catch a specified train from London to Ciifton. He reached the platform where the specified train was ready to leave. The train consisted of p carriages each of which will hold q passengers and in all (pq - m ) passengers had occupied their seats. If waston is equally likely to get any vacant palace, find the chance that he will travel in the same carriage with Sherlock Holmes. [10]

10.

If p and q are chosen randomly from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} with replacement, find the probability that the roots of the equation x2+ px + q = 0 are real and distinct. [10]

fi|TJ€€ Ltd. ICES House (Opp. Vijay Mandal Enclave), Sanupme. >/lhart New Delhi-110016. Ph:6854102, 6865182, Fax: 6513942

fiitjcc Rankers Study Material IIT - JEE 2002 PHASE - IV


Maximum Marks : 100

NOTE : 1.


2.

There is NO NEGA TIVE MARKING.

3.


4.


5.

Useful Data :

i


g

=

10m/s2

Plank's constant

h

=

6.63 x 10"34 J-s

Mass of electron

me

=

9.1 x 10~31 kg



RSM12-P-3-T(M)-PH-42

Light of two different wavelengths 4000 A0 and 6000 A0 is incident on two identical slits, the slit separation being 0.1 mm. A screen is kept at a distance of 2m (see figure).

V V VY

(a) Find the fringe width of the interference pattern formed on the screen due to each wavelength. (b) Find the path difference and the corresponding phase difference for light of both wavelengths (4000 A0 and 6000 A0) arriving at a point A, which is at a distance of 4 mm from the central point P.

4 mm

(c) If the ratio of the intensities lA (at the point A) and l P (at the point P) is lA : IR = 2 : 9, find the ratio of the intensities of the light of 4000 A0 to that of 6000 A0 incident on the slits. [4+4+7=15] Electromagnetic Radiation consisting of a mixture of three wavelengths 4000 A0 5000 A0 & 14000 A0 is incident on a metallic sample. It is observed that the emerging photoelectrons having maximum energy could just cause the excitation of H-atoms from n = 2 to n = 3 level. Find (a) the de-Broglie wavelengths of the photoelectrons, of maximum energy, emitted by each kind of photons. (b) The photocurrent from the material, assuming that the efficiency of conversion is 20 % and that the total intensity of 1.44 x 102 W/m2 is distributed evenly among all wavelengths. (Area of the sample is 2 cm2). (c) If the work-function of the material was 50 % lower, what will be the photocurrent and the stopping potential of the photoelectrons? [5+5+5=15] The figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated . The new distance is measured to be 30 .0 cm. What is the refractive index of the liquid ?

[10] (a) In a Young's double slit experiment , the slit are 2mm apart and are illuminated with a mixture of two wavelengths X-i = 750nm & X2 = 900nm . The screen is placed at a distance of 2m from the plane of slits. If a bright fringe of one interference pattern coincides with a bright fringe from the other at a distance x from the common central bright fringe, find the minimum value of x. [7] (b) A slit is located at infinity in front of a lens of focal length 1m and is illuminated normally with light of wavelength 600nm. The first minima of either side of the central maxima of the diffraction pattern observed in the focal plane of the lens are separated by 4mm. What is the width of the slit ? [8]

fISTJii,

ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Ph: 6865182, 6854102. Fax: 6513942

RSM12-P-3-T(M)-PH-43

5.

(a) A piano - convex lens has a thickness of 4 cm, when placed on a horizontal table , with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table , the apparent depth of the centre of the plane face of the lens is found to be 25/8 cm. Find the focal length of the lens. (b) A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. (i) How far from the objective should an object be placed in order to obtain the final image at the least distance of distinct vision (25cm), (ii) infinity ? What is the magnifying power of the microscope in each case ? [5+5=10]

6.

(a) !n a double slit arrangement, the separation between the slits is three times the width of each slit. Find how many interference maxima can be observed in the principal diffraction maxima. (b) A neutron at rest decays as given : n -» p + e + v Assuming the resulting proton to remain at rest, find the kinetic energy of the electron and the energy of the anti-neutrino. Take, mn = 1.0087 u, m P = 1.0072 u, me = 0.00055u, Antineutrino is massless. [7+8=15]

7.

The K-absorption edge of an unknown element is 0.171 A (a) Identify the element. (b) Find the average wavelengths of the K-series lines. (c) If a 100 eV electron struck the target of this element, what is the cutt off wavelength ?

o

[10] 8.

(a) The half life of 215At is 100ns. If a sample initially contains 6 mg of the element, what is its activity (i) initially (ii) after 200|is. (b) If a radioactive nuclide with half life period T is produced at the constant rate of n per second, find (i) the number of nuclei in existence t seconds after the number is N0 (ii) the maximum number of these radioactive nuclei [5+5=10] *

fISTJii,

*

*


FIITJ€€ RANKERS STUDY MATERIAL IIT-JEE, 2002 PHASE- III-IV CHEMISTRY Time: Two Hours

Maximum Marks : 100

Note: i)

There are THIRTEEN questions in this paper. Attempt ALL questions.

ii)


iii)


iv)



R =

8.314 J mor1KT1 0.0821 lit atm moP1 K~1 2 Cal moP1

Avogadro's Number

N =

6.023 x 1023

Planck's constant

h

=

6.625 x 10~34 J sec.

Velocity of light

c

=

3 x 10s m sec-1

1 electron volt

ev =

1.6 x 10~19 J

F

96500 C

Atomic No:

=




FIITJCC,

ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Ph: 6865182, 6854102. Fax:

6513942


1.

Predict products -of the following reactions a) O BASE

C 6 H 5 - c - CH3 + CI - CH2 - COOC 2 H 5 CHO

b)

O2N c)

M

+

d)

C = CH - C - CH3 Me

>

HoO

o

CH^ CH3

BASE

(CH 3 C0) 2 0

>

NAOCI

>

Me

Ph_i

-d).—Me

He!)

— ^

I

e)

5\-C-/oV-Me

MeO-

RC 3H

°

> [5 X2]

Which of the following two ketones is more acidic. Give a reason

ff . [4]

Write an appropriate mechanism for each of the following reactions. (a) O H \ /OCH3 CH3OH/H+ (b) ^

sP

OH (i) BR—CH2COOC2H5

:H 2 COOC 2 H 5

(ii) Zn (iii) H 2 0 (c)

(d)

ÔH

(e)

H+

^ [5 x3]

Predict A to E

O

CH3 — C— CH3

FIITJCC,

dil -> A Ba(OH)2

CH2(CQ2Et)2 Base

ICES House, (Opp. Vijay Mandal Enclave),

>B

Sarvapriya

-OC 2 H 5

) C


H+

)

P —A—^ ^

[10]

Ph: 6865182, 6854102. Fax:

6513942


5.

Write the products of following reactions, (a) Br Br

O

-—i

•

Ethanolic ) Solution of kOH (1 .equiv.)

NO2 B

( )

I L

L I I

N

CHCI3 —

K

V

^

A

-

5

member ring

B

6

member ring

H CH 3 CH 2 CH 2 CH-CH3

EtONa

) Major product.

1= [3 x3] 6.

An organic compound C10H10O(A) which do not decolourise bromine water, reacts with MeMgBr to form an alcohol. CnHi 4 0(B) which an dehydration gave CnH12(C). Compound (C) on ozonolysis gave a compound CNH1202(D) which on oxidation with [Ag(NH3)2]+ gave C H H 1 2 0 3 ( E ) . E on decarboxylation gave C I O H 1 2 0 ( E ) F is one of the isomers formed by Friedel Craft acetylation of ethyl benzene. Deduce the structures of A,B,C,D,E and F. Give reactions also? [12]

7.

Write the structures of all the isomeric hydroxy acids having the formula C 4 H 8 0 3 . How do they behave towards heat? [6]

8.

A solution containing 2.665g of CrCI3-6HO is passed through a cation exchanger. The chloride ions obtained in solution react with AgN0 3 and give 2.87g of AgCI. Determine the structure of the compound. [5]

9.

Explain the stability of oxides of alkali metals.

10.

The gas liberated on heating a mixture of two salts with NaOH gives a raddish brown precipitate with an alkaline solution of K2Hgl4. The aqueous solution of the mixture on treatment with BaCI2 gives a white precipitate which is sparingly soluble in concentrated HCI. On heating the mixture with K 2 Cr 2 0 7 and conc. H 2 S0 4 , red vapours (A) are produced. The aqueous solution of the mixture gives a deep blue colouration (B) with K3[Fe(CN)6] solution. Identify the radicals in the given mixture and write the balanced equations for the formation of (A) and (B). [10]

11.

Explain the following facts a) Among the halides of lithium, the melting point of LiF is maximum. b) Bond order of N2 and N2 are same but bond energies are different.

12.

a) What are the hybridisations of the central elements into the following species. [VA + VA]

(i) POCI3 13.

^

[5]

[3 + 4]

(ii) CH2Me

Balance the following redox reaction with proper explanation. a) MnO^" + H+ >Mn0 2 +Mn0 4 b) As2S3 + N0 3 + H + + H 2 0

fllTJ€€ ICES House (Opp. Vijay Mandal Enclave),

—> H3As0 4 + NO + S

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Vihar, New Delhi - 16. Ph:6854102,

[2 x 2]

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Fax:

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fllTJ€€ RANKERS STUDY MATERIAL IIT - JEE, 2002 PRACTICE TEST - PHASE-III, IV MATHEMATICS Time: Two hours Note:

Maximum Marks:10G i

i).


ii).


iii).


iv).



Candidate Number


RSM12-PT-PH-III, IV(M)-MA-2

1.

Consider the equations : 2 cos"1 x + 4(sin"1 y)2 = pn2 tan"1 x + tan"1 y = [p] — V x, y e R+ & xy < 1 4 1—x show that y = where [x] denotes the greatest integer less than or equal to x. 1+x

2.

In a right angled triangle ABC right angled at A, a point E is so taken on the side AC such that AE : EC = 1: 2. The median AD on BC intersects BE at F. If BE : AD = 2 : V3 find the measure of Z ABC.

3.

[10]

Find the general value of 0 satisfying the equation V l 6 - 4sin 3 9 - 4 s i n 2 9 +sine + ^ 5 - s i n e + 4COS2(1 +sine) = 7

4.

[10]

If (1 + x)n = C0 + C-|X + C ^ 2 + \ ± ( - r c

2

J

+ \ i ( - r c

[10]

+ Cnxn where x is any variable then prove; 2

=xcr1whereco.ci,c2,......,cn

J r=0

have their usual meanings.

5.

6.

[10]

1 1 7 Find the sum of the series given by:- S = 1 + - + — + + 6 18 324

Evaluate the determinant:

1

sin 39

sin3 9

2cos9

sin69

sin3 29

4cos2-1

sin99

sin3 39

oo

[10]

[10]

7.

Find the complexl' numbers satisfying the equation 2|z|2 +z2 - 5 + i V3 =0.

8.

Show that the number of ways of selecting n objects out of 3n objects, of which n are alike and rest are different is 2 -1 +. (2n - 1 ) n! (n - 1 ) '

9.

[10]

[10]

Find the distance of the point P(i + j + k j in the plane n which passes through the points A(2i + j + k), §(i + 2j + k) and c(i + j + 2k). Also find the position vector of the foot of the perpendicular from P on the plane. [10]

10.

Watson received a telegram from Sherlock Holmes to catch a specified train from London to Clifton. He reached the platform where the specified train was ready to leave. The train consisted of p carriages each of which will hold q passengers and in all (pq - m ) passengers had occupied their seats. If waston is equally likely to get any vacant palace, find the chance that he will travel in the same carriage with Sherlock Holmes. [10] ieieie

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FIITJCC RANKERS STUDY MATERIAL IIT - JEE, 2002 PRACTICE TEST PHASE III, IV PHYSICS Time: Two hours

Maximum Marks: 100

Note: i).


ii).


iii).


iv).


Name of the Candidate Enrollment

FIITJCC,

:

Number


R S M 1 2 - P T - P H -I! I - I V ( M ) - P H - 2

1.

In the given circuit diagram, find the charge which will flow through direction 1 and 2 when the key is closed.

[10] Shown in the figure is a rod of mass rrh kept on a smooth conducting horizontal frame fitted with resistor R. It is attached to a hanging mass m2 by an inextensible string. A vertical magnetic field of induction B is applied. On releasing the rod, find its speed in the function of time and its terminal speed.

nrn

[10] In Young's double slit experiment the slits are 0.5 mm apart and interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that the 9th bright fringe is at a distance of 8.835 mm from the 2nd dark fringe from the centre of fringe pattern. Find the wavelength of light used. [10]

Find an expression for the magnetic dipole moment and magnetic field induction at the center of a Bohr's hypothetical hydrogen atom in the nth orbit of the electron in terms of universal constants. [10] A bi-convex lens L having radii of curvature 40 cm and 30 cm rests horizontally with the face of larger radius of curvature in contact with a horizontal polished metal plate M. A little water is held by capillary action between L and M, thus forming a water lens. Calculate the focal length of the combination. [M« = 4/3, Ho = 3/2]

L M

[10] A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 20 mV is added to the base-emitter voltage, the base current changes by 20 Â and the collector current changes by 2mA. The load resistance is 5 kQ. Calculate (a) the factor p, (b) the input resistance RBE , (c) the trans conductance and (d) the voltage gain. [2.5x4=10]

A ray of light incident normally on one of the faces of a right angled isoceies glass prism is found to be totally reflected. What is the minimum value of the refractive index of the material of prism ? When the prism is immersed in water, trace the path of emergent ray for the same incident ray taking the refractive index of the material of the prism to be equal to the minimum value of the refractive index as calculated in this problem earlier, indicating the values of ail angles ( j ^ = 4/3). [10]

F I ! T J € € . ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 1 6 . Ph: 6865182, 6854102. Fax: 6513942

RSM12-PT-PH-IH-IV (M)-PH- 3

8.

Two identical point charges each equal to Q are fixed at point A and B as shown in the figure. The separation between the charges is 2m and OA = OB. The electric field and potential at O is E = -2CW6 (i+j) NC"1 and V = 80 Volt respectively. Calculate (i) the charge Q. (ii) the coordinates of A and B.

O

&&

~ X [5+5=10]

9.

A perfectly absorbing surface intercepts a parallel beam of monochromatic light of X = 500 nm, incident on it normally. If the power through any cross-section of beam is 10 w, find (i) the number of photons absorbed per second by the surface, (ii) the force exerted by light beam on the surface. [10]

10.

(a) An X-ray tube with a tungsten target is found to be emitting lines other than those due to tungsten. The k a line of tungsten is known to have a wavelength 21.3 pm and the other two k a lines observed have wavelength 71 pm and 198 pm.Find the atomic number of the impurity . [7] (b) A Silicon diode requires a minimum current of 2mA to be above the knee point (0.7 V) of its V -1 characteristics. Assume that the voltage across the diode is independent of current above the knee point.

WWVVR +

HH

10 V

(i) Find the maximum value of R so that voltage across the diode is above the knee point. (ii) Find the power dissipated in the resistor when current through the circuit is 4mA. [2+1]

•k -k -k

FHTJ€€. ICES House, (Opp. Vijay Mandal Enclave), St -aprjya Vihar, New Delhi - 1 6 . Ph: 6865182, 6854102. Fax: 6513942

f IITJCC RANKERS STUDY MATERIAL IIT-JEE, 2002 PHASE -1 to IV


Maximum Marks : 100

Note: i)

There are ELEVEN questions in this paper. Attempt ALL questions.

ii)


iii)


iv)



R

=

8.314 J mor 1 K" 1 0.0821 lit atm mol -1 K"1 2 Cal mor 1

Avogadro's Number

N

=

6.023 x 1023

Planck's constant

h

=

6.625 x 10-34 J sec.

Velocity of light

c = ev =

3

F

96500 C

1 electron volt

Atomic Masses

=

x

1.6

10s m sec -1 x

10-19 J

Ag = 108, Mn = 55, Cr = 52, Ca = 40, C = 12, O = 16, K = 39, CI = 35.5, N = 14, S = 32, Na = 23, H = 1, P = 31, I = 127, As = 75, Fe = 56, Ag = 108




RSM12-PLIVto IV - T ( M ) - C H - 2

1.

a) A sample pitch blende is found to contain 50%. Uranium and 2.425% Lead. Of this lead only 93% was Pb206 isotope. If the disintegration constant is 1.52 x10~10 yr"\ how old could be the pitch blende deposit. [6] b) The I Pi of Hydrogen is -13.6 eV. It is exposed to electromagnetic waves of 1028A and gives out induced radiations. Find the wavelength of these induced radiations. [8]

2.

a) A sample of AgCI was treated with 5 ml of 1.5 M Na 2 C0 3 solution to give Ag 2 C0 3 . The remaining solution contained 0.0026g of CP per litre. Calculate the solubility product of AgCI? (Ksp Ag 2 C0 3 = 8.2x10~ 12 ) [7]

3.

4.

b) A 10 gm mixture of Cu2S and CuS was treated with 200 ml of 0.75 M Mn04~ in acid solution producing S0 2 , Cu2+, Mn2+. The S0 2 was boiled off and the excess of Mn04~ was titrated with 175ml of 1M Fe2+. Calculate the percentage of CuS in original mixture. [8] Write down the hybridisation state of the central atom and indicate the shape. (i) XeOF2 (ii) PBr3CI2 (iii) XeF6 (iv) \C\+2 (v) S 2 0 2 " [10] a) Explain why: i) Among trimethylamine and trisilyl amine, which one is more basic ii) The electron affinity of Ci2 is higher than that of F2

[ 6 * 4 = 10]

b) Write balanced chemical equation i) K 2 Cr 2 0 7 + HCI > KCI + CrCI3 + H 2 0 + Cl2 ii) P + HN0 3 >HP0 3 + N 0 + H 2 0 5.

The pressure in bulb dropped from 2000 to 1500mm of Hg in 47 mins when the contained 0 2 leaked through a small hole. The bulb was then completely evacuated. A mixture of 0 2 and another gas of molecular weight of 79 in the molar ratio 1:1 at a total pressure of 400mm of Hg was introduced. Find the mole ratio of two gases remaining in the bulb after a period of 74 mins. [8]

6.

An unsaturated organic compound (A) MF C9H9CI decolourises Br2 water and when heated with NaNH2 in liquid NH3 produced another unsaturated compound (B) having M.F. CgH8. The compound (B) on treatment with dil. H 2 S0 4 containing mercuric sulphate gives a compound C which on vigorous oxidation with acidified KMn0 4 produces 1, 4 benzendicarboxylic acid. Identify (A), (B), (C) and show how (C) is formed from B. Treatment of C with MeMgl followed by acidification produces a compound which on dehydration followed by catalytic hydrogenation produces D. Identify (D). An isomer (E) of (B) on reduction with Na/NH3(liq) produces a compound F which exhibits trans stereochemistry. Write the structures of (E) & (F). What reagent would you use for converting (E) to (G) which is a geometrical isomer of (F) give structure of G. [7]

7.

a) Account for the formation of 1, 2-dimethyl cyclohexene (A) and cyclopentane (B) on dehydration of 2, 2-dimethyl cyclohexanol.

isopropylidene [4]

b) Compound (A) C5H8 gives red ppt. with ammonical cuprous chloride. (A) on treatment with NaNH2 followed by n-propyl bromide gave B. (B) on ozonolysis followed by hydrolysis gave only one carboxylic acid. Identify A. B and C giving reason. [6] 8.

a) What happens when CH3C = C" reacts with CH3CHO in presence of CH30~ as catalyst. b) Me

V

[ O ]

+ CH3 -
An

"'3

>A

H Identify A c) Write a balanced equation for the reaction of N14 with a - particle. FI17J€€ ICES House, Sarvapriya

[2 + 2 + 2 = 6]


RSM12-PLIVto IV - T ( M ) - C H - 2

9.

a) Explain the following i) The central carbon-carbon bond in 1,3-butadiene is shorter than that of n-butane. ii) Nitrobenzene does not undergo Friedal Crafts alkylation. b) What will be the resultant pH when 200mL of an aqueous solution of HCI (pH = 2) is mixed with 300mL of an aq. solution of NaOH (pH = 12.0). [4 + 3]

10.

The decomposition of a compound A at temperature T according to the equation. 2P(g) > 4Q(g) + R(g) + S(l) is a first order reaction. After 30 min. from the start of the decomposition in a closed vessel, the total pressure developed is found to be 317 mm Hg and after a long period of time the total pressure observed to be 617mm Hg. Calculate the total pressure of the vessel after 75 min. Given vapour pressure of S(l) at temperature T = 32.5 mn Hg [7]

11.

An alkane (A) C5H12 on chlorination at 300°C gives a mixture of four different monochloro derivatives (B), (C), (D) and (E). Two of these derivatives gives the same alkene (F) on dehydrohalogenation. Give structures of (A) to (F) with proper reasoning. [6]

FI17J€€ ICES House, Sarvapriya


FIITJCC Rankers Study Material IIT - JEE 2002 PHASE - 1 - IV MATHEMATICS Time : Two hours

Maximum Marks:100

NOTE : 1.

This paper consists of twelve questions only

2.


3.


4.


PERMITTED.


Number

FIITJCC Ltd ICES House (Opp. VijayMcmdol Enclave), Sanxtprya Vihar, New Delhi -110016. Ph:6854102, 6865182, Fax: 6513942

RSM12-PI-IV-T(M)-MA-2

1.

A tower standing at the point O is being observed from two stations A and B. The angles of elevation of the top of the tower from A and B are 30° and 45° respectively. If AB subtends angle 60° at O and the area of the triangle AOB is 75 sq. m. Find the distance between A and B. [8]

2.

Let f be real a valued function satisfying f

'V

= f(x) - f(y) and Lim

v-ry area bounded by the curve y = f(x), the y - axis and the line y = 3. _x

If f (X) =

a +va

2n-1

(a > 0), evaluate £

/

2/

+

X

= 3. Find the

[10]

_

[8]

\2n

4.

A box contains 'a' white balls and 'b' black balls and besides the box lies a large pile of black balls. Two balls chosen at random are taken out of the box. If they are of the same colour, a black ball from the pile is put into the box; otherwise, the white ball is put back into the box. The procedure is repeated until the last two balls are removed from the box and one last ball is put in. What is the probability that this last ball is white? [8]

5.

Solve the equation x3 - [x] = 5 where [x] denotes the integral part of the number x. [8]

6.

Prove that the square of the length of the tangent drawn from a point on one circle to another circle is equal to twice the product of the perpendicular distance of that point from the radical axis of the two circles and the distance between their centres. [8]

7.

Let zu z2, z3 be three distinct complex numbers satisfying

|zi - 1| = |z2 - 1| = |z3 - 1|.

Let A, B and C be the points represented in the Argand plane corresponding to z^ z2 and z3 respectively. Prove that z^ + z2 + z3 = 3 if and only if AABC is an equilateral triangle.

[10] 1/V3 COS"

8.

Evaluate

I -

2x 1+x

+ tan

2x

1-x 2

e x +1

-1/V3 9.

2

dx.

[8]

If (a, b, c) is a point on the plane 3x + 2y + z = 7, then find the least value of a2 + b2 + c2, using vector method.

10.

[8]

There are exactly two points on the ellipse the same and is equal to

11.

a 2 + 2b2

<2 r

y2

+~

= 1 whose distance from its centre is

. Find the eccentricity of the ellipse.

[8]

Let f be a real valued function satisfying f(x) + f(x + 4) = f(x + 2) + f(x + 6). Prove that X+8

jf(t)dt is a constant function.

[8]

f(x + a) f(x + 2a) f(x + 3a) 12.

Let A(x)

f(a) f(a)

f(2a)

f(3a)

f'(2a)

f'(3a)

(the prime denotes the derivative with respect to

x, for some real valued differentiate function f and constant a. Find lim x-»0

A(x)

[8]

fllTJ€€ Ltd. ICES House (Opp. Vijay Mandal Endace), Sarvapriya Vibar, New Delhi -110016. Ph:6854102, 6865182, Fax: 6513942

FIITJCC Rankers Study Material

NT - JEE 2002 PHASE - I - IV

PHYSICS Time : Two hours

Maximum Marks:100

NOTE : 1

This paper consists of twelve questions

2


3

Marks for questions are shown in the right hand margin.

4


5

Useful Data :

only.

PERMITTED.

Planck's constant

h

6.63x10" 34 J-s

Boltzmann's constant

k

1.38x10" 23 J/K

Charge of an electron

e

Universal gas constant

R

8.314 Jmol"1 K' 1

Wien's constant

b

0.29 cm K.

=

1.6X10"19C

Permeability of free space M-o

=

4tc x 10" 7 H/m

Permittivity of free space

=

8.8 x 10"12 F/m

=

10m/s 2

E0

Acceleration due to gravity g Gravitational Constant

Name of the student

:

Enrollment

:

Number

G

6.67 x10" 1 1 Nm 2 /kg

FIITJCC Ltd. ICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 110016. Ph:6854102, 6865182, Fax: 6513942

RSM-12-P-l-IV-T(M)-PH-2

1.

An ice cube of mass 0.1 Kg at 0°C is placed in an isolated container which is at 227°C. The specific heat S of the container varies with temperature T according to the empirical relation S = A+BT, where A = 100cal/Kg-K and B = 2 x 10~2cal/Kg-K2. If the final temperature of the container is 27°C, determine the mass of the container. [5]

2.

An ideal massless spring can be compressed 1 m by a force of 100 N. The same spring is placed at the bottom of a frictionless inclined plane which makes an angle 9 = 30° with the horizontal. A 10 kg mass is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring 2 meters. (a) Through what distance does the mass slide before coming to rest ? (b) What is the speed of the mass just before it reaches the spring ?

[10]

3.

A circular loop of radius R is bent along a diameter and given a shape as shown in the figure. One of the semicircles (KNM) lies in the xz plane and the other one (KLM) in the y-z plane with their centres at the origin. Current I is flowing through each of the semicircles as shown in figure. (a) A particle of charge q is released at the origin with a velocity v = - v 0 i . Find the instantaneous force f on the particle. Assume that space is gravity free. (b) If an external uniform magnetic field B j is applied determined the forces F.) and F 2 on the semicircles KLM and KNM due to this field and the net force F on the loop.

4.

A thin uniform metallic rod of length 0.5 m and radius 0.1cm rotates with an angular velocity 400 radian/second in horizontal plane about a vertical axis passing through one of its ends. Calculate the tension in the rod and elongation of the rod. The density of the material of the rod is 104 kg/m 3 and Young's modulus is 2 x 1011 N/m 2 . [5]

5.

(a) We know that velocity of a wave travelling along a stretched string is given by VT/V where T is the tension in the string and (i is the mass per unit length of the string. Find the time taken by the wave travelling along a vertically suspended string of m a s s ' m ' and lengthT from the free end to the fixed end. [5] (b) A band playing music at a frequency f is moving towards a wall at a speed v b . A motorist is following the band with a speed v m . If v is the speed of sound, obtain an expression for the beat frequency heard by the motorist. [5]

6.

A plank of mass M rests on a smooth horizontal plane. A sphere of mass m is placed on the rough upper surface of the plank and the plank is suddenly given a velocity v in the direction of its length. Find the time after which the sphere begins pure rolling, if the coefficient of friction between the plank and the sphere is \i and the plank is sufficiently long. [10]

7.

A f.i-meson particle moves in a circular orbit around a very heavy nucleus (of infinite mass) of charge + 3e. Assuming Bohr's model is applicable to this system, (a) derive an expression for the radius of nth Bohr orbit. FIITJCC Ltd. ICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 110016. Ph:6854102, 6865182, Fax: 6513942

RSM-12-P-l-IV-T(M)-PH-3

(b) find n for which radius of orbit is approximately same as that of 1st Bohr orbit for a hydrogen atom. (c) find wavelength of radiation emitted when ji.- meson jumps from 3 rd orbit to I st orbit - meson is a particle, whose charge = that of an electron, mass = 208 times that of an electron], [10] 8.

A rocket is fired vertically and ascends with constant vertical acceleration of 20m/s2 for 1 minute. Its fuel is then all used and it continues as a free particle. Find the (a) maximum height reached by the rocket. (b) total time elapsed from the take off till the rocket strikes the earth.(g=10m/s 2 ) [10]

9.

Consider two small balls connected to a rigid rod of length '21' which in turn is suspended by a thread & rotated about the thread at angular velocity co. What would be the magnitude & direction of total force exerted by rod on one of the balls?

10.

11.

12.

[10] Find an expression for the magnetic dipole moment and magnetic field induction at the center of a Bohr's hypothetical hydrogen atom in the nth orbit of the electron in terms of universal constants . [5] An inductor of inductance L = 400 mH and resisters of resistances Ri = 2Q and R 2 -2C1 are connected to a battery of e.m.f. E = 12V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at time t = 0. What is the potential drop across L as a function of time? After the steady state is reached, the switch is opened. What is the direction and magnitude of current through F^ as a function of time?

Ri-

[5]

A point source is placed at a distance d/2 below the principal axis of an equiconvex lens of refractive 3 index — and radius 20-cm. The emergent light from lens fall on the slits ST and s 2 placed symmetrically with the principal axis. The resulting interference pattern is observed on the screen kept at a distance D = 1 m from the slit plane. Find (a) the position of central maxima and its width (b) the intensity at point O

SI

& s

V [5+5=10]


ALL INDIA TEST IIT-JEE, 2002 PART TEST -1

CHEMISTRY Maximum Marks: 100

Time: Two Hours

Note: t

i) ii) iii) iv) v) vi)

This paper has FIFTEEN questions. Attempt ALL questions. Answer to each new question should begin from a new page Answer all sub parts of a question at one place. Use of logarithmic table is PERMITTED. Use of calculator is NOT PERMITTED.


Avogadro's Number Planck's constant

R

=

Na

= =

h

=

8.314 J K"1 mol"1 0.0821 Lit atm K"1 mol"1 1.987 * 2 Cal K mol 23 6.023x 10 - 1

34

6.625 x 10~

1-1

J•s

27

6.625 x 10~ erg • s 1 Faraday

=

96500 Coulomb

1 calorie

4.2 Joule

1 amu

1.66 x 10"27

kg

Atomic No:

H = 1, D = 1, Li = 3, Na = 11, K = 19, Rb = 37, Cs = 55, F = 19, Ca = 20, He = 2, O = 8, Au = 79.

Atomic Masses

He = 4, Mg = 24, C = 12, O = 16, N = 14, P = 31, Br = 80, Cu = 63.5 Fe = 56, Mn = 55, Pb = 207, Au = 197, Ag = 108, F = 19, H = 2, CI = 35.5 Sn = 118.6


FIITJCC,

ICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi- 110016. Ph:6854102,

6865182, Fax:

6513942

AITS2002-PT-I-CH-2

1.

Bromophenol blue is an indicator with Ka value of 5.84 x 10~5. What percentage of the indicator is in its acidic form at a pH = 5.1 [4]

2.

At 90°C (the vapour pressure in mm of Hg) of CCI4 and SnCI4 solution can be given by the relation P = 750 XA + 362 where XA = mole fraction of CCI4. a) Find the vapour pressures of the pure components at this temperature. b) Calculate the % by weight of CCI4 in the vapour in equilibrium with a liquid containing equimolar mixtures of the two liquids at this temperature [6]

3.

Equal masses of two samples of charcoal A & B are burnt separately and the resulting C0 2 is collected in two vessels. The radioactivity of 14C is measured for both the gas samples. The gas from charcoal A gives 2100 counts per week and the gas from charcoal B gives 1400 counts per week. Find the age difference between the two samples? Half life 14 C = 5730 years. [6]

4

A polystyrene having formula Br3C6H2 (C8H8)n was prepared by heating styrene with tribromobenzoyl peroxide in absence of air. If it was found to contain 10.46% bromine by weight. Find n. [3]

5.

Calculate the Rydberg's constant if He+ ions are known to have wavelength difference between the first (of the longest wavelength) line of the Balmer and Lymann series equal to Ak= 133.7 nm? [6]

6.

A complex compound of unknown molar mass is known to dissolve in benzene without association or dissociation. A solution of this compound in benzene has a vapour pressure of 100 torr at 300.15 K. Pure benzene boils at 353.15 K at 760 torr pressure and its entropy of vapourisation is 87.03 JK~1 mol-1. What would be the mole fraction of the complex compound in this solution and boiling pt. of this solution? [8]

7.

In a face centered lattice with all the positions occupied by A atoms, the body centered octahedral hole in it is occupied by an atom B of appropriate size. For such a crystal, calculate the void space per unit volume of unit cell. Also predict the formula of the compound. [7]

8.

In the presence of chloride ion Mn2+ can be titrated with MnCXf, both reactants being converted to a complex of Mn(lll). A 0.545 g sample containing Mn 3 0 4 was dissolved and all manganese was converted to Mn2+. Titration in pressure of chloride ion consumed 31.1 ml of KMn0 4 that was 0.117 N against oxalate. Calculate the percentage of Mn 3 0 4 in the sample. [8]

9.

What H 3 0 + must be maintained in a saturated H2S solution to precipitate Pb2+, but not Zn2+ from a solution in which each ion is present at a concentration of 0.01 M? Ksp H2S = 1.1 x 10' 22 KspZnS = 1 x 1 0 " 2 1 [6]

10.

Calculate the energy required to excite one litre of hydrogen gas at 1 atm and 298 K to the first excited state of atomic hydrogen. The energy for the dissociation of H - H bond is 436 kJ mol"1. [6]

11.

A silver electrode is immersed in saturated Ag 2 S0 4 (aq). The potential difference between the silver and the standard hydrogen electrode is found to be 0.71 V. Determine Ksp (Ag 2 S0 4 ). Given E° „ = 0.799 V [6] Ag' / Ag

12.

FIITJCC.

a) The equilibrium constant K for the reaction

ICES House (Opp. Vijaij Mandal Enclave), Sarvapriya


p 6865182,

Fax: 6513942

yO°

AITS2002-PT-I-CH-3 1

is 300 at 280°C. if the enthalpy of polymerisation is - 27.216 kJ mol" . Calculate K at 250°C? [6] b) At 540 K, 0.10 ml of PCI5 are heated in a 8 L flask. The pressure of the equilibrium mixture is found to be 1 atm. Calculate Kp and Kc for the reaction. [6] 13.

A mixture in which the mole ratio of H2 and 0 2 is 2 : 1 is used to prepare water by the reaction 2H2(g) + 0 2 ( g ) — » 2H20(g) The total pressure in the container is 0.8 atm at 20°C before the reaction. Determine the final pressure at 120°C after reaction assuming 80% yield of water. [8]

14.

A sample of lead weighing 1.05 g was dissolved in a small quantity of nitric acid to produce 2+ + an aqueous solution of Pb and Ag (which is present as impurity). The volume of the solution was increased to 350 ml by adding water, a pure silver electrode was immersed in the solution and the potential difference between this electrode and a standard hydrogen electrode was found to be 0.503 V. What was the percentage of silver in the lead metal. E°Ag / Ag = 0.79 V. [61 1

15.

J

Calculate the value of logKp for the reaction N2(g) + 3H2(g) ^ 2NH3(g) at 25°C. The standard enthalpy of formation of NH3(g) is - 46kJ and the standard entropies of N2(g) 1 H2(g) and NH3(g) are 191, 130, 192 JK mol" respectively. [8]

FRITJCC ALL INDIA TEST SERIES IIT-JEE, 2002 4

MATHEMATICS (PART TEST - i) Time : Two Hours

Maximum

Marks : 100

Note: (i)

There are TEN questions in this paper.

(ii)

Attempt ALL questions,

(iii)

Number in brackets on the right hand margin indicates the marks for the corresponding question.

'(iv)

Answer all parts of a question at one place.

(vi)

Use of logarithmic table and calculator is NOT PERMITTED.

¥

5!

Name of the candidate Enrollment

Number

fllTJCC, Ltd ICES

House (Opp. Vijay Man dal Enclave), Sarvapriya Vihar, New Delhi-16. PH: 6865182, 6854102, Fax: 6513942

AITS2002-PT1 -MA-2

1.

Use mathematical induction to prove that for n > 1 n+1

C3 n+2 C3 n+3

2.

n-1G

np

3 n+1/^ 3

_ (n -1) n (n +1)

np n+1

C3

[10]

6

c

In a triangle ABC, a cosA + b cosB + c cosC = s. Prove that the triangle is equilateral. [10] 4

2

, 1 sec 9-3tan 9 , Prove that - < 4 2— < 1. 3 sec 0 - tan 9

3. 4.

[10]

A vertical flagstaff of length a is posted on the top of a vertical tower. When the sun is at an elevation a, the flagstaff is tilted away from the sun so as to make an angle 9 e (0, n/2) with the vertical. Prove that the shadow of the flagstaff on the ground is enlongated through a

. e distance 2a sin a — s i n - coseca. 2 V 2y

[10] n

5 (a).

If a and a + 1, for some value of a, are the roots of the equation

(x + k - l)(x + k) ~ 10 = 0, k=1

find n.

[5]

(b). If a1f a2, ... ,an are in A.P. with Sn as the sum of first n terms, prove that n

2

V"CkSk=2"- [na1+Sn].

[5]

k=0

6.

Inside a semi- circle of radius 1 unit, two circles of radii r, and r2 are drawn, each touching the circumference and the diameter of the semi-circle, and also touching each other externally. Prove that max.(r, + r2) = 2( V2 - 1 ) .

[10]

7.

Complex numbers z1f z2, z3 are represented by the points of contact D, E, F of the incircle of triangle ABC, with the centre O of the incircle taken as the origin. If BO meets DE at G, find the complex number represented by G. [10]

8.

A straight line cuts three concentric circles at A, B, C. If the distance of the line from the~ centre of the circles is p, prove that the area of the triangle formed by the tangents to the 1 circles at A, B, C is — .BC.CA.AB . [10]

2p

9.

2

The normals to the parabola y = 4ax at points P and Q meets the curve again at R. If T is the point of intersection of the tangents at P and Q to the parabola, prove that the locus of 2 the centroid of the triangle TRQ is y = a(3x + 2a). [10] 2

10.

2

x y Prove that the length of that focal chord of the ellipse — + -^-==1 which touches the a" b 2

2a(l-e H1 + e) parabola y = 4ax is —^ ^— - , where e is the eccentricity of the ellipse. J 1+ e - e 2

[10]

ick'k

FIITJ€€, Ltd ICES

House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi- 16. PH: 6865182, 6854102, Fax: 6513942

i

PllfJ€€ ALL INDIA TEST SERIES PART TEST -1 NT - JEE. 2002 PHYSICS Time : Two Hours

Maximum Marks : 100

NOTE : (i)

Attempt all questions.

(ii)

Start each question on a fresh page.

(iii)

There are twelve questions in this paper.

(iv)

Use of logarithmic table is PERMITTED

(v)


(vii)

Useful Data: 2


g

=

10 m/s


R

=

8.31 J/mol-K.

v

Name of the candidate Enrollment Number

FIITJ€€

ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Pb: 6865182, 6854102. Fax: 6513942

AITS2002-PT-1-PH-2

A man wants to reach point B on the opposite bank of a river flowing at a speed u as shown in the figure. What minimum speed relative to water should the man have so that he can reach point B ? In which direction should he swim ? 2.

B

[5]

Bob B of the pendulum AB hanging from a rigid support A is given an initial velocity *J2>Lg in horizontal direction. Mass of the rod AB is equal to the mass of the bob. Find the maximum height of the bob from the starting point.

[5] In a disc arial mass density varies directly proportional to the square of the distance from the centre of the disc. The moment of inertia of the disc about an axis passing through its circumference and perpendicular to the plane of the disc is I. Find the moment of inertia of the disc about the diameter of the disc. [8] In the figure shown there is a massless rigid rod at one end connected to a thin plate of mass m = 1 kg and area A = 10 cm2 which is covering a hole in a container filled with water of density p = 1 gm/cc to a height h = 1m. The other end of the rod is connected to a massless spring of stiffness k = 40 N/m on to which a block of same mass m = 1 kg is placed in equilibrium. Find the maximum compression from equilibrium position that can be given to the block so that after releasing it, liquid does not come out of container, (g = 10 m/s2)

J water

[8] On to a sphere of radius R/2 and density p2 with Centre at C2 a second sphere is moulded with density p-i radius R and centre Ci. Find the force experienced by a point mass m at point P at a distance y from the combination as shown.

[8] A wire is stretched between two fixed points separated by a distance of 2m such that tension in it is equal to 30.25 kgwt. The wire vibrates in its first overtone. A closed organ pipe of length 2m is brought near the wire. The temperature of the gas in the pipe is 27°C. When the organ pipe is made to vibrate in second overtone, five beats are heard every second. If the tension in the string is reduced slightly then the number of beats heard per second is reduced to three. Find the linear mass density of the wire. Given Cp/Cv of gas in the organ pipe = 1.44 and its mean molar mass is 27.7 gm. [8]

FIITJCC. ICES House, (Opp. Vijqy Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Pb: 6865182, 6854102. Fax: 651394-2

AITS2002-PT-I-PH-3

7.

A ball is sliding on a 80 cm high horizontal table with a speed of 2 m/s. It falls off the table, hits the ground and bounces to a height of 45 cm as shown. Find the distance x, if coefficient of friction between ball and ground is 0.2 2m/s

A 80 cm

-^X

n

45 cm

[8] A container of volume 4V0 made of a material perfectly non conducting is divided into two equal parts by a fixed rigid wall whose lower half is non conducting and upper half is pure conducting. The right side of the wall is divided into two equal parts (initially) by means of a massless non-conducting piston free to move as shown. Section A (Please refer to fig.) contains 2 moles of a gas while the section B and C (fig.) contains 1 mole each of the same gas (y = 1.5) at pressure P0. If the heater in left part is switched on till the final pressure in section C (fig.) becomes (125/27)P0. Calculate final temperature in each section and heat supplied by the heater (Express your answers in terms of gas constant R).

[10] A block A of mass m resting on a smooth surface has a rectangular slot in which a second block 'B' of same mass m can move freely in contact without friction. The string connected to the block B is passing through the massless frictionless pulley and is wrapped on solid cylinder again of same mass m and radius R. If the system is released from rest calculate the time in which the block B will hit the bottom of the slot. Friction is sufficient between the cylinder and the block A to prevent any sliding. 10.

11.

A container of empty mass m is pulled by a constant force in which a second block of same mass m is placed connected to the wall by a mass less spring of constant k. Initially the spring is in its natural length. Calculate the velocity of the container at the instant the compression in the spring is maximum for the first time.

[10]

m -T3W5TT

1 [10]

A cyclic process A - > B - > C - » A a s shown in figure is performed for 1 mol monoatomic gas (y = 5/3). Calculate the work done by the gas in the entire process and the value of Max and min pressure. From B to C curve follows the Vn2 parabolic equation (V - V0)2 = — ( T - T 0 ). Tn [10]

FIITJCC, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vibar, New Delhi -16. Ph: 6865182, 6854102. Fax: 6513942

AITS2002-PT-I-PH-4

12.

In the figure shown is a tank of negligible mass filled with a liquid of density p to a height h placed on a frictionless surface. The area of crosssection of the tank is A. Suddenly an orifice of area s ( s « A ) opens at the bottom of the tank as a result an automatic source S starts emitting sound of frequency f 0 and the tank recoils. Find the initial frequency of sound detected by the detector D. (Speed of sound in air = c)

[10] *

*

*

FIITJCC, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vibar, New Delhi -16. Ph: 6865182, 6854102. Fax: 6513942

FIITJCC ALL INDIA TEST SERIES IIT- JEE, 2002 PART TEST - II


Maximum Marks:

100

Note: i) ii) iii) iv) v) vi)

This paper has SIXTEEN questions. Attempt ALL questions. Answer to each new question should begin from a new page. Answer all sub parts of a question at one place. Use of logarithmic table is PERMITTED. Use of calculator is NOT PERMITTED.


R

Avogadro's Number

N, =

8.314 J K"1 mol"1 0.0821 Lit atm K"1 mol"1 1.987 » 2 Cal K"1 mol"1 6.023x 1023

Planck's constant

h

6.625 x 10"34 J s

=

=

6.625 x 10"27 erg s 1 Faraday

96500 Coulomb

1 calorie

4.2 Joule

1 amu

1.66 x 10~27 kg

Atomic No:


Atomic


Masses:

Name of the Candidate

Enrollment Number fllTJ«e. ICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 110016. Ph:6854102,

6865182, Fax:

6513942

AITS2002-PT-II-CH-2

1.

The treatment of 3, 3-dimethyl-but-1-ene with HCI gives three isomeric alkyt chlorides. Explain. [3]

2.

Write down the configuration of the geometrical isomers of the oxime of 2-bromo-5nitroacetophenone. How would you determine their configuration? Indicate the products obtained from each of them on treatment with PCI5 with mechanism. [8]

3.

a) Write down the structures of the following compounds with hybridisation i)

CI0 2

ii)

BrF5

iii) X e 0 3

b) The B - F distance in BF3 in 1,3A but that in [(CH3)N 4.

[6]

> BF3] is 1.42A. Explain

[3]

a) Write down the product obtained when sodium ethoxide reacts with the following compound CI-CH2-CH-CH2

\/ O

5.

[3]

Identify A - D in the following sequences EtCH - CO \ EtCH - CO C CH2 - CH

CH2 -

CH3 — C — CH2

\

CH /

EtCH - CH 2 OH (

CH2-CHOH

Me-CO

Me - C H O H D + C«-

HIO4

[8]

a) Write down a balanced chemical equation when sodium thiosulfate reacts with excess of mercuric chloride. [2] b) Write down the IUPAC nomenclature of the following i) NH 4 [Cr(NCS) 4 (NH 3 ) 2 ] ii) K 3 [Cr(C 2 0 4 ) 3 ] iii) K[Fe(C 2 0 4 ) 2 N 2 0 2 ] [6] Identify the following a) PhCHO

->A-

UAIH4

OH"

b) PhCHCI2

t-BuOK

PhC^CPh

->B-

H30+ [6]

-»E-

OEt~

->F

[6]

Two bottles A & B contain a solution of Sn2+ and Sn4+ respectively. To both of them H 2 S is passed followed by addition of yellow ammonium sulfide. Explain the observations in the two bottles. [4] 9.

Conversions i) Nitrobenzenep-nitrophenol ii) Acetophenone • —> phenyl acetic acid

10.

[ 3 x 2 = 6]

A compound of molecular formula C 5 H 12 0 while dehydrogenated produces B (C5H10O). B gives colour test with Schiff's reagent. (B) on treatment with concentrated alkali gives two disproportionation products (C) and (A). Alkane of the corresponding alkyl group present in (A) has got minimum boiling point as compared to other isomeric alkanes. Identify A, B and C. [6]

fllTJ«e. ICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 110016. Ph:6854102,

6865182, Fax:

6513942

AITS2002-PT-II-CH-3

11.

a) Electron affinity of SF5 is among the highest known but that of SF6 is quite modest. Explain. [3] b) Write down sequentially the action of heat on orthophosphoric acid [3]

12.

Site two reaction illustrating reducing property of H 2 0 2 one in acid medium and the other in alkaline medium. [4]

13.

Write down the reactions involved in the extraction of copper from copper pyrites by self reduction. [3]

14.

A metallic chloride A when treated with sodium hydroxide and H 2 0 2 gives a yellow solution due to the formation of compound B. The colour of this solution changes to orange yellow when dilute H 2 S0 4 is added to it due to formation of (C). Compound (D) when heated with (C) in presence of conc. H 2 S0 4 a red volatile liquid E is formed. E dissolves in NaOH giving yellow solution B, which changes to yellow ppt. F on treatment with lead acetate. (C) on reaction with NH4CI gives G which on heating gives colourless gas (H and a green residue (I). Identify (A) to (I). [8]

15.

a) Balance the following equations i) CI0 2 + Sb0 2 > CI0 2 " + Sb(OH)6 (alkaline medium) ii) KOH + K4[Fe(CN)6] + Ce(N0 3 ) 4 » Fe(OH)3 + Ce(OH)3 + K 2 C0 3 + KN0 3 + H 2 0 [4] b) Find the missing products or reagents i) A

)

B

LiAIH4

>

C

[3] 16.

Identify the following C6H10O4

+

(D)

^ C = O + CH 3 C0 2 H /

Optically inactive hot KMn0 4

CnHao < (B) Optically active

Lindlar s

'

calalyst

CiiH18

Li;EtNH2

>4 ethyl, 2,4 dimethyl 2,5 heptadiene

(A) Optically active 0 3 /Zn

C6H10O3 Optically active

+

N ; = o + CH3CO2H /

,C)

[5]

gym

.

fllTJ«e. ICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 110016. Ph:6854102,

6865182, Fax:

6513942

FIITJ€€ ALL INDIA TEST SERIES I I T - J E E , 2002 MATHEMATICS (PART TEST - II) Time : Two Hours

Maximum

Marks : 100

Note: (i)

There are TEN questions in this paper.

(ii)


(iii)


(iv)


(vi)



f l l T J « , Ltd ICES House (Opp. Vijay Mandal Enclave),

Sarvapriya

Vihar, New Delhi-16.

PH: 6865182, 6854102,

Fax:

6513942

AITS2002-PT2-MA-2

1.

I have six friends and during a vacation, I met them at several dinners. I found that I dinned with all of them exactly once, with every five of them on two days, with every four of them on three days, with every three of them on four days, with every two of them on five days. Each one was present for seven dinners and each missed seven dinners. How many dinners did I have alone.

2.

[10]

O is a point inside a triangle ABC, and OA = a, OB = p, OC = y. The bisector of angles BOC, COA, AOB meet the sides BC, CA and AB at P, Q and R respectively. Use vectors to Area of APQR 2oc(3y prove that [10] Area of AABC (a + p)(p + y)(y+ oc)'

3 (a).

If F(x) = - + / ,+ , —-v + x 2x(x + 1) 3x(x + 1)(x + 2j

to oo, prove that F(x) - F(x+ 1) = \ x

•

[5]

(b). X is the circumcentre of a triangle ABC. A point Y is taken on the side BC (inside the circle). Find the circumradius of the triangle CXY in the limit when Y tends to coincide with C. 4.

[5]

A curve is drawn in such a way that the distance of a given point (a, b) from the normal at any point of the curve bears a constant ratio k to the distance of the origin from the normal. Find all of such curves.

5 (a).

secx dx Evaluate J 2tanx + s e c x - 1 V 3

(b).

.

. - r

-

- 1

/v

V

[5]

2x

,

Prove that fj — -sin J 1 +x 0

6

[10]

1

+

X

\

-7

dx = 2

2

[5]

72

j

Let a function f be such that f(x) f(y) + 2 = f(x) + f(y) + f(xy) for all real x and y, with f (0) = 0, f(1) = 2, f(0) * 2. Prove that 3 Jf(x)dx-x(xf(x)+2) is constant.

[10]

i

7.

The vertex of a parabola lies on a circle of radius a and its axis is along a diameter of the circle. Find the latus rectum of the parabola if the area of the segment bounded by the parabola and its common chord with the circle is maximum.

8(a).

[10]

2

The equation t + 2xt + 4 = 0 does not possess distinct real roots. Find the equation of the 3

2

tangent of greatest slope to the curve y = x - 2x + x. (b).

[5]

If a, b, c, p and q are non- zero vectors, prove that a x (q x c ) x (p x b) = b x [(p x c ) x (q x a)]+ c x [(p x a ) x (q x b ) .

9.

[5]

A bag contains 5 bails and it is not known how many of these are white. Two balls are drawn and they are found to be white. What is the probability that all the balls in the bag were white.

[10] 2

2

10 (a). Find the area of the region bounded by y = (x - 4) , y = 16 - x and the x - axis. '

n

1 (b). Prove that lim _ L y " C n - > o c

vn

kTo

k

^ J L = e.

n

[5] 15}

k +1,

icicic

fill J€€? Ltd ICES House (Opp. Vijay Muncial Enclave),

Sarvapriya Vihar, New Delhi- 16. PH: 6865182, 6854102, Fax: 6513942

FIITJCC ALL INDIA TEST SERIES IIT-JEE, 2002 PART TEST - II


Maximum Marks : 100

Note: ) i) ii) v) v) vi)

This paper has ELEVEN questions. Attempt ALL questions. Answer to each new question should begin from a new page, Answer all sub parts of a question at one place. Use of logarithmic table is PERMITTED. Use of calculator is NOT PERMITTED.

Useful Data: Charge of an electron

e

1.6X10"19C

Acceleration

10m/s2


g R

8.314 Jmor1K"1

Gravitational

G

6.67 x10"11 Nrrv

Permittivity of free space

So

8.8 x 10"12 F/m

Planck's

h

6.625 x 10"34 J •

due to gravity

Constant

constant

Name of the Candidate

Enrollment Number

fllTJ«e. ICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 110016.Ph:6854102, 6865182, Fax: 6513942

AITS2002-PT-II-CH-2

1.

in the figure shown, a screen is placed normal to the line joining the three point coherent sources S-,, S2 & S3, emitting waves of equal amplitude. The separations S ^ & S2S3 are equal to a. Light of wavelength X is incident on the slits as shown. a) If the phase difference between the waves reaching a point P (as shown in the diagram) from St & S2 is 4>, find <> j in terms of a and 6,and also in terms of y and D b) Assuming that waves of equal amplitude A from S 1f S2 and S3 interfere at P, those from Si advanced by (f> and those from S3 retarded by <> j w.r.t. S2, find the resultant amplitude at P. c) Find the location of the points where the intensity is zero (minima). Write the expression in terms of y. [2+3+5=10]

2.

A point charge Q is fixed at the centre of an insulated disc of mass M. The disc is resting on a rough horizontal plane. An another charge - Q is fixed vertically above the centre of the disc at a height h. If the disc is displaced slightly in the horizontal direction. Friction is sufficient to prevent slipping. Find the period of oscillation of disc. A Bi

radioactive nuclei decays according to the following chain

Bi210

/il

po210

[10]

yL2 Pb206(stable)

Where decay constants are Xi and X2. Calculate the a and p activities of m0 gm of Bj 210 preparation at time t after its manufacture. [10] 4.

A wire loop carrying a current i is placed in the x-y plane as shown in figure. A particle of mass m and charge q is placed at origin v and given a velocity v = ~ ( i + j) m/s. Findinstantaneous acceleration a) b) If an external magnetic field B = B0 i is applied, find the force and torque acting on the loop due to this field.

[7+3=10]

A circuit consists of a permanent source of e.m.f. 'e' and a resistor R and a capacitor C connected in series. The interna! resistance of the source is negligible. At the moment t = 0, the capacitance of the capacitor was abruptly decreased by a factor n. Find the current in the circuit as function of time. [8] 6.

Figure shows a rod PQ of length I and an infinitely long current carrying wire in the same plane of the paper. If the rod is rotating with a constant angular velocity © about an axis normal to the plane of the paper and passing through P, find the emf induced across its ends P and Q at the moment it is normal to the wire and also determine which terminal is at higher potential. Current in the infinitely long wire is equal to i0.

P e- I

©

[8]


AITS2002-PT-II-CH-3

7.

For a singly ionized helium atom a) Compute the four lowest energy levels and construct the energy level diagram, b) What is the minimum energy of photon that can be absorbed by such atom in the stage n = 2? o c) What wavelengths ( A ) can be emitted when these atoms in the ground state are bombarded by electrons that have been accelerated through a potential difference of 50 V ? d) If these atoms are in the ground state, can they absorb radiation having a wavelength of 350A° ? 12+2+3+2 = 9]

8.

A light source of frequency v illuminates a metallic surface and ejects photoelectrons. The photoelectrons having maximum energy are just able to excite the hydrogen atoms in ground g state. When the whole experiment is repeated with an incident radiation of frequency — v , 5 the photoelectrons so emitted are able to excite the hydrogen atoms which then emit a radiation of six different wavelengths. a) Find the work function of the metal. b) What is the frequency of radiation . [6+2=8]

9.

An equiconvex lens of focal length 4/3 m is lying on a plane mirror placed at the bottom of a clear lake. A bird flies through its own image 1 meter above the surface of the lake find the depth of the lake. Given refractive index of water is 4/3, refractive index of the material of the lens is 1.5. ' [9]

10.

Consider the circuit shown in figure. Prior to t = 0, the switch is in position A for a long time and the capacitor is uncharged. At t = 0, the switch is instantaneously moved to position B. a) Determine the current in the L-C circuit for t > 0. b) Find the charge q on the lower capacitor plate for t >0.

11.

In the figure AB and OD are conducting rails and PQ is a conducting wire of mass m and length L The ends A & O are connected by a capacitor of capacitance C. There exists a magnetic field of induction B(x), where

B(x)

R

r V W W 4

B-

VO_

[6+ 4 = 10]

-1-

C

1+ 0. I) perpendicular to the plane of the rails, x-axis is in the direction OD with origin at O. The wire PQ is pulled by a constant force F. Find the acceleration of the wire PQ when it is at a distance 21 from O and has a velocity v0 Bo

[8]

*

*

*


FIITJCC RANKERS STUDY MATERIAL IIT-JEE, 2002 PRACTICE FULL TEST -1 CHEMISTRY Time: Two Hours

Maximum Marks: 100

Note: i)

There are TEN questions in this paper. Attempt ALL questions.

ii)


iii)


iv)



R

=

8.314 J mor 1 K- 1 0.0821 lit atm moP1 K~1 2 Cal mor 1

Avogadro's Number

N

=

6.023 x 1023

Planck's constant

h

=

6.625 x 10~34 J sec.

Velocity of light

c

=

3 x 108 m sec -1

1 electron volt

ev =

1.6 x 10~19 J

F

96500 C

Atomic No:

=




FIITJCC, ICES HOUSE (Opp. VIJAY MANDAL ENCLAVE), SARVAPRIYA VIHAR, NEW DELHI -16. Ph:6854102, 6865182, FAX: 6513942

RSM12-PT-ll(M)-CH-2

1.

2.

A finely powdered mineral of calcium 'A' was boiled with Na 2 C0 3 solution and the precipitate was filtered off. The filtrate was cooled and after cooling another precipitate 'B' produced which normally exists in the hydrated form. The aqueous solution of 'B' is alkaline. When 'B' is strongly heated, it gives a compound 'C' along with a glassy bead 'D'. 'C' was also present initially into filtrate. 'D' is fused with one of the metal sulphate 'E' to give green colour both in oxidising flame and reducing flame. Identify A to E with reaction and give the explanation. [12] (a) Two non reacting gases are taken in a vessel and allowed to effuse out through a pinhole into vacuum. At a particular time it is found that total 1.5 moles of gas have been effused out from an equimolar mixture ( 1 mole each initially) of two gases. If one gas is N0 2 and the molecular weight of other gas is 1.5 times of the average molecular weight of mixture of gas effused out then calculate the average molecular weight of gas mixture left in the vessel. [7] (b) Write the IUPAC name of the following complex. [Pt(NH3)(H20)(C6H5N)(N02)]N03 [2] (c) What is the hybridisation and shape of TeF5. [2] (a) Predict the major products in the following reaction with proper mechanism. + tt-BuOLi - R . i O l i+ ether

_

I [1+3] (b) Identify the organic products in the following reactions. H+ HzO

Ph-C03H CHCIg (ii) CH 3 —CH—COOH |

(i)LAH

(ii) H30+

>C

H+

)B

>D

NH2 4.

[1.5 x 4] (a) Let Mg Ti0 3 exists in pervoskite structure. In this lattice, all the atoms of one of the face diagonals are removed. Calculate the density of unit cell if the radius of Mg2+ is 0.72A and the corner ions are touching each other. [Given atomic mass of Mg = 24, Ti = 48] [6] (b)A cell is represented as follows Zn| Zn2+ J|HOI | Pt,H,(g) E° 2+ - -0.76V Zn /Zn 0.1M iatm If the volume of HCI solution in the R.H.S. half cell is 1.5 litre then calculate the weight of NaOH (70% pure by weight) required to be added in the H+ containing half cell to consume all H+ from that half cell. Given e.m.f of this cell is 0.701 volt at 25°C before its use. What is the change in e.m.f of the cell after addition of NaOH. [6]

5.

Equilibrium constants of T 2 0 (1T 3 is isotople of IH 1 ) differ from those of H 2 0 at 298K. Let at 298K pure T 2 0 has pT (like pH) = 7.62. Find out the pT of a solution prepared by adding 10 ml of 0.2M TCI to 15 ml of 0.25 M NaOT. [5]

6.

(a) An weak acid HA exists partially as trimer. If freezing point of a solution of this acid having 0.03 mole fraction of acid in benzene is 276.9K, then calculate the equilibrium constant for trimerisation of HA. Freezing point of benzene is 278.4K and latent heat of fusion of benzene is 10.042 K Joule mof 1 . (Assume molality of the solution is same as molarity). [7] 2+ 9 2+ 2 (b) Cu is'd ' system. But [Cu(NH3)4] is a dsp complex - explain. [3] FIITJCC, ICES HOUSE (Opp. VIJAY MANDAL ENCLAVE), SARVAPRIYA VIHAR, NEW DELHI -16. Ph:6854102, 6865182, FAX: 6513942

RS M12-PT-L (M )-C H -3

7.

An acid (A) when heated gives another acid (B). (B) When treated with Ca(OH)2 and heated (C) is obtained. Another compound (D) having molecular formula C 9 H 18 gives (C) and (E) due to ozonolysis. (C) didn't reduce Fehling solution but (E) did. (E) after catalytic hydrogenation gives (F) which treated with conc H 2 S0 4 to give (G) which again due to ozonolysis gives (I) and (H). (I) can reduce Tollen's reagent but (H) can't. Identify (A) to (I)

8.

(a) Eyes pass a signal to the brain when the visual receptors are struck by photons of wave length 850nm. If a total energy of 3.15 x 10~14 J is required to trip the signal, what is the minimum number of photons that must strike the receptor to pass a signal? [4] (b) To determine the concentration of HCN in blood of a patient a doctor decided to titrate a dilute sample of the blood with Kl solution of i2. A diluted blood sample of 15 ml was titrated with 5.21 ml of an l~ solution. 10.42 ml of this l3 was needed 1.222 gm sample of AS4C>6 for complete reaction. Calculate the molar concentration of HCN in blood? Atomic weight of As = 75, I = 127 It is also given As 4 0 6 coverts into H3ASO3 in solution [7]

9.

(a) Convert the benzene to p-methylbenzoic acid through formation of a Grignard reagent.

[9]

[3]

(b) Identify A and B in the following reaction PhS CH 2 =CH—CMe 2 CI ~" > A + B [2] (c) A compound having molecular formula C7H7CI exists into three isomeric form A,B,C, in which C has maximum dipolemoment. When they are treated with NaNH2 in liq.NH3 they yield three different isomeric product X,Y,Z as follows [8] A in presence of NaNH2 in liq. NH3 gives all the three products X,Y,Z B in presence of NaNH2 in liq NH3 gives only X and Y C in same condition gives Y and Z. Identify A,B,C and X,Y,Z with proper explanation.

10.

0.0852 gm of an organic halide (A) when dissolved in 2gm of camphor, the melting point of the mixture was found to be 167°C. Compound (A) when heated with Na in dry ether gives a gas (B). 280 ml of gas (B) at STP weighs 0.375 gm. Showing different steps give structure formulae of (A) and (B). [Kf for camphor = 40, melting point of camphor = 179°C] [7]


FIITJCC RANKERS STUDY MATERIAL , IIT - JEE, 2002 PRACTICE FULL TEST -1 MATHEMATICS Time: Two hours

Maximum Marks:100

Note: i).


ii).


iii).


iv).



Candidate Number

FIITJCC. ICES House, (0pp. Vijay Mandal Enclave), Sarv'.priya Vihar, New Delhi - 1 6 . Ph: 6865182, 6854102. Fax: 6513942

RSM12-PT-II (M)-PH- 81

1.

Given two curves y = f(x)

passing

through

(0,1)

and

y =

Jf(t)dt passing through -oo

(0 ,1 / n ) . The tangents drawn to both the curves at the points with equal abscissa intersect on the x- axis. Find the curve y = f(x). [10] 2.

If the tangent at (x^y^ to the curve x3 + y3 = a3 meet the curve again in (x2, y2), then prove that —2- + X2. = _ i . yi

3.

[10]

A tangent to the parabola y = x2 is drawn so that the abscissa x0 of the point of tangency belongs to the interval [1, 2], Find x0 for which the triangle bounded by the tangent, the y-axis, and the straight line y =x 2 has the greatest area.

4.

5.

6

[10]

Find the length of the focal chord of the parabola x2= 4cy which touches the ellipse

4a + b£ = 1
[10]

If x and y are real variables satisfying x2 + y2 +8x -10y +40 = 0 , and a = max.[(x+2)2 +(y -3) 2 ], b = min.[(x +2)2 +(y -3) 2 ] then prove that a +b = 18.

[10]

If z-i + z2 = 0 and Zi z 2 +z3 z4 = 0 then prove that the points representing zu z2, z3, z4 are concyclic. [10]

7.

sin3 f) cos 3 ft If — 7 , = : v, prove that tan 20 = 2tan (39 + a); sin(29 + a) cos(29 + ct)

8.

/ ^ / , ' 1 1 Prove that n Ci -[ 1 + - | n C 2 + |1 + - + - ] n C 3 +.... + (-l) n " 1 1 + — + — + ... + —n C n = - V n eN. 2 2 3 ) 2 3 n n V

1[10] J

[10] 9.

10.

Let f : R -> R be a function defined as f (x) =

1-1 x |, | x | < 1

[0, |x|>1 and g (x) = f (x - 1 ) + f (x + 1) V x e R. Determine g (x) in terms of x and discuss it's continuity and differentiability. [10] Show that x^ + y j + z ^ , x 2 i + y 2 j + z2k and x 3 i + y 3 j + z3k are non-coplanar if N

> lyil + !zil.

lyzl > |x2| +|z2| and |z3| > |x3| +|y 3 |.

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[10]

PlITJii RANKERS STUDY MATERIAL IIT - JEE, 2002

PRACTICE FULL TEST -1 PHYSICS Time: Two hours

Maximum Marks: 100

Note: i).


ii).


iii).


iv).



Candidate Number

f I I T J € € , ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar. New Delhi - 16. Ph: 6865182, 6854102. Fax: 6513942 i

RSM12-PT-l-(M)-PH-2

A rectangular tank of height 10 m filled with water, is placed near the bottom of an incline of angle 30°. At height x from bottom a small hole is made (as shown in figure) such that the stream coming out from hole, strikes the inclined plane normally. Find x.

[10]

A sound source generating a natural frequency of n0 is lying on the incline at a distance 2h along incline. The source starts sliding with initial velocity 2V0 up the plane. The coefficient of friction between the inclined plane and the sound source is 1/V3. After t = V3Vo/2g sec, the source emits a pulse, and it is found that its apparent frequency equals — n0 as observed at the point O. Calculate,the :velocity of 11 • " " sound. 3.

4.

5.

[10]

Two identical adiabatic vessels A and B each containing n mole of a rhonoatomic and diatomic ideal gas respectively. Both the vessels are connected by a rod of length I and crossectional area A. Thermal conductivity of rod material is k and lateral surface of the rod is insulated. At any time t = 0, the temperature of the. gases in the vessel are J-i and T2 respectively (T-, > T2). Neglect heat capacity of the rod and the vessels, find the time when the temperature difference of the vessels becomes half the initial -temperature difference. Assume that there is no loss of heat from the sides of the rod. ' : " A wire loop ABCDE carrying a current i is .. . placed in the x-y plane as shown in figure. A particle of mass m and charge q is projected y from origin with velocity v = - ~ ( i + j) m/s. v2 Find the (a) instantaneous acceleration (b) If an external magnetic field B = B0i is applied, find the force and torque acting on the loop due to this field.

"

;

-

' '

'

[10]

[6+4=10]

The resistance each of 16 Q and capacitance of each 100 |aF are arranged as shown in the figure. A battery of emf 12 V is joined across A and B. Find the (i) reading of the ammeter just after key is closed and after long time. (ii) charges in each capacitors when steady state is achieved. 16 n

D

16 Q

[5+5=10]

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RSM12-PT-I-{M)-PH- 3

Figure shows three concentric thin spherical shells A, B and C of radii a, b and c respectively. The shells A and C are given charges q and - q respectively and the shell B is earthed. Find the charges appearing on the surface of B and C, [10] A monochromatic point source S is emitting light of wavelength X = 5400 A0 is placed at a distance 1.6m from a screen. A equiconvex lens of focal length f =/15 cm is cut along a diameter into two 0.5 mm identical halves. The two halves are s placed symmetrically about the line So with a gap 0.50 mm between the source and screen at a distance 30 cm from the 0.3 m < source. A thin transparent plate of < 1.6 m thickness 4.5 ^m is inserted on the path of ray emerging from one of the halves. It is found that the intensity at point O on the screen is now (3/4) of the maximum intensity on the screen. It is further observed what used to be fourth maxima earlier lies below O while sixth maxima lies above O Neglecting absorption of the light by transparent plate. Calculate (a) fringe width (b) refractive index of the transparent plate. [5+5=10] Assume that the images formed by the lens act as secondary sources of light. A long rectangular slab of transparent medium of thickness d is placed on a table with its length parallel to the X-axis and width parallel to Y-axis. A ray of light travelling in air makes a near normal incidence on the slab as shown. Take the point of incidence as origin (0, 0, 0) and jj. =

1-(x/r)

. Where (j0 and r(> d) are constants. ^

YA

=1

air medium air

> X

[10] Determine the X-coordinate of the point A, where the ray intersects the upper surface of slab - air boundary. A nucleus X - initially at rest, undergoes alpha - decay, according to the equation -> z Y 2 2 8 + a (i) Find the value of A and Z in the above process. (ii) The a - particle in the above process is found to move in a circular track of radius 0.11 m in a uniform magnetic field of 3T. Find the energy (in MeV) released during the process and binding energy of the parent nucleus X. Given : my = 228.03 amu m a = 4.003 amu m (0n1) = 1.009 amu m(iH1) = 1.008 amu [10] 92 X

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RSM12-PT-I-{M)-PH- 3

10.

A gas of identical hydrogen -like atoms has some atoms in the ground state and some atoms in a particular excited state and there are no atoms in any other energy level. The atoms of the gas make transition to a higher energy state by absorving monochromatic light of wavelength 304 A 0 subsequently, the atoms emit radiation of only six different photon energies. Some of emitted photons have wavelength 304 A0, some have wavelength more and some have less than 304 A0, then (i) find the principal quantum number of the initially excited state (ii) identify the gas. (iii) find the ground state energy (in eV) (iv) find the maximum and minimum energies of emitted photons. [4+3+1+2=10]

*

-k

*

I

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FIITJCC RANKERS STUDY MATERIAL IIT-JEE, 2002

PRACTICE FULL TEST - II CHEMISTRY Time: Two Hours

Maximum Marks: 100

Note: i)

There are FIFTEEN questions in this paper. Attempt ALL questions.

ii)


iii)


iv)



R

=

3.314 J mor 1 K" 1 0.0821 lit atm moP1 K"1 2 Cal mor 1

Avogadro's Number

N =

6.023 x 1023

Planck's constant

h

=

6.625 x 10~34 J sec.

Velocity of light

c

=

3

1 electron volt

ev =

1.6 x 10~19 J

F

96500 C

Atomic No:

=

x

108 m sec"1





RSM12-PT-ll(M)-CH-2

1.

2.

A certain metal was irradiated with light of frequency 3.2 x 1016 Hz. The photoelectrons emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with light of frequency 2 x 1016 Hz. Calculate the threshold frequency for the metal, ' [6] (a) Draw the structures of the following indicating their shape and hybridisation (i) CI0 3 (ii) ICI2- (iii) BrFs [ 3 x 2 = 6] b) Complete the following equations with proper balancing i) Ca 3 (P0 4 ) 2 + Si0 2 + C > ii) (NH4)2S208 + H 2 0 + MnS0 4 > iii) NaHS0 4 + Al + NaOH > [ 3 x 2 = 6]

3.

What mass of Pb2+ ions is left in solution when 50 m! of 0.2 M Pb(N0 3 ) 2 is added to 50 ml of 1.5 M NaCI? Ksp(PbCI2) = 1.7 x 10""4. [6]

4.

Fallout from nuclear explosion contains 1311 & 90Sr. Calculate the time required for the activity of each of these isotopes to fall to 1.0% of its initial value. Radioiodine & radiostrontium tend to concentrate in the thyroid and the bones respectively, of mammals which ingest them. Which isotope is likely to produce the more serious long term effects? (Half life of 1311 = 8 days and t1/2 of 9 0 Sr= 19.9 years. [6]

5.

Explain the following i) CH3CH2CI hydrolyses slowly in aqueous medium but the reaction is rapid in presence of catalytic amount of Kl. ii) Acetone gives iodoform with NH4OH & l2 but C2H5OH does not, C 2 H 5 OH gives iodoform only when it reacts with aq. NaOH & l2. [ 3 x 2 = 6]

6.

An ideal solution was prepared by dissolving some amount of cane sugar (non-volatile) in 0.9 moles of water. The solution was then cooled just below its freezing temperature (271 K) where some ice get separated out. The remaining aqueous solution registered a vapour pressure of 700 torr at 373 K. Calculate the mass of ice separated out, if the molar heat of fusion water is 6 kJ. [7]

7.

i)

8.

At 300 K and 1 atm pressure the density of gaseous HF is 3.17g/L. Explain this observation and support your explanation by calculations. [3]

An alloy containing two metals A ' & B is treated with dilute HCI. A ' dissolves with evolution of hydrogen leaving behind B. B is separated from the solution C. ii) The residue B dissolves in concentrated nitric acid giving a blue solution D. iii) Mercuric chloride solution with solution C gives a silky white precipitate which turns grey on addition of excess C. iv) Addition of NH3 solution to D gives a blue precipitate which dissolves in excess of NH3 giving a deep blue coloration. Name two metals A & B and alloy. Give equations for the reaction (i) to (iv). [8]

[3x2]


RSM12-PT-ll(M)-CH-2

10.

a) Citral an organic compound present in lemons and oranges having M.F. C10H16O is an unsaturated octa-aldehyde derivative with no adjacent side chains. It, on treatment with hot KMn0 4 gives a compound (A), levulic acid (M.F. C 5 H 8 0 3 ) containing carbon atoms in a straight chain including a carboxy group, compound (B) and oxalic acid. Compound (A) & (B) both gives positive iodoform test. Give structures of citral, (A) & (B). [7] b) A hydrocarbon was found to have a molecular weight between 80 - 85. A 10.02 mg sample took up 8.4 ml of H2 gas measured at 0°C& 760 mm pressure. Ozonolysis yielded formaldehyde and glyoxal only. What was the hydrocarbon? [4]

11.

At 46°C K p for the reaction N 2 0 4 (g) 2N0 2 (g) is 0.66 atm. Compute the percent dissociation of N 2 0 4 at 46°C and a total pressure of 380 torr. What are the partial pressures of N 2 0 4 & N0 2 at equilibrium? [6]

12.

1.1g CH3(CH2)nCOOH was burnt in excess air and the resultant gases (C0 2 & H 2 0) were passed through a solution of NaOH. The resulting solution was divided into 2 equal parts. One part required 30 m! of 2.5 N HCI for neutralization using phenolphthalein as indicator. The other part required 40 ml of 2.5 N HCI for neutralization using methyl orange as indicator. Find the value of n. [8]

13.

Molar conductances of MCI 4 xNH 3 complexes at a dilution of 1024 litres are 7, 97, 229, & 523 Ohm"1 cm2 of x = 2, 3, 4 and 6 respectively. Rationalise the modes of ionisation of these complexes according to Werner's theory of coordination. [6]

14.

The unit cell of TIAI(S0 4 ) 2 nH 2 0 is face centered cubic with a = 1.221 nm. If density of unit cell is 2.32 g cm"3 find the value of n? [4]

15.

Identify A, B, C & D -COPh i) PhMgBr / CuCI ii) E t 2 0 iii) H

ii)

-» A (C19H2OO)

+

Na/Liq. NH3

-»B-

NBS

(CH3)2CuLi

[2 + 3 = 5]


FIITJCC RANKERS STUDY MATERIAL IIT - JEE, 2002 PRACTICE FULL TEST - II PHYSICS Time: Two hours

Maximum Marks: 100

Note: i).


ii).


iii).


iv).



Candidate Number

fllTJ«e. ICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi -110016.Ph:6854102, 6865182, Fax: 6513942

RSM12-PT-I! (M)-PH- 2

Four spherical masses of radius R/4 each are taken out of a spherical, body of mass M and radius R. Find (i) gravitational force on a particle placed at origin (ii) gravitational potential at A, co-ordinate (R/2, 0, 0) > x

[10] A disc of mass M and radius R is kept and sandwiched between two other discs each of mass 2M and radius 2R in such a manner that they are coaxial and their common axis is parallel to the horizontal plane. A string is wound over smaller discs and pulled along a line upward at an angle 0 to the horizontal and the system is purely rolling over the rough horizontal surface. Find the acceleration of centre of system of discs. [10]

A bead A is attached to the mid point of a wire B. If the wire is cooled down to 0°c and a transverse wave is sent, it takes a time t to travel across it from point 1 to point 2. Assuming gravity free space and ratio of the masses of the wire and the bead be n«1,find the time period of small oscillations of the bead at 0°c.

A thermally insulated vessel is divided into two parts by a heat insulating piston which can move in the vessel . The left part of the vessel contains one mole of an ideal mono atomic gas, and the right part is empty. The piston is connected to the right wall of the vessel through a spring whose length in free state is equal to the length of the vessel. Determine the heat capacity C of the system neglecting heat capacity of the vessel, piston and spring.

A wire frame ABCDEA of resistance r, consist of a semicircular wire BC of radius 2a, a quarter circular wire DE radius a and a straight wire AE are connected as shown in the figure. A magnetic field B exist in the space such that B is equally inclined with positive direction of all the three axis. The magnitude dB of magnetic field increases at a constant rate — dt a. Calculate (i) magnitude of induced emf in the mesh (ii) magnitude and direction of induced current in the mesh.

[10]

[10]

[8+2-10]

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6.

In the figure shown a parallel plate capacitor is connected across a source of emf s. The plates are square shaped with edge T and separated by a distance d. A dielectric slab of dielectric constant k and thickness d is inserted between the plates with constant speed v. Find the current in the connecting wires, ignore the resistance of connecting wires. A current of equal strength is passed through a coil of resistance R and dipped into water kept in a bucket at 0°C. What would be the temperature of the water at the end of 5 minutes if mass of the water is m and specific heat of the water is c ? [10] Parallel rays from a monochromatic source are incident on a piano convex lens containing indentical parts 1&2 having refractive indices nearly equal to n and having a small difference to be taken into account say An. A screen is situated at a large distance D from the focus of the combination. Find the value of the y for first maxima. Assume that the radius of the curvature of each surface of the lens as R and a real image formed may be considered as a source.

8.

[10]

The arrangement of the Lloyd's mirror experiment is shown in the figure. 'S' is a point source of frequency 6x1014 Hz. A and B represent the two ends of a mirror placed horizontally, and LOM represents the screen. L

. s lmii^

Mirror

O 5cm

•« -

5 cm

M

190cm

M

Determine the position of the region where the fringes will be visible and calculate the number of fringes. 210

206

4

[10]

It is proposed to use the nuclear reaction 84Po » + 2He to produce 2kW 82Pb electric power in a generator. The half life of polonium (Po210) is 138.6 days. Assuming efficiency of the generator be 10 % calculate (a) how many grams of polonium (Po210) required per day at the end of 1386 days (b^ initial activity of the material Given : (Mass of nuclei: Po210 = 209.98264 a.m.u., Pb206 = 205.9440 a.m.u. 4 = 4.00260amu, 1 amu = 931 MeV) [5+5=10] 2 He 10.

A Bi210 radioactive nuclei decays according to the following chain A1 pGû TT^ Pb (stable) Bf" where decay constants are X-i and X,2. Calculate the a and p activities of m 0 gm of B,210 preparation at time t after its manufacture. [10]

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ALL INDIA TEST SERIES 2002

I

FULL TEST - III


Maximum Marks: 100

Note:


This paper has THIRTEEN questions. Attempt ALL questions. Answer to each new question should begin from a new page Answer all sub parts of a question at one place. Use of logarithmic table is PERMITTED. Use of calculator is NOT PERMITTED.


R

=

= = Avogadro's Number Planck's constant

Na = h

=

1

1

8.314 J K~ mol" 0.0821 Lit atm IC'mol 1 1 1.987 « 2 Cai K" mo!" 23 6.023 x 10 - 1

34

6.625 x 10"

J• s

27

6,625 x 1Q~ erg • s 1 Faraday 1 calorie 1 amu

=

96500 Coulomb 4.2 Joule 1.66 x 10~27 kg

Atomic No:


Atomic Masses



FIITJCC Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Terminalj, New Delhi - 16, Ph : 6515949,

6865182,

6854102, Fax : 6513942

AITS2002-FT-III-CH-2

1.

An organic compound (A) of m.f. C4H8 was reacted with an aqueous bromine solution. A colorless solution (B) was obtained. (B) when treated with a base gave a compound (C) of m.f. C 4 H S O . ( C ) on reaction with MeMgl followed by acid treatment gave(D). ( D ) on reaction with conc. HCI gave (E). (E) on reaction with ale. NaOH gave (F). (A) can show geometrical isomerism, but (F) can't. Identify (A) to (F). [6]

2.

Identify the missing reagents (or products) CI a)

(A)

-»(B)

Ha

°'

excess

CI

> (C) Ph'

x

Ph

[5]

CH2OH b)

(A)

NBS (2eq.)

->(B)

. KOH (excess)

CH2OH

PCC

->(C)

NH2-NH;

•»(D) [4]

An organic compound (A) reacts with H2NOH to give (B). (B) on treatment with P 2 0 5 produces only one product (C). (C) on hydrolysis produces (D) and (E). (D) on reduction with DIBALH gives (A) back. Identify (A) to (E). [5] Effect the following conversions a) [3]

b) [3]

c)

(in 3 steps) [4]

Give reasons for the following or a) CICH2CHCI2 ~ > CH2 CCI2 and not CICH = CHCI. b) Vinylchloride doesn't give SN reactions but allyl chloride gives c) Carbonyl compounds are not formed in good quantity by using Grignard reagent and nitriles. [ 3 x 2 = 6]

a) Diphenyl ether is not very easily prepared using Williamsons synthesis. Provide an useful synthetic route for the same. b) a-keto acids on treatment with warm dilute sulphuric acid undergoes decarboxylation. Explain a suitable mechanism for this reaction. [ 2 * 3 = 6] i)

An ore (A) on roasting with sodium carbonate and lime in the presence of air gives two compounds, (B) and (C). ii) The solution of (B) in conc. HCI on treatment with potassium ferrocyanide gives a blue colour (or) precipitate of compound (D). iii) The aqueous solution of (C) on treatment with conc. H £ 0 4 gives a yellow coloured compound (E). iv) Compound (E) when treated with KCI gives an orange red compound (F) which is used as an oxidising agent. v) The solution of (F) on treatment with oxalic acid and then with an excess of potassium oxalate gives blue crystals of compound (G). Identify (A) to (G) and give balanced chemical equations for the reactions. [10] FIITJ€€ Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Terminal), New Delhi - 16, Ph : 6515949 , 6865182, 6854102, Fax : 6513942

AITS2002-FT-II1-CH-3

8.

a) In steel manufacture, manganese metal is used as a scavenger to reducetraces of iron oxide and iron sulphide. Suggest reasons why manganese is effective for this purpose, b) 100.0g of P 4 0 6 is reacted with 100g of KMn0 4 in HCI solution to form H3P04 and MnCI2. Determine which reagent is in excess and by how much. [ 3 ^ 3 = 6]

9.

a) A balloon whose diameter is 20 meter weighs 100 kg. It is filled with He at 1.0 atm and 3 27°C. Calculate its pay load if density of air is 1.2 kg m" b) Calculate the energy required to excite one litre of hydrogen gas at 1 atm and 298 K to the first excited state of atomic hydrogen. The energy for the dissociation of H - H bond 1 is 436 kJ mol" . Also calculate the minimum frequency of photon to break this bond. 13 c) Disintegration of radium takes place at an average rate of 2.24 x 10 a-particles per minute. Each a-particle takes up 2 electrons from the air and becomes a neutral helium 3 atom. After 420 days, the He gas collected was 0.5 * 10" L measured at 300 K and 750 mm of mercury pressure. From the above data calculate Avagadro's number. d) The H - O - H angle in the water molecule is 105°, the H - O bond distance being 0.94A. The dipole moment for the molecule is 1.85D. Calculate the charge on the oxygen atom. e) If the temperature co-efficient for the hydrolysis of isopropyl propionate by caustic soda is 1.75, calculate the activation energy. [5*3-15] 2+

2+

10.

The analytical technique used for the removal of Cu ions from a solution of Cu (aq) is to 2+ add NH3(aq). A blue colour signifies the formation of a complex [Cu(NH3)4] having 13 2+ formation constant as 1.1 x 10 and thus confirms the presence of Cu in solution. 0.25 L of 0.1 M aqueous copper sulphate solution is electrolysed by passing a current of 3.512 ampere for 1368 seconds, subsequently aqueous ammonia whose concentration is 2 maintained to be 0.1 M is added to the electrolysed solution. If [Cu(NH3)4] * is detectable 5 upto its concentration as low as 1 x 10~ , would a blue colour be shown by the electrolysed solution on addition of NH3? [5]

11

An organic liquid, A , immiscible with water, when boiled together with water, has a boiling point of 90°C at which the partial vapour pressure of water is 526 mm Hg. The superincumbent pressure is 736 mm Hg. If mass ratio of liquid and water collected is 2.5 : 1, what is the molecular mass of the liquid? [5]

12.

a) A reaction carried out by 1 mole of N2 and 3 mole of H2 show at equilibrium the mole fraction of NH3 as 0.012 at 500°C and 10 atm pressure. Calculate Kp. Also report the pressure at which mole % of NH3 in equilibrium mixture is increased to 10.4. [5] 2+ b) A solution contains 0.1 M Mg and 0.8 M NH4CI. An equal volume of NH3 is added to the solution. A solid substance just starts settling down the reaction vessel. Given that Ksp of 11 5 Mg(OH)2 = 1.4 x 10" and Kb of NH4OH = 1.8 x 10~ calculate [NH3] in solution. [5]

13.

a) The magnetic moment of [Mn(CN)6f~ is 2.8 B.M and that of [MnBr4f~ is 5.9 B.M. What are the geometries of these complex ions. [4] b) In the equation A + 2B + H 2 0 > C + 2D, (A = HN02, B = H2SCh, C = H2NOH). Identify D and draw the structures of A, B, C and D. [3]

H8?J€€ Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Terminal), New Delhi - 16, Ph : 6515949 , 6865182,

6854102, Fcvc : 6513942

FIITJCC RANKERS STUDY MATERIAL IIT - JEE, 2002 PRACTICE FULL TEST - II MATHEMATICS Time: Two hours

Maximum Marks: 100

Note: i).


ii).


iii).


iv).



Candidate Number

FIITJ€€, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi -16. Ph: 6865132, 6854102. Fax: 6513942

RSM12-PT-II (M)-MA-2

1.

Let z h z2, z3 be three distinct complex numbers satisfying

|z-, - 1| = |z2 - 1| = |z3 - 1|. Let A,

B and C be the points represented in the Argand plane corresponding to zu z2 and z3 respectively. Prove that Zi + z2 + z3 = 3 if and only if AABC is an equilateral triangle. 2.

[10]

Find the points on the x-axis from which exactly three distinct chords (secants) of the circle x2 + y2 = a2 can be drawn which are bisected by the parabola y2 = 4ax, a >0.

3.

[10]

Suppose f(x) = x3 + ax2 + bx + c. a, b, c are chosen respectively by throwing a die three times. Find the probability that f(x) is an increasing function.

4.

[10]

If the sides of a triangle are in G.P. and its biggest angle is twice the smallest angle, then prove that the common ratio (>1) lies in (VV2 !A /(V5 +

5.

J.

[10]

If (a, b, c) is a point on the plane 3x + 2y + z = 7, then find the least value of a2 + b2 + c2, using vector method.

6.

[10]

There are exactly two points on the ellipse same and is equal to J

7.

a 2 + 2b2

x2 a

+

v2 b

= 1 whose distance from its centre is the

. Find the eccentricity of the ellipse.

[10]

Find the number of ways of selecting 5 coins from coins, three each of Rs.1, Rs.2 and Rs.5. [10]

8.

2

Show that the area of the quadrilatral, formed by the common tangents of the circle x +y = c2 x2 v2 and the ellipse — + a b

b < c < a is

2

2c2 (a2 - b 2 ) ^ • 7(a - c 2 j ( c 2 - b 2 ) u o 2

f(x + a) f(x + 2a) f(x + 3a) 9.

Let A(x) =

f(a)

f(2a)

f(3a)

f'(a)

f'(2a)

f'(3a)

(the prime denotes the derivative with respect to x),

for some real valued differentiate function f and constant a. Find lim ^ ^ . *->o x 10.

[10]

ABCDE is a regular pentagon. A tower stands at the point A. The angles of elevation of the top of the tower at B and C are a and p respectively. Show that (3 + V5 )cot2a = 2cot2p.

It-kick

HIYJ€€, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi -16. Ph: 6865182, 6854102. Fax: 6513942

[10]

ALL INDIA TE IIT - JEE, 2002 FULL TEST - III (MAINS) PHYSICS Time : Two Hours

Maximum Marks : 100

NOTE : (i)


(ii)


(iii)

There are Twelve Questions in this paper.

(iv)

Answers of ail parts of a question should be given at one place.

(v)

Use of calculator is NOT PERMITTED.

(vi)

Use of logarithmic tabie is PERMITTED.

(vii)

Useful Data : Acceleration due to gravity Permittivity of free space Planck's constant

g e0 h

= = =

10 m/s

2 12

8.8 x 10" F/m 34

6.625 x 10"

J•s

r


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AITS-2002 - FT-Iil- PH-2

1

Two identical beads connected by an inextensibie massless string can slide along the two arms AB and BC of a rigid smooth wire frame in a vertical plane. If the system is released from rest, find the speeds of the particles when they have moved by a distance of 0.1 m

0.4m

i

0.3m

v/y/A

2.

3.

2

A uniform rod of cross-sectional area a = (0.2 cm ) is bent to form a square of side £ = 10cm. Three of the ends of the square A, B and C are maintained at 100°C, 40°C and 0°C respectively. If the conductivity of the material of the rod be k = 385 W./m°C. Find the net rates of heat supplied or absorbed by the reservoirs at A, B and C. The lateral surfaces of the rod are insulated.

\

1Q0X

[6] B 40 °C

Jk o°c

[6] Partially silvered

A ray is incident on a glass sphere as shown. The opposite surface of the sphere is partially silvered. If the net deviation of the ray transmitted at the partially

1

silvered surface is — rd of the net deviation suffered by J 3 the ray reflected at the partially silvered surface (after emerging out of the sphere). Find the refractive index cf the sphere. 4

6]

Two ends of a Sight rigid rod having length I = 60cm are connected to two identical uniform discs A & B. The wall in the shown diagram is smooth and the floor is sufficiently rough to ensure pure rolling. The system starts from the position 8 = 0. Find the velocity of the mid point of the rod when 9 - 60°. [8]

5

A gas containing hydrogen like ions with atomic no Z3 emits photons in transition n + 2 -> n, where n = Z. These photons fall on a metallic plate and eject electrons having minimum de-Broglie wavelength X of 5A°. Find the value of lZ if the work function of metal is 4.2eV.[8]

6.

AB is a vertical rigid infinite wire carrying a linear charge of density X = 10|iC/m. A particle having mass m = 2 gm. and charge Q = 1jaC is attached to the wire by means of a light, insulating and inextensibie string having length ( = 2V2m. Find the vertical velocity u with which it should be projected under gravity from the shown position so that the string slacks when its anale with the vertical becomes 45°.

A

u X.

m

g V

B

[8]

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AITS-2002 - FT-III- PH-3

7.

X x x x x x x x x x x x x Two infinite parallel wires, having the A cross sectional area 'a' and resistivity 'k' are connected at a junction point 'P' (as X X X XMX X X X X X X X shown in the figure). A slide wire of p F-FoV negligible resistance and having mass X x x xkflx X X X X X X X 5 'm and length T can slide between the parallel wires, without any frictional X X X X X X X X X X X X X resistance, If the system of wires is introduced to ' a magnetic field of (0,0) 4 intensity B' (into the plane of paper) and X X X X X X X X X X X X [83 the slide wire is pulled with a force which varies with the velocity of the slide wire as F = F0V, then find the velocity of the slide wire as a function of the distance travelled. (The slide wire is initially at origin and has a velocity v0)

e.

A fighter plane is flying horizontally at a height of 250 m from ground with constant velocity of 500 m/s. It passes exactly over a cannon which can fire a sheli at any time in any direction with a speed of 100 m/s. Find the duration of time for which the plane is in danger of being hit by a cannon shell. [10]

9.

A particle having charge q~10|j,C and mass m~3 mg. has a velocity, v = (10 cm /s) (i+2j) at t = Oat origin. There exists a uniform magnetic field B=0.67i:Ti. Another uncharged particle is moving with a constant velocity along negative x-axis. At t = 0 its x co-ordinate is -H5G cm. The two particles collide and stick together and the combined mass goes in a circular path. Find the possible values of the mass of the uncharged particle. [10] 'n' moles of a diatomic gas undergo a cyclic process as shown in the figure, if the process D-E is adiabatic then find (a) The value of V c , VD and VE [Take (1.5)^=1.33] (b) The heat absorbed in the process DE, and 1.6 P the net change in interna! energy during the entire cyclic process, (c) the heat supplied to the gas in tne process AB, BC and CD. (d) the work done by the gas in the process BC and DE. 2 (The values of V c and VD are roots of the equation V* - 3.5V V c + 3 V0 = 0, and express the answer in terms of PQ & V0.) [2-f-l4-44-3]

11.

The figure shows a cylinder having length 15cm, cross 2 3 sectional area 100cm and density 500kg/m . The base of the cylinder is connected with a thread of length 35cm and a spring of natural length 40cm. (The spring constant of spring is 150 N/m) The arrangement is kept in a large water tank (The tank has depth 40cm and the 40 c m 3 density of water is 1000 kg/m ). Find (a) tension in the thread, Now, if the thread snaps at t = 0. Find (b) maximum length of the cylinder out of the water (c) the value of t at which the maximum length of the cylinder will be out of water for the first time.

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i

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AITS-2002 - FT-III- PH-4

12.

in a usual Young's Double Slit Experiment using monochromatic visible light the distance between the plane of slits and the screen is 1.7 meter. At a point (P) on the screen which is directly in front of the upper slit nth maximum is observed. Now the screen is moved 50cm closer to the plane of slits. Point P now lies between third and fourth minima above the central maxima and the intensity at P is one-fourth of the maximum intensity on the screen. Find the (a) vaiue of n. (b) wavelength of light if the separation of slits is 2mm. [7+3]

*

it

-k

\

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ALL INDIA TEST SERIES

IIT-JEE, 2002 FULL TEST - IV


Maximum Marks: 100

Note:


This paper has FOURTEEN questions. Attempt ALL questions. Answer to each new question should begin from a new page. Answer all sub parts of a question at one place. Use of logarithmic table is PERMITTED. Use of calculator is NOT PERMITTED.

Useful Data:

R =

Gas Constant

1

1

8.314 J K" mol" 0.0821 Lit atm K"1 mol' 1 1.987 « 2 Cal K"'mol 23 6.023x 10

Avogadro's Number

Na

= =

Planck's constant

h

=

6.625 x 10"34 J - s

= = =

6.625 x 10~27 erg • s 96500 Coulomb 4.2 Joule 1.66 x 10~27 kg

1 Faraday 1 calorie 1 amy

- 1

Atomic No:


Atomic Masses



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k


Photoelectrons are emitted when 400 nm radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region containing a-particles. A maximum energy electron combines with an a-particle to form a He+ ion, emitting a single photon in this process. He+ ions thus formed are in their fourth excited state. Find the energies in eV of the photons, lying in the 2 to 4 eV range that are likely to be emitted during and after the combination (h = 4.14 x 10~15 eVs). [8] a) Carry out the following conversions O O -C0 2 Et J-L / E t

b)

[3 + 3 = 6] 3.

The solubility product of a sparingly soluble salt X2Y3 is 3.5 * 10~8 at 30°C and the vapour pressure of its saturated solution in water is 31.8 mm of Hg at 40°C. Calculate the enthalpy change of the reaction X2Y3 5==^ 2X3+ + 3Y~2. Vapour pressure of pure water = 31.9 mm of Hg at 40°C. [7] Identify the unknown alphabets in the followings a) R' x = O + CICH 2 C0 2 Et

EtO" _ > A _ c 5 H § N _ > b A

[5] b) Work backward and identify the starting materials of the following product to be derived by aldol condensation and comment on the feasibility of the reaction. i) M'j 2 C = CHC0 2 Et ii) PhCH = CHCOCH = CHPh [5] a) The standard Gibbs energy of reaction for the decomposition of H 2 0 (g)

1 H2(g) + — 0 2{g )

is 118.08 kJ moP1 at 2300 K. Calculate the degree of dissociation of H 2 0 at 2300 K 1 atm? " b) Calculate the weight of 0 2 necessary to fill up a cylinder of 5 litre capacity at 0°C and atm pressure when the compressibility factor is 0.96

and [6] 100 [5]

a) 100 litres of a gas (Cv = 3 cal/mole) was initially at 0°C and 10 atm. Calculate the work when it undergoes adiabatic expansion reversibly till the final pressure is 1 atm. [5] b) Write down the products obtained when CH2OH I

c=o

I CH-OH CHO

_H!04_,

[2]

a) Write down the structure of P 4 Oi 0 and then show sequentially the action of water on P4O10, indicating the intermediates formed and the final product. b) Give one example to show that XeF2, behaves as a F~ ion donor. [4 + 1 = 5 ] FIITJ€€ Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Terminal), New Delhi - 16, Ph : 6515949 , 6865182, 6854102, Fax : 6513942


a) Identify A, B, C, D, E and F in the following reactions electrolysis Bauxite + Cryolite + CaF2 > A + 02 980° C

A + N2 —

H

2

°

>C + D HCI

F <

K0H

E + H20

[6]

b) We often hear a term 'Lead free petrol' which basically means that petrol is devoid of any lead. In earlier days some quantity of lead compound were used along with petrol for automobiles. Write down the formula of that lead compound that used to be present in the petrol along with its necessity. How can this lead compound be prepared from lead? [4] 9.

a) A compound (M) when mixed with coke and heated on a charcoal block in a reducing flame gives a residue (N). The residue (N) when treated with HCI liberates a gas (P). which when passed through chlorine water gives a ppt. (R). The ppt. is found to be soluble in CS2 and on burning in air a pungent smelling gas (S) evolves. This gas when passed through acidified K 2 Cr 2 0 7 , turns the solution from orange to green. The metal in the compound (M) gives dazzling yellow colour in flame. Identify (M) to (S) with balanced chemical reactions. [8] b) Identify A [4]

10. a

)

14

OCH 2 -CH = CH2 OMe^^l^/OMe i rSY L ^ J

is heated at 200°C to form A. A when treated with K0H/Me2S04 gives B which when treated with 0 3 /Zn gives C Ex |ain t h e ^ P formation of C & D indicating the fate of radioactive carbon. [3]

b) Potassium permanganate liberates the halogen gases from their respective halide salts (KCI, KBr, Kl) at pH = 1 (except fluoride salt) but at pH = 6 it only liberates iodine from iodide salt. Explain briefly (EMno-/Mn^ M n / M n °"

11.

=1

-51V'

E

°

1 2

Cr/-CI2

= - 1-36V). E° = 1,05V & E° = 0.54V. —Br2 / Br" -\ /r 2 22

[5]1

Write down a plausible mechanism for the following reaction. Br

+

\ j - H

»

N

^

x

R R/ Account for the fact that the compound in which R = H reacts 35 times as fast as the one in which R = CH3. [4] 12.

Write down a balanced chemical equations i) Potassium iodate solution is treated with excess sulphur dioxide. ii) Lead acetate solution is reacted with excess of bleaching powder.

[2] [2]

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13.

Compound (A) C13H10O on treatment with NH2OH and HCI gives (B) C 13 HnON which on treatment with PCI5 gives (C), a neutral compound. (C) on treatment with H 2 S0 4 gives an acid (D) C 7 H 6 0 2 & C6H7N which gives diazo coupling reaction with alkaline p-napthol. (A) on treatment with Zn/KOH gives (E) C 13 H 12 0 which liberates hydrogen on treatment with metallic sodium. A solution of A in Zn/HOAc when treated with Mg followed by acidification forms (F) C 26 H 22 0 2 , which also liberates H 2 with metallic sodium. On treatment with HI0 4 (F) regenerates A. Identify (A) to (F) with relevant reaction. [6]

14.

During the extraction of pure alumina by electrolysis mention the composition of the electrolyte and nature of electrodes used. [2]

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ALL INDIA TEST SERIES NT - JEE, 2002

FULL TEST - IV (Mains) MATHEMATICS Time : Two Hours

Maximum Marks : 100

Note: (i)

There are Twelve questions in this paper.

(ii)


(iii)


(iv)


(vi)



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AITS2002-FT-4-MA-2 1

1 (a).

io r 2 Evaluate I = Jx • e L - m (where [.] denotes the greatest integer function).

[4]

-10

r\

(b).

2

Evaluate I (k) = J[x]{x }dx, where k e f - {1}. (where [.] denotes the greatest integer -k

function and {.} represents fractional part).

[6]

A superman goes either to moon or to mars in the years listed as {1510, 1520, 1530, ,2000}. In a leap year amongst the years listed, he throws a fair die and if he gets a prime or odd number he goes to moon, otherwise to mars and in an ordinary year he goes to mars when he gets a prime or odd number, otherwise to moon. In an year he is found to be on mars then find the probability that it is a leap year. A century is leap if it is divisible by 400 and the rest years (other than centuries) are leap if they are divisible by 4. JX [8] 2

y.

4.

2

x y Let L : y = x + c be the variable chord of ellipse S : — + = 1 (where c is a real a b^ parameter). Now let L intersects with S at A and B. Find the locus of the point P on L such ih that PA ~ PB = - . Also sketch the locus of the point P. [10] 2

Let P and Q be two points on xy = c such that abscissa of P = ordinate of Q and normals at P and Q intersect at R (h, k). Find the locus of point R. [8] m

2

2

Let f (x) = x + 2x - t , where T is a real parameter. Now let a, (3 be the roots of f (x) where p

a < p. If F (t) = jf(x) dx, then find the minimum and the maximum value of F (t) and the a

corresponding t.

[8]

6.

Let P (z) = |2z - 1 - i| + |3z - 2 - 2i| + |4z - 3 - 3i|. Then find the minimum value of P (z) and also find the corresponding z. [8]

7.

The area of a given triangle ABC is A. Points Ci and Bj are the mid points of the sides AB and AC respectively. Two distinct points D and E are taken on the side BC such that DE = EC. Lines CiD and AE meets the median BBi at points G and F respectively. If A0 be \ 1 A0 1 the area of quadrilateral DEFG, then using vector method prove that —< — < — . [10] 6 A 5

8.

A line through P(-1, p) cuts the curve y = {x}, 0 < x < 4 (where {.} denotes the fractional part), at 4 distinct points such that distances of these four points from P are r1} r2, r3, r4 (r4>r3>r2>r1). 2r — r If the line satisfies the condition — - — - < 0 , then find the range of p. [10] 3r 2 -r 4

9.

k

A function f is defined as fk (x) = jsin 2 x|; V x e R and k e I. n

n

f

/x\

Show that ^ f k ( x ) > - ^ / , where n e I. k=i

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[8]

AITS2002-FT-4-MA-3

10(a). Find all the real values of x which satisfy the following inequalities simultaneously, tan x > 1; sin x < (b). 1¥ ^

Vs and^ cos x < - J— (2 2 v3 8

[4]

8

2

2

Find the general solution of sin x + cos y + 2 - 4 sin x cos y = 0.

[4]

|x2 6x+81

Find the intervals in which f (x) = e " (a) increases

(b) decreases

In a triangle ABC, where A and B are acute angles, A * B and Vsin2AcosB +sinBV2sinA = sinAV2sinB + Vsin2BcosA . Show that A ABC is a right angled triangle.

• ••

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[6]

FIITJ€€ ALL INDIA TEST SERIES IIT - JEE, 2002 FULL TEST - IV (MAINS)

PHYSICS Time : T w o Hours

M a x i m u m Marks : 100

NOTE : (i)


(ii)


(iii)

There are ten questions in this paper.

(iv)

Answers of ali parts of a question should be given at one place.

(v)


(vi)

Use of logarithmic table is PERMITTED.

(vii)

Useful Data : 2


g

=

10 m/s

Radius of earth

R*

=

6400 km

Stefan's constant

a

=

8

2

1

5.67 x 10~ Jm~ sec" K*


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Fax : 6513942

AITS-2002-FT-IV-PH-2

1.

A man is standing on a high cliff and he drops a sonic device (at t = 0) of mass 5.4 kg from there. It is found that the air resistance is equal to kv; where k = 1.7 and v is the instantaneous velocity of the body. The sonic device emits frequency of 700 Hz. The man hears a frequency of 660 Hz after some time. What was the speed of device and time when it has emitted the note corresponding to this frequency. Also find the minimum frequency heard by the man. (Given, natural logarithimic exponent e = 2.7 and velocity of sound = 330 m/s) [10] The P-T curve of a cyclic process undergone by one mole of a monatomic gas is as shown. The process AB is given by equation P = aT1/2. (a) Find work done and heat exchange in each process. (b) Also, find the molar specific heat of the gas for process AB. 4T 0

T

[8+2=10] AB. BC and CA are identical rods of length

2£

forming a V3 triangular frame (A, B, C are joints) kept on a smooth horizontal table. A particle of mass m and velocity 2v0 normally hits the rod BC as shown in the figure. The particle collides elastically and come to rest immediately after the collision. (Assume that the particle is immediately removed after the collision). [4+3+3=10] (a) Find ratio —; where M = mass of each rod, m = mass of incident particle. M (b) Find velocity of vertex A immediately after impact. (c) Find the time after which the frame regains its original orientation for first time after impact and the distance moved by 'O' (Centroid of the AABC) in that time. A block of mass m is attached to one end of a long elastic rope. The other end of the elastic rope is fixed to the roof of a building. Initially the block is in contact with the roof at the point where the rope is fixed and is allowed to fall freely from rest. The unstretched length of the rope is L and has a force constant k. (i) If the block just reaches the floor find the height of the room. (ii) what is the maximum speed of the block during this drop. (iii) The time taken during the drop before coming to rest for first time. (Take acceleration due to gravity as g) [3+3+5=11] An open cylinder is rigidly fixed to a floor. AB is a piston of mass 9 kg and of cross-sectional area A = 5n2 x 10"4 m 2 (equal to the cross-sectional area of the cylinder). Piston is free to move on smooth inner surface of the cylinder as shown in the figure. Spring of stiffness 7c2N/m is attached to piston and to a rigid wall as shown. It is initially compressed by 1m. The thread which initially holds piston AB at rest snaps at t = 0. < 2^5m i. A /rmwr

t

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(i) Find the time for all resonances till the spring is at its maximum elongation for first time, if the air column is made to vibrate with a tuning fork of frequency 320 Hz. Velocity of sound = 320 m/s. (ii) If P0 (N/m2) is the atmospheric pressure present inside and outside the cylinder and AP0 (N/m2) is the maximum amplitude of pressure variation in the air column inside the cylinder then find the maximum pressure on the piston and its position from open end of tube when this occurs. [6+2=8] A thin convex lens of focal length 30 cm and having aperture diameter d = 1 cm is used to focus sun rays on a metal surface. The absolute temperature of sun is 6000 K and assume that rays fall on the lens normally. The work function of the metal is 1.48 eV. Assume that sun radiates like a black-body and it radiates a single wavelength corresponding to its maximum spectral radiancy (a) Find the intensity on the metal surface. (b) What is the de-Broglie wavelength of the photoelectrons emitted. (Given Wein's constant = 3 x 10"3 mK). [7+3=10] 7.

There are two fixed identical rings A and B of radius a = 3m. Their centres O and O' lie along y axis as shown. Given H = 4 m. Coordinates of O' are (0, 0). Charge on ring A is +q and on B is +nq. A positive charge q0 of mass m is dropped from O and it just reaches O'. (a) Find the value of n. given m =

t q

' ° 2ne0g (b) Plot a rough sketch of potential energy of the falling particle as aûnction of its height above O' In the shown figure PQ is a semi-circular arc of radius r. QR and RP are straight wires and OR = OQ = OP. They form a mesh PQR which is not in a plane. A current I flows through circuit along the direction shown in the figure. A magnetic field B exists in the space such that B is equally inclined with positive direction of all the three axis. (a) Calculate moment of force (x) acting on the circuit. (b) Now suppose if there is no current in the circuit and the magnitude of magnetic field starts increasing at a constant dB rate = a. Calculate magnitude of emf induced in the dt mesh and show direction of flow of induced current in the mesh.

Ay

C 3' O 3

B [6+2=8]

» X

[6+6=12]

(a) A satellite is orbiting the earth in equatorial plane in a circular orbit having radius 2R and same sense of a rotation as that of the earth. Find the duration of time for which a man standing on the equator will be able to see the satellite continuously. Assume that the man can see the satellite when it is above the horizon. [5] (b) A wooden block (having cross-sectional area A), with a coin (having volume V and density d) placed on its top, floats in water as shown in figure. If the coin is lifted and then dropped into water, find (i) change in the submerged length (£) of the block. (ii) change in the water level (h) in container. Cross sectional area of the container is & density of water is p. '

[3+3=6]



10.

In a Young's double slit expe r iment the region between the slits and the screen is filled with a liquid whose concentration starts changing at t = 0 and because of that o its refractive index also changes with time 5 _t_ as jum = The final value of refractive 2~ 4 5 index is found to be The separation 4 between the slits is d = 2 mm and between the slit and screen is D = 1m. The thickness of glass plate shown in the figure is 36 urn and its refractive index is 1.5. (a) Find the position of central maxima as a function of time and the time when it is at O. (b) Find the speed of central maxima when it is at O. [6+4=10] *

*

*


ALL INDIA TEST

IIT-JEE, 2002 FULL TEST - V

CHEMISTRY Time : Two Hours

Maximum Marks: 100

Note i) ii) iii) iv) v) vi)

This paper has FIFTEEN questions. Attempt ALL questions. Answer to each new question should begin from a new page Answer all sub parts of a question at one place. Use of logarithmic table is PERMITTED. Use of calculator is NOT PERMITTED.

Useful Data: R =

Gas Constant

Avogadro's Number

Planck's constant

Na

h

= =

=

1

1

8.314 J fC moP 0.0821 Lit atm K"1 mol"1 1.987 « 2 Cal K mol 23 6.023x 10 - 1

6.625 xlO"

34

J s

27

1 Faraday 1 calorie 1 amu

= = =

6.625 x 10~ erg • s 96500 Coulomb 4.2 Joule 27 1.66 x 10~ kg

Atomic No:

H = 1, D = 1, Li = 3, Na = 11, K = 19, Rb = 37, Cs = 55, F = 19, Ca = 20, He = 2, 0 = 8, Au = 79.

Atomic Masses



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AITS2002-FT-V-CH-2

1.

i / Calculate the pH of a solution obtained by mixing 0.03 moles of sodium acetate to 500 ml of an aqueous solution containing 0.02 moles of CH3COOH? What will be the pH if 5 0.02 moles of HCI is added to the above solution. (Ka of acetic acid 1.8 x10" ). [5] ii) Write down the products obtained when CaC2, AI4C3 & Mg2C3 are separately hydrolysed. Give the reactions that take place. PI

2.

Write down the missing alphabets. 7crc

i) ZnS0 4 .7H 2 0 ii) Na(NH4)HP04 —

100 c

>A ° >B F + G + H20

28Q C

°

>C

770 c

°

> D+E+O [7]

3.

What happens when i) Tin is treated with conc. HN0 3 of sp. Gravity 1.25 and then products are ignited. ii) Sodium bromate solution is treated with XeF2 [2 ** 2 ~ 4]

4.

It is observed that statues which contain copper like the "statue of liberty" in New York gets covered with a green layer on its surface. Discuss the reaction that leads to the formation of the green layer. [3]

5.

E.M.F. of the cell Ag / AgCI(s) / 0.1 N KCI / Hg2CI2(s) / Hg is 0.0455 V at 25°C. Calculate the approximate + concentration of Ag ions in a saturated solution of AgCI in 0.1 N KCI. EpCI /Hg rCIi /H = 0.334 V & E° = 0.799V [6] +A /Hg Ag /Ag 2 2

6.

a) In a photoelectric effect an absorbed quantum of light results in the ejection of an electron from the absorber. The kinetic energy of the ejected photons is equal to the energy of the absorbed photons minus the energy of the largest wavelength photons that causes the effect. Calculate the kinetic energy of a photoelectron produced in cesium by 400 nm light. The critical wavelength for photoelectric effect in cesium is 600 nm. [5] 3 3 b) Calculate the volume of 10~ M NaOH needed for complete reaction with 30 cc 10~ M pyrophosphoric acid. [3] 7

-1

1

1

7.

The rate constant of a reaction is 1.5 x 10 sec Evaluate the Arrhenius parameters A & Ea.

at 50°C and 4.5 x lO sec" at 100°C. [6]

8.

a) A sample of pure p-sulphur is melted at 119.25°C but after a few minutes the melting point fell to 114.25°C. The molten sulphur was cooled and immediately treated with CS2, when 3.6% of the resultant solid was found to be insoluble. What is the molecular formula of this new variety of sulphur insoluble in CS2? (Latent heat of fusion of sulphur = 9 cal/gm) [6] b) At 1000°C the pressure of iodine gas is found to be 0.112 atm whereas the expected pressure is 0.074 atm. The increased pressure is due to dissociation l 2 21. Calculate Kp. Also find out pressure at which l 2 will be 90% dissociated at 1000°C. [6]

9.

Write down the unknown alphabets in the following ')

(CH2 H)2/H+

4-bromobutanal

°

>A

Mg/Et2

° ->B CH3CHO/H+ H2O

ii)

Q[_j _

PhLi 1 eq v.

10.

[T

D

C

CH3COCH3

>

CH3OH/H+

p

Hg 2 +/H +

C

>

q

PhCHO/OH"

>

p

NH4CI

Carry out the following conversion i) OHCH2CH2CH2Br > OHCH2 - CH2CH2C ^ CH

H8?J€€

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i

AITS2002-FT-V-CH -3

N)

CH

COOH

o

o

NO

NH

[ 3 x 2 = 6]

11

An optically active aromatic alkene C I 3 H 1 8 on ozonolysis gives acetophenone as one of the products. Write down the structure of the alkene and show the different geometrical isomers of this compound. [4]

12

Write down the products in the following CH3CHO + DCDO

OH"

(excess)

+ CH3MgX

» A

»B

CO »C H,0

N

HN0 2 1 eqv.

COOH COOH

•» D

18

CH? - OH 18

C H , - OH

[5 x 2 - 10]

13

Compound (A) CgH^Br gives iodoform test (A) on treatment with aqueous KOH followed by acidification gives (B) C 9 Hi 2 0 2 . (B) on reaction with HI04 gives benzaldehyde and (C). Identify A, B & (C). [5]

14

a) Write down the reactions involved in the extraction of lead by self reduction process. [2] b) Write down the reaction involved in the preparation of 'triple superphosphate' from fluorapatite. [2]

15.

You have been given o-nitroaniline. Suggest the methods with relevant chemical reactions by which you will be able to identify both the functional groups in the above compound. [4]

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y

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FIITJCC ALL INDIA TEST SERIES SIT - J E E , 2 0 0 2

FULL TEST-V (Mains) MATHEMATICS Time : T w o Hours

M a x i m u m SVlarks : 100

Note: (i)

There are Fourteen questions in this paper.

(ii)


(iii)


(iv)


(vi)


Name of the candidate Enrollment

£SI¥J€€

Number

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AITS2002-FT-5-MA-2

x,

-2 < x < -1

2

I.

Let f (x) = <

x + 2x, 2

-1 < x < 0

2x-x ,

0
2 - x,

1< x < 2

(a).

Draw the graph of f (x) and discuss the continuity and differentiability of f (x) in [-2, 2]. Also find the maximum value of f (x) in the given interval.

(b).

Find out f (f (x)) and discuss the continuity of f (f (x» in [-2, 2].

[5+5]

st

2.

Let z1 = 1 + i and z be a point in I quadrant of argand plane satisfying |z| < 1. Find the 2 2 maximum and minimum value of f (z) = (Im (z )). (Im (z - z^ ) and the corresponding z. [7]

3.

Let P be any point on x + a = 0. Find the locus of the reflection of the point P in its chord of 2 contact with respect to y = 4ax. [5] 2

2

4

Let A: (-3, 0) and B: (0, -3) be two fixed points and C: x + y - 4x = 0 be a fixed circle. If a r 9 2lS variable line L passing through intersects with C at two distinct points P and Q, T t 4' 4 v~ - J then prove that A, B, P and Q are concyclic points. [6]

5.

Let A (z0 be a fixed point on a circle whose centre is O (z0). From an external point P (z) two tangents PA and PB are drawn to the given circle, such that the area of triangle BAO is the maximum possible area in the given circle. Find the complex numbers associated with P and B. [6]

6.

Three bags A, B and C are given. Bag A contains m blue balls, bag B contains n green balls and bag C contains p red balls, and min {m, n, p} > 7. A fair dice is rolled. If outcome is 1 or 2 then one ball from A is drawn. If outcome is 3 or 4, one ball from B is drawn and otherwise a ball from C is drawn. This drawn ball is put into a bag D which is empty, initially. This process is repeated six more times and after that one ball from D is withdrawn and found to be blue. Then find the probability that D contains balls of each colour and the number of green and red balls are equal. [10] 2

7.

Draw the approximate graph of |yj + (1 - |x|) = 5 and find the area enclosed by this curve. [5]

8.

Let P be any point in the first quadrant on a curve passing through (3, 3) and the slope of the curve at any point in the first quadrant is negative. The foot of the perpendicular from P to xaxis is the point A and to y-axis is the point B. Tangent at P intersects with x-axis at point R 3 and with y-axis at point CL Now if 2(Area (APAR)) + Area (ABPQ) = —(Area of rectangle 2 OAPB), for all positions of P, where O is the origin, then find all possible curves on which P can lie. Also find the minimum distance of P from the origin on each curve. [8] (jy 7

y H _ XV(X2 + V2 J J z 2

9

Solve the differential equation

10,

If the side length of a regular pentagon is 'a' and its diagonal is of length 'b' then prove that 3

2

x(1 + xy(x +y ))

u2

b

Q 2 + —25" = 3 . b a

FIITJCC

dx

[6]

[6]

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AITS2002-FT-5-MA-3

11. 12.

Find out the maximum and minimum value of f (x) = (3 - V 4 - X corresponding x. Show that 3 x x 1-x

2

1-x

x 6

5

1

1-x

x 10

7

'1-x

x CO

14

=

1+x

x 2

1

3

2

x 6

1-HX

1

2

) + (1 +

3

X ) and the 2

[5]

5

x 10

1+x

7

1

1+x

14

oo;

|x| < 1. [10]

2

2

13.

A pair of variable straight lines 5x + 3y + axy = 0 (where a is a real parameter) cut the 2 parabola y = 4x at two points P and Q. Find the locus of the point of intersection of tangents at P and Q. [8]

14.

A vertical tower PQ is standing on level ground and a flagstaff is mounted on it with an inclination of 30°from the vertical towards north. A man starts walking towards north from the base of the tower .After walking a distance 'a' he notices that the flagstaff is subtending an angle '0' at that point and after walking a distance 'b' from the base of the tower in the same direction, the angle subtended by flagstaff is again '9'.Find the length of the flagstaff. [10]

• ••

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FIITJCC ALL INDIA TEST SERIES

IIT-JEE, 2002 FULL TEST - V (MAINS)

PHYSICS Time: Two Hours

Maximum Marks: 100

NOTE : (i)


(ii)


(iii)

There are eleven questions in this paper.

(iv)

Answers of all parts of a question should be given at one place.

(v)


(vi)

Use of logarithmic table is PERMITTED.

(vii)

Useful Data : Acceleration due to gravity

g

=

2

10 m/s

Name of the candidate : Enrollment Number

FIITJC6

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AITS-2002-FT-V-PH-2

1.

A ring of mass 2m is placed horizontally on a frictionless table. A bullet of mass m travelling horizontally on the table with velocity v0 gets embedded into the ring. Calculate the velocity vector of point P just after the collision.

m

[5]

2

10pF, 10nC +

Six capacitors A, B, C, D, E & F are charged initially and connected in a circuit as shown in the figure. Their capacitances, initial charges and polarities are also shown in the figure. If all the keys are switched on simultaneously, find the final charge with polarities on all the capacitors.

5jiF, 5|iC

[5]

3.

(Screen)

(Double slit)

Two source and s2 emitting wavelengths and X2 are placed to the left of the double slit in Young's Double Slit Experiment. Calculate the distance between the central maximas due to two wavelengths on the screen.

[6] 4

5

A thin convex lens whose focal length in air is known to be 50 cm is divided into two parts 0.5cm above the principal axis. The two parts are now placed on the x-axis as shown. Calculate the co-ordinates of the image thus formed. Object is placed at (-100, 0) cm.

075 cm (-100, 0) 30 cm

f = 50 cm

[8] 0.7V

Shown in the figure is a circuit consisting of three P-N junction diodes (having knee voltages 0.7V each) and some resistances connected to a battery. Calculate the current through the battery. All diodes are considered to be ideal.

\ *

v

•o-

0.6£1 y w v w v .

0.4Q "

W

W

0.3H

7

- W —

0.7V

yWsAAA

0.7V 1V

FIITJC6 Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 686 5182,6965626,

m

[8]

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AITS-2002-FT-V-PH-3

6.

A rope has its mass per unit length X changing from bottom to top in the manner X = —

\

0

x=L

x

1+ — V LJ

where X0 is a constant; x = 0 at the bottom, x = L at the top, L being the length of the rope. The highest point is fixed to the ceiling. A transverse pulse is given at the bottom. Find out the time taken by the pulse to cross the length of the rope. 7

8

9

10

A radioactive battery, consists of N0 atoms of a radio-element emitting a-rays with a disintegration constant X, of which a fraction f is captured at the anode A and converted into a current. A charged capacitor C initially charged as shown in the figure and a resistance R are connected across the battery. Find the charge across the capacitor as a function of time t, assuming that the battery contained N0 atoms at t = 0. At what time does the charge on the capacitor become zero?

Shown in the figure is a chimney of mass m = V3 kg placed on a frictionless floor constrained to move along the line OP. The chimney is pulled with a constant force by means of an ideal string connected to it starting from rest at t = 0.

10 mIs

[8}

-qo

[12]

A sphere of mass M and radius R is placed on a frictionless surface. A small block of mass m is placed on it at its topmost point. The contact between the sphere and the block is frictionless. At t = 0, the sphere is pulled by a force acting at the centre with constant acceleration a = g in horizontal direction. Calculate the velocity with which the block hits the ground w.r.t. the ground frame of reference. A thin walled cylindrical tank of radius R is filled with water of density D at 0° C. The upper and lower flat surfaces are covered with non conducting material and tank is placed in surrounding whose temperature - 60 in °C. As a result water starts freezing on the curved wall of the cylinder. Calculate the time in which half of the volume of water in the tank freeze. Ignore variation in density due to freezing. Latent heat of fusion is L and thermal conductivity of ice is k.

x =0

Y

M\ r\ I (O, R) J

?

\

x

[12] -e°c

h

[12] »

F

0

Simultaneously a bullet is fired from a point O at an angle of 60° with the horizontal with an initial velocity of 10m/s. Calculate the distance OP so that the bullet hits the bottom of the 27 chimney without hitting its surface from inside. Height of chimney is m. When bullet 20 reaches the top of the chimney, the string connected to the chimney melts due to heat. [12] 11.

FIITJC6

A thin non conducting ring of mass m having total charge q can rotate freely about its own axis. Initially, when the ring is stationary, a magnetic field B directed perpendicular to the plane is switched on. Find the angular velocity cd acquired by the ring. Find also the induced magnetic moment. [12] Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 686 5182,6965626, 685 4102, 6515949 Fax: 6513942

ALL INDIA TEST SERIES

IIT-JEE, 2002

OPEN TEST (Mains) CHEMISTRY Time: Two Hours

Maximum Marks: 100

Note: i) ii) iii) iv) v) vi)

This paper has EIGHTEEN questions. Attempt ALL questions. Answer to each new question should begin from a new page Answer all sub parts of a question at one place. Use of logarithmic table is PERMITTED. Use of calculator is NOT PERMITTED.


R

Avogadro's Number

Na =

Planck's constant

h

=

=

1

1

8.314 J K" mor 1 1 0.0821 Lit atm K~ mor 1 1 1.987 * 2 Cal K" mol" 23 6.023 x 10 -34

6.625 x 10

J-s

-27

=

6.625 x 10""' erg • s

1 Faraday

=

96500 Coulomb

1 calorie

= =

4.2 Joule 27 1.66 x 10~ kg

1 arnu

Atomic No:


Atomic Masses:



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AITS2Q02-OT-CH-2

1.

The compound Co(en)2(N02)2CI has been prepared in a number of isomeric forms. One form undergoes no reaction with either AgN0 3 or ethylene diamine. A second form reacts with AgN0 3 but not with ethylenediamine and the third form reacts with both AgN0 3 and ethylenediamine. Identify each of the these forms by their IUPAC names and discuss the above reaction in light of Werner's theory. [6]

2.

In an experiment conducted at 300K, 51.5 ml of PCI5 along with its dissociation products diffused through an orifice in the same time as taken by 100 ml of 0 2 under similar conditions. Calculate the percentage dissociation of PCI5. [6]

3.

a) The following compound is a strong base. Explain 2(C2H5)^ N(C2H5)2 CHsO

OCH3

[3] b) Compound (X) + H2/Pt 1. o3

> 1-isopropyl-4-methyl cyclohexane

+

2. Zn/H30

HCHO + Find out the structure of X

[2]

4.

a) There are two substances Y & Z, one is hypochlorous acid and the other is chlorine. Site a chemical reaction by virtue of which you can identify the above two substances. [3] b) A solution contains a mixture of sodium nitrate and sodium carbonate. Is it possible for you to confirm the presence of nitrate (N0 3 ") radical by brown ring test? If yes why and if no, then site a reaction by which you can confirm the presence of nitrate radical in this mixture. [4]

5

a) Which of the following compounds has a higher dipole moment and why? 1, 2-dibromoethane and 1, 2-ethanediol b) Which of the following has got the greatest affinity for water. Explain with reasons i) P4O10} Cl 2 0 7

[3] [3]

6

There are two flasks containing two different solutions. Flask 'A' contains 100 ml of pure water, pH = 7, and the flask 'B' contains 100 ml of a solution containing 10 m mol of acid HA, pka = 7.00 and 10 m mol of conjugate base A". To each flask 1 m mol of NaOH is added. Calculate the pH of the resulting solutions and the change in pH? [5]

7.

Write down the products in the following i) 02N

- lO

c=o

CF3CO3H

ii) conc H 2 S 0 4

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AITS2002-QT-C H -3

8.

What happens when one equivalent of MeCH = CHCOMe in Et 2 0 is separately treated with one equivalent of the following reagents i) EtMgBr in ether ii) EtLi in ether [3]

9.

Iron crystallises in two body centered cubic lattices, the a form below 750°C and the 8 form above 1400°C, and in a face centered cubic y form between these two temperatures. Of these 3 crystalline phases, only the y form can dissolve appreciable amounts of carbon. i) What is the maximum ratio of a foreign particle radius occupying a hole in the face of a body centered cubic unit cell to the host ion radius? ii) What is the ratio of a foreign particle radius occupying a hole whose coordinates are (0, a/2, a/4) in a bcc unit cell to the host ion radius? [7]

10.

Explain the following a) LiF is the least soluble of the alkali metal halides while Agl is the least soluble of the silver halides. CH3 b) The SN reaction of EtS CH2CHCIwith ethanol proceeds at a rate many fold faster than the similar reaction of EtOCH2CH2CI. Also write down the products obtained. [3 x 2 ~ 6]

11.

The standard cell potential of Pt, H2(g) | HBr (aq) | AgBr (s) | Ag(s) was measured over a range of temperatures and the data were fitted to the following E 4 polynomial. ^ = 0.07131 - 4.99 x 10"

-6

-298

- 3 . 4 5 x 10

vK change in Gibbs free energy, enthalpy and entropy at 298K. 12.

Find out the missing alphabets in the following sequence CH3 - C H - C 0 2 E t CH (CQ Et) CH CH-CN EtOH 2

2

OEt~

2

>

A

2=

>

b

>

,T vK

-298

,

. Evaluate the [6]

Q

OEt

Lr OEt' •

E< ,

1. 0 H " . H 2 0 , A 2. H 3 0

3A

-

13.

+

„

D

(C15H2207)

[8]

ya) 3.476 gm of KMn0 4 is dissolved in water and the solution is made upto 1 litre. An +2 unknown salt containing 5.88 gm Fe ion was dissolved in water and solution was made upto 100 ml. It was found that 20 ml of salt solution decolourised 27.25 ml of the above permanganate solution. What was the % of ferrous ion in the salt? [4] b) Compounds having the formulas CsBr3 and CsBrCI2 are stable below 100°C and crystallise in a cubic lattice. Is the existence of these compounds a contradiction of the statement that alkali metals have only one positive oxidation state? Explain [3]

14.

a) Write down the mechanism of the acid catalysed esterification of 2, 4, 6 trimethylbenzoic 18

acid with CH 3 OH, explaining the ultimate fate of labelled oxygen.

[3]

b) Between sodium chloride and sodium bromide in dimethylsulfoxide solution which anion (Cl~ or Br") behaves as a better nucleophile and why? [2] 15.

A sample of 0.1 mol of H2 and 0.05 mol of 0 2 at 25°C in a sealed bomb is ignited by an electric spark. Calculate the final temperature of the gaseous water produced. Ignore the energy of the spark and any heat loss to the surroundings. AHf for H 2 0(g) = - 57.79 kcal/mol Cp for H 2 0(g) = 5.92 cal/mol.K [6] H8?J€€

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AITS20Q2-OT-CH-4

16.

a) Compare the acidity of p-chlorophenol and p-flurophenol. [3] b) You have been given a sample of carbon monoxide collected from a busy area in a city like Delhi. Discuss a method by which you can trace the amount of CO in the sample. [3]

17.

Calculate the number of photons emitted by a 100 W yellow lamp in 1 sec. The wavelength of yellow light is 560 nm, assuming 50% efficiency. [3]

18.

a) Write down the structure of the monomer of natural rubber. b) Write down sequentially the action of heat on boric acid.

[2] [2]

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k x '

\

"^mr \

FIITJ€€ RANKERS STUDY MATERIAL

IIT-JEE, 2002 PHASE -1

CHEMISTRY SOLUTIONS a) Using normality equation 25 x N = 20 x 0.01 20x0.01,, N K1 N= N = 25 125 Amount of HCI in 2500 mL of N/125 = Amount of HCI in 700 mL of N/10 =

36 5 x 2 5

125 36 5 x 0 7

= 0.73ga

= 2.555 g a 10 Amount of HCI consumed = (2.555 - 0.73) = 1.825 g Let amount of CaC0 3 in the mixture is 'x' g Amount of MgC0 3 = (2.36 -x)g CaC0 3 + 2HCI > CaCI2 + H 2 0+ C0 2 100g 73g 73 xg xg 100 a MgC0 3 + 2HCI > MgCI2 + H 2 0 + C0 2 84 g 73g (2.36 -x)g

[2]

||(2.36-x)g

But, — + —(2.36 - x) = 1.825 => x = 1.63 yg 100 84 - 6 3 x 1 ° ° = 69.06 2.36 % of MgC0 3 = 100 - 69.06 = 30.94 % of CaC0 3 =

[31

1 1

1

b) Eq. weight of KMn0 4 = 1/5 x molecular weight of KMn0 4 M/50 KMn0 4 = N/10 KMn0 4 Meq. of KMn0 4 = 20 x 1/10 = 2 Meq. of N 2 H 6 S0 4 in 10 ml solution = m.eq. of FeCI3 reacting with N 2 H 6 S0 4 = m.eq. of Fe2+ formed = m.eq. of KMn0 4 used Meq. of N 2 H 6 S0 4 in 10 ml solution = 2 i. of N H S0 = mol. wgt. r Eq. wgt. — = 130 = 32.5 2 6 4 4 4

[2]

[3]

,„. [2]

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RSM12-P1-T(M)-CH(S)-2

[N22H6S04 —> Nj + 4e~] weight of N 2 H 6 S0 4 in 10 mL = 32.5 x 2 x 10~3 g = 0.065 g weight of N 2 H 6 S0 4 in one litre = 6.5 g 2.

[2]

Let the amount of NaCI in the mixture be 'x' gms. Hence the amount of MCI in the mixture = (1-x) g The equation of the reactions are as follows NaCI + AgN0 3 > AgCI+NaN0 3 58.5g

MCI

(W+35.5)g

143.5g

+ AgN0 3

> AgCI + MN0 3 (W

atomic wgt. Of M)

143.5g

Since 58.5 g NaCI produces 143.5 g AgCI 143: 5x x g NaCI produces = — g of AgCI 58.5 143 5(1 - x) Similarly, (1-x) g MCI produces =

^

y

g of AgCI

[3]

Since of the precipitate is 2.567g, hence 1 4 3 jthe x + weight 143-5(1-x) = 58.5 (W + 35.5) According to the (ii) part of the question one component is volatile. Since NaCI is not a volatile compound hence MCI may be considered as volatile. This compound forms AgCI according to the equation where it is equal 11.341 g. Thus, 143.5(1-x) _ ^ 241 W + 35.5 143.5x + 1.341 =2.567 =>x = 0.5 ag [4] 58.5 (143.5)(1 - 0 . 5 ) Also, = 1.341 W + 35.5 (143.5X0.5) 1.341 Hence molecular weight of MCI = 18 + 35.5 = 53.5

[3]

a) 25cm3 of remaining oxalic acid solution = 32cm3 of 0.1N KMn0 4 ^ N ^ V ^ 0.1 x 32 =>N, — N 25 32 Unreacted oxalic acid = 250 cm3 of — N 2.5 M.eq. of unreacted oxalic acid = 32 M.eq. of total oxalic acid = 1 x 50 = 50 M.eq. of used oxalic acid = 50 - 32 = 18 Mn0 2 + H 2 S0 4 + H 2 C 2 0 4 — > MnS0 4 + 2H 2 0 + 2C0 2 Meq. of H 2 C 2 0 4 = m.eq. of Mn0 2 m.eq. of Mn0 2 = 18 Wgt. Of Mn0 2 = 18 x 43.469 x 10~3 = 0.7824 g % of MnOz = ° - 7 8 2 4 x 100 = 48.9 1.6 M.eq. of Mn0 2 = m.eq. of 0 2 = 18 Wgt. Of 0 2 = m.eq. of 0 2 x its eq. wgt. x 10~3 = 18 x 8 x 10"3 = 0.144g 0 144 % of available oxygen = - ^ x 100 = 9

[2]

[2]

[2]

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b) Let the mixture contain 'x' g of K 2 Cr 2 0 7 Amount of KMn0 4 = (0.561 - x) g The reactions involved are i)

2KMn0 4 + 8H 2 S0 4 + 10KI > 6K 2 S0 4 + 2MnS0 4 + 8H 2 0 + 5I2 5I2 + 10Na 2 S0 3 >10Nal + 5Na 2 S 4 0 6 2KMn0 4 = 10(Na 2 S 2 0 3 .5H 2 0) = 101 s 10H

ii)

K 2 Cr 2 0 7 + 6KI + 7H 2 S0 4 » 4H 2 S0 4 + Cr 2 (S0 4 ) 3 + 7H z O + 3I2 3I2 + 6Na 2 S 2 0 3 > 6Nal + 3Na 2 S 4 0 6 K 2 Cr 2 0 7 = 6(Na 2 S 2 0 3 .5H 2 0) = 6i = 6H From (i) we have that 316 g of KMn0 4 will require hypo = 10 x 248 g , _ ... . . 10x248x(0.561-x) /ACC, (0.561 - x) g of KMn0 4 will require hypo = — -g 316 Similarly from (ii), we have 249 g of K 2 Cr 2 0 7 will require hypo = 6 x 248 g ... . 6 x 248 x x A x g of K 2 Cr 2 0 7 will require hypo = — ^ ^ — - g

... [4]

We also know, that 100 mL of 0.15 N hypo = .-. accordingly 10 x 248 x (0.561-x)

100 X

°q1qX

6 x 248 x x

248

9 hypo = 3.720 g hypo

O. I Cm

316 294 7.848 (0.561 - x) + 5.061 x = 3.72 7.848 x 0.561 - 7.848x + 5.061x = 3.72 2.787x = 0.68 x = 0.244 g Amount of K 2 Cr 2 0 7 = 0.244 43.5% Amount of KMn0 4 = 0.317 g => 56.5% 4.

[2] [2]

a) AE = B [ l / n 2 - l / n 2 ] B = 2.179x 10~18 J Electron falls from n = 3 to n = 2 (n2 = 3, n-i = 2) Then AE3_>2 = B [ l / 2 2 - l / 3 2 ] If electron falls from n = 2 to n = 1 (n2 = 2, n-i = 1) Then A E 2 ^ = B [ l / l 2 - l / 2 2 ] AE total = AE3_>2 + AE2->1 = B [l/2 2 - l / 3 2 ] + B [ l / 1 2 —1/22] = B [l/l 2 — 1/32 ] (i) Electron falls from n = 3 to n = 1 (n2 = 3, ^ = 1) Then AE ( 3 ^) = B[l/12 - 1 / 3 2 ] (ii) Thus from equations (i) and (ii) we have AE (3 ^ 2) + AE(2^D = AE(3_»i) b)

[2]

[2] [2]

v = AE/h v cc AE Since AE values are additive hence frequencies are also additive. V(3-»i) = V(3^.2) + V(2^-i) Also AE = hc/A. X = 1/AE Higher the energy of the given transition, smaller the wavelength. Thus wavelengths are not additive.



5.

a) By Heisenberg's uncertainty principle Ax,Ap > h/4% given that, Ax = Ap, hence Ax = Ap = yjh/4n = 0.726 x 10"17 Also, Ax.Av > h/47im Thus Av = h/47tm.Ax

[2]

= h/47im y/h/An = y]h/4rc x 1/m _ 0.72 x10' 17 " 9.1x 10 31 = 7.98 x 1012 ms"1 [2] We find that uncertainty in the measurement of position (or) momentum is negligible, but not in velocity. b) Using de Broglie equation X = h/mv v = h/Am _ 6.626 x10' 34 0.09x10" 1 ° x9.1x10 = 8.09 x 107 ms~1 KE = - mv 2

31

[2]

2

= - x 9.1 x 10"31 x (8.09 x 107)2 = 2.98 x 10' 15 J 2 2.98 x 1 0 1 5 w = eV 19 1.6 x 10"

= 1.86 x 104 eV 6.

[2]

a) Solution: PV = nRT nRT Volume occupied by 1 mol of gaseous water at 100°C = — — =

1 x 0 0 8 2 1 x 3 7 3

1

-15

= 30.64L

volume of 1 mol liquid water =

[2] mas$

density

= 18/0.958

= 18.79 mL = 18.790 x 10"3 L 18 79x10" 3 .-. percentage of volume occupied by liquid water = — ^ percentage of free volume = 100 - 0.0613 = 99.9386% 0.789x0.100 . b) n„02 = = 0.0032 mol 0.0821x300 1.053x0.250 , nMn = — = 0.0107 mol N0 0.0821x300 By the reaction 2NO + 0 2 > 2N0 2 > N204 We find that given mol of 0 2 will completely react with NO. Hence moles of NO reaction with 0 2 = 2 x 0.0032 = 0.0064 mol Moles of NO left unreacted = 0.0107 - 0.0064 = 0.0043 mol.

^ — x 1 0 0 = 0.0613% [2]

roi

[2]

[2]



Since final temperature is 220K, hence N 2 0 4 will exist in solid state. Only, gaseous species left will be NO in 350 ml flask. p = n ^ R T = 0.0043x0.0821x220 V 0.350 7.

PH,oV a) Water in vapour state = — - — RT (27/760)x2.5>10' = 0.0821x300 = 36.06 x 18 gm 36.06x18 . = mL = 0.656L 0.99 Water left in liquid state = 5 - 0.656 = 4.344L

[2]

b) Applying Garham's law, r r

fcTîJsO

(so3)

V M

V M

80

t=0 t=t nT =

= 59.5 1.16x1.16 Cl2(g) 2CI(g) 1 mol 0 - X mol 2X mol

[2]

(1-X) mol 2X mol (1+X) mol 2x(35.5) + (1-»)71 1+X

71 = 59.5 1+X or, 1 + X = 1.193 X = .193 i.e., 19.3% of molecular chlorine dissociated

a) t=0

• [2]

N 2 0 4 (g) ^ = ^ 2 N 0 2 ( g ) 1 mol +2 mol - a mol 2 a mol

t =t ( 1 - a ) mol 2 a mol nT = (1+a)mol Let a be degree of dissociation. From equation, a = .483 PT = .0787 atm Kp =

= Pn 2 0 4

[2] P

T ^N 2 0 4 .P t

=

4a2

x

p

Substituting the values of a and P T and on solving K P = .096 KP = K c (RT) An ,K

c

[2]

096 = ^ ( v A n = 1) = = 4 x 10-3 RT .082x294.5

Again, KP =

An2

1-a

rxPT

4 y / 1\2

= - ^ x P 1-(-1)

PT = 2.376 atm « 2.4 atm

T

= .096 [2]



b)

PCI3(g) + Cl2(g) At equilibrium 2mol 2mol 2mol Total pressure = 30.3975 Kpa = 3 atm = P(say) P P PCI 5 (g)

K

_ Ppci3

x p

c,2 _ 3

Ppas

~

X

3" P _ 3

Kp= | = 1

...(1)

[2]

When chlorine is added to the system. The system will behave to nullify the effect and hence formation of PCI5 will be preferred. Since P and T are constant are V2

= H l = m = 6 V1 = V V 2 = 2V n2

.. n2 = 12 moles Say a moles of Cl2 were added

PCI5

PCI3 + Cl2

Initial 2 2 2 Final 2 + x 2-x 2+a-x nT-12 = 6 + a - x .-. a - x = 6 2+ax-x 2-x X Kp = ~ l 2 ~ ! r , x P = (2 + a - x X 2 - x ) P 2jo< 4(2 + x) 12 Solving equation (1) and (2) we get 20 a = — moles 3 20 Hence — moles of Cl2 was added 3 n 9.

[2]

= 1

(2)

[2]

[2]

1 t * -j sample 1 activity .. -, = 1„ x 10 ar\-i Ci x 3.7x 10 10 dis/sec = 3.7 - _ x 10 ._ 3 d/sec. a)\ Infected 1Ci In sample withdrawn : 2 Q c j i s x 1 m i n _ q 33 dis/seo [2] min 60 sec. The ratio of total activity to activity of sample withdrawn is equal to the ratio of volumes, where V is original volume of blood. 3.7x10 3 dis/sec _ (V + 0.20),) Thus 0.33dis/sec 0.10ml 3 V = 1.1 x 10 ml = 1.1L [2] b)

[N 2 0 5 ] = — N = 0.5M 5 K = 1 . 6 8 x 10~ 2

t = 1 min = 60 sec [A], = ? K = ^ | o g [ N 2 0 5 ] t

0

[N 2 0

0.5 10.2=Z303, 1.62 x 1 0 = — log—60 n on solving, n = 0.185M FIITJee ICES House, Sarvapriya Vihar (Near Hauz Khas Term.), New Delhi - 16, Ph : 686 5182, 685 4102. Fax : 6513942

RSM12-P1-T(M)-CH(S)-7 n

N2o5

= MV = 0.1825M x 5L = .9215 mol.

nN2o5 decomposed = (2.5 - .9215) mol = 1.5875 mol n

x n

o2 =

N2o5 decomposed

1

= - X1.5875 mol = 0.7938 mol 2 c) The optical activity remained = 35% 65% optical activity has been lost It means that 35% of dextro isomer has been converted to laevo isomer. Now, applying rate law for first order reaction t ^ _ 2.303, 100 10"° sec"' = loga t 65 on solving, t = 4.31 x 107 seconds = 1.37 years 1

[3]

[21 J

[2]

x

[4]


Rankers Study Material

FIITJCC

IIT - JEE 2002 PHASE-II (CHEMISTRY) SOLUTIONS 1.

The volume of the orthorombic unit cell is

= 10.46 x 12.87 x 24.49 x 10~24 cm3 = 3296.85 x 10"24 cm3.

o«,• contains • • 4128 , o atoms . . . . Since this of* sulphur per atom = 3296.85 — x10"

This volume per mole =

3 2 9 6 8 5 x 1 0 24

128

x 6 02

24

x 1023 = 15.51 cc

32 32 .. 15.51 = — . . - . p = — — = 2.063 gm/cc p 15.51 2.

[3] [2]

[4]

(0.9777 - 1 ) AH2 = 1.288 or AH2 = -57.67 kcal/mol AH3 = 0.9777 X (-57.76) = -56.47 kcal/mol CuS0 4

+8Q0 moles

h2o

CuS0 4 • H 2 0 CuS0 4 -5H 2 0

> CuS0 4 (aq)

+800 moles

h2o +800moles

h2o

AH, = -15.90 kcal

[2] [2]

> CuS0 4 (aq) AH2 = -9.33 kcal

[2]

> CuS0 4 (aq)AH3 = +2.80 kcal

[2]

4Hg0 CuS0 4 > CuS0 4 ,5H 2 0 AH = (-9.33 - 2.80) kcal = -12.13 kcal

4.

[2]

C2H4(g) + H 2 0(g) = C2H5OH(g); AH = -11.21 kcal If the enthalpies of formation of C2H4(g), and C2H5OH(g) are AH!, AH2 and AH3 respectively. AHs-AHS-AHt = 11.21 1.288 - A H , = - 1 1 . 2 1 [4] AH, = 12.498 kcal/mol = 12.5 kcal/mol -——— = 0.9777 ; AH3 - AH2 = 1.288 AH2

3.

[3]

[4]

CH4(g) + 20 2 (g) = C0 2 (g) + 2H 2 0(I) AH, = - 212.79 kcal / mol CH3CI(g) = | o 2 ( g ) = C0 2 (g) + H 2 0(l) + HCI(g) AH2 = -164 kcal/mol H2(g) + ~ 0 2 ( g ) = HzO(l) AH3 = -63.317 kcal/mol ^H 2 (g) + 1 Cl2(g) = AH4 = -22.06 kcal/mol

[2]

Adding to the first equation and the reverse of the second equation, twice the fourth equation and the reverse of the ttiird equation, CH4(g) + 20 2 (g) > C0 2 (g) + 2HzO(l) [2] FIITJ€« ICES House, Sarvapriya Vihar (Near Hauz Khas Term.), New Delhi - 16, Ph : 686 5182, 685 4102, Fax : 6513942


HCi(g) + H 2 0(l) + C0 2 (g) H2(g) + CI2(g)

We get .-.AH

> CH3CI(g) + 1 0 2 ( g )

[2]

> 2HCI(g)

H 2 0(l)

=

H2(g) + ± 0 2 ( g )

CH4(g) + Cl2(g)

=

CH3CI(g) + HCI(g)

= (-212.79) + (164) + 2 x (-22.06) + (68.317) kcal = - 24.59 kcal

5.

The equation fnp =

dlnp = AHvap ^ = , gives on integration dT " RT2

[2]

+constant

[2]

— AH ^ RT

AH

-

i.e., log p =

[4]

vao

+ constant 2.303RT Comparing this with the given equation, AH 798 5 = » • •'• AHvap = (798.5 x 2.303 x 1.987) cal/mol Z303R AH = 3654 cal/mol = 3.654 kcal/mol the normal boiling point corresponds to p = 760 mm

[2]

log 760 = 6 . 8 5 7 798.5 T

= 6.857-2.8808 = 3.9762

... J = J ^ L k

3.9762

= 200.8K

w

6.

[2] [2]

i W

=

Po

=

N+n

=

W2

w

M2

M,

y

[2]

.

M, 1 2 vM2y

w. R= — = W1+Ml w2

— r +m

[2]

M2

— = 1 + —. When R = 0.22, m = = 1.333 R r 46 1 . 1.333 1.333 — = 45.45 = 1 + — — . = 44.45 R r r ... r = ^ ^ = 0.03 = - ^ L 44.45 1000 7.

W1

[2] [2]

= 30g

[2]

Under the conditions stated, both solutions have the same value for — i.e. the same value Po

of molality Solution 1 has molality =

2x1000 Mx 98

[4]



Solution 2 has molaHty =

60x95 2x1000 _ 5x1000 2x60x95 98M ~ 60x95 " 5x98

[41

1 1

r2i

= M = 23 3

Weight of copper deposited = (0.175 x 0.728)g = 0.127 g

[1]

Weight of nickel deposited = (0.175 x 0.013)g = 0.007525g Weight of zinc deposited = (0.175 x 0.029)g = 0.041 Og In the conversion of K3[Cu(CN)4] to Cu, the equivalent weight of copper = 63.6.

[1] [1]

number of faradays involved =

0-1274

63.6

58 7 Equivalent weight of nickel = — = .. Number of faradas =

[2]

29.35

[1]

0 00753

= 2.566 x 10^ 29.35 654 Equivalent weight of zinc = - 32.7

[2]

= 1-226 x 10"3

[1]

total number of coulombs = 96500 [2.003 + 0.2566 + 1.226] x 10"3 = 336.36

[1]

Number of faradays =

9.

= 2.003 x 10~3

i)

The electrode potential for the half cell, Pt E U =0.77 +

Fe 3 + Fe2+

[1]

IS

0.059, [Fe 3+ ] log-. 2+ 1 ~ [Fe ]

Similarly, the electrode potential for the half cell Pt

[2] Cu+ Cu+

is

„ n . 0.59 [Cu ++ ] E" = 0.17 + log1 [Cu + ]

[2]

, E ' - E " = ( 0 . 7 7 - 0 . 1 7 ) + ^ l o g ^ ++ lx.[Cu+] 1 [Fe ] [Cu ++ ]

[2]

c

-p. [Fe 3+ ] Thus if „ =r » 1 [Fe 2+ ] We have AE = 0.60 + (2 x 0.059) log r which is positive. It is clear that E' has a higher positive value than E". Thus electrons would flow in the exteranl circuit from right side to the left side. Hence. Fe+++ + e -> Fe++ on the left side. Cu+ -» Cu++ + e on the right side i.e., Fe+++ is an oxidant for Cu+ ii) Under equilibrium conditions E' - E" = 0 0.60 = - 2 x 0.059 log r .-. log r = -5.0847 0 059 Thus E' = 0.77 + — log r = (0.77 - 0.059 x 5.0847)V 1 E' = 0.47 V = E' 10.

[4]

H2S contribute negligible H+ compared with HCI0 4 [H ]2[S2 ] _ - KtK2 [H2S]



1.3x10-x1.0x10-xQ.1Q

2

(0.003)2 Given [Mn2+] = [Cu2+] = 2 x 10"4 mol Now [Mn2+] [S 2 1 = 2.8 x 10~21 [Cu2+] [S 2 1 = 2.8 X 1CT21 i) CuS will precipitate (2.8 x 10"21 > Ksp (Cus) ii)

[Cu2+]=i|^=6x10-20 IP

J

1 1

[1] [1]

The amount of Cu2+ remaining unprecipitated 6x10 2

" °-x100 =3x10-14%

2x10

-4

2

(1.3x10-)0.1Q=1

(10"7)2 Now, [Mn ] [S2T = [2x10^] [1.3 xlO - 8 ] = 2.6 x 10~12 Hence MnS will precipitate 2+

[Mn2*] =

[1]

3x1 °"'4 = 2.3 x 10 -6 1.3x10 Percentage precipitation 2 3x10"® = 100 x 100 ~ 98.85% 2x10

[2] [1]

1 J

[2]

1 J


fBITJCC RANKERS STUDY MATERIAL IIT JEE, 2002 PHASE-1II CHEMISTRY SOLUTION 1.

(a)

The mass of the platinum ball and of the liquid it displaces are determined from the respective densities and the volume of the ball. Volume is given by V = —Ttr3 = —3.14 x (0.25)3 = 0.0654 cc 3 3 Mass of Pt ball = 0.0654 x 21.4 = 1.4 g Mass of liquid = 0.0654 x 3.2 g/cc = 0.21 g (1-4-0.21)980 v = (m-mB)fl = = 67i-nr 6x3.14x0.25x10 1.00 1= d _ ~ = ~ = 40.5 sees v ~ 2.47x10"

... (b)

1Q_2 c m / s

[2]

_ 2.303RT C. E= — l o g — 1 (Ci > C2) nF C2 For Ag+, n = 1 and at 18°C — — — = 0.058 F Since AgCI in sparingly soluble, concentration of Ag+ ion will be more in N/10 AgN0 3 than in AgCI. Hence Ci = [Ag + ]=

0 813

10

= 0.0813 mol/lit

Let C2 be the concentration of Ag+ ion in saturated solution of AgCI in 1N KCI 0.51 = 0.058 log — — C

C2 = 1.31 x 10"10

[2]

2

Because 1N KCI is 76% dissociated .-. [CI1 = 1x0.76 g. ion/lit ••• [Ag+][C|-] = 1.31 10-10 x 0.76 = 0.995 x lO -10 Solubility product of AgCI = 0.995 x 10_1° Let the solubility of AgCI in aqueous solution be S mol/lit ••• [Ag+] = [Crj = S Ksp = S2 = 0.995 x 10~10 S = 0.9974 x 10~5 Hence solubility of AgCI in water = 9.974 10-6 mol/lit

FIITJCC.

ICES House (Opp. Vijay Mandal Enclave),

Sarvapriya


[2]

[2]

Ph:6854102,

6865182,

Fax:

6513942

RSM12-P1-2-T(M)-CH(S)-150

2.

(a)

H2(g) At equilibrium 0.1-x

+

C0 2 (g) 0.2-x

H 2 0(g) + CO(g) x x

Mole percent of H 2 0 = — x 100 = 10 0.3 x = 0.03 • [H^CO^ (0.03)' [H 2 ][C0 2 ] 0.07x0.17 Now after introduction of CoO(s) and Co(s), the equilibria's with equilibrium concentrations are shown as H2(g) + C0 2 (g) H 2 0(g)+. CO(g) (0.1-x-y) (0.2-x+z) x+y x-z CoO(s) + H2(g) Co(s) + H 2 0(g) 0.1-x-y x+y CoO(s) + CO(g) Co(s) + C0 2 (g) x-z (0.2 - x+z) _ ( Now, mole percent of H 2 0(g) =

x

+ y x x100 = 30 0.3

/. x + y = 0.09

= 0.06

Now equilibrium constant for I reaction is (x + y ) ( x - z ) Ki = ( 0 . 1 - x - y ) ( 0 . 2 - x + z)

[2]

0.0756 = ° 0 9 ' 0 0 3 - Z » 0.01(0.17+ z) z = 0.028334 . [HgO]= x+ y = 0 ^ 9 = 9 [H2] 0.1-x + y 0.01 [CQ1] = 0 . 2 - x + z = 0.198334 [CO] x-z 0.001666

(b)

B = ® a P = 2 6 5 r(CO) V M ( H e ) * 4 Helium escapes at a rate 2.65 times as fast as CO. r(He) = — 6.4mmoi x 1 = 2.4 _ . mmol. CO/hr __ r(CO) = ——2.65 hr 2.65 Time taken for 10 mmol of CO to leak through = 10 mmol CO x

2H+ +

— =4.2hr. 2.4 mmol CO

[2]

[3]

S2"

At pH = 0 K K

= t H + ] 2 [S 2 1 [H2S]

... io~7 x io~14 =

fx

ts2"l 100x10 3

[S 2 1 = 10"22 M The ionic product of Cd2+ and S2_ is given as [Cd2+] [S 2 1 = 10-22 x 10-3 = 10"25M2 The ionic product cannot exceed the solubility product CdS will not get precipitated at pH = 0

[3]

FIITJCC ICES House (Opp. Vljay Mandal Enclave), Sarvapriya Vihar, New Delhi - 150. Ph:6854102, 6865182, Fax: 6513942

RSM12-P1-2-T(M)-CH(S)-3

At pH = 2 (10 2 ) 2 [S 2 ] 100 x 1 0 3 2 [S ~] = 10~18 M ionic product = 10"3 x 10"18 = 10~21 M2 Thus ionic product exceeds solubility product CdS will get precipitated at pH = 2

[2]

Process: (+2)(-1) (+7) Fe S522 +Ba(Mn0 + Ba(Mn044) 22

[1]

10 -2i =

4.

(+6) (+2) 3+ 2 > Fe + SO " + Mn

Hence, remaining permanganate reacts with Kl. (+7) 2+ Ba(Mn0 4 ) 2 + Kl > l2 + Mn iBa(SCN) 2 SOj"+CN"+r > SO2" is 6.

Here, n - factor of Ba(SCN)2 So number of equivalent of Ba(SCN)2

=

[2]

50x0 12x6

= 0 .036 1000 = number of equivalent of l2 = number of equivalent of Ba(Mn0 4 ) 2 remaining. So, number of Ba(Mn0 4 ) 2 consumed by FeS2 . _ _ 10x0.35x5 _ . = equivalent of FeS2 = = 0.0175 [21 1 1 1000 Total number of equivalent of Ba(Mn0 4 ) 2 in 100 ml = 0.036 + 0.0175 = 0.0535 Moles of Ba(Mn0 4 ) 2 in 100 ml = 0 "0535

[as 'n' factor is '10' for Ba(Mn0 4 ) 2 ]

[2]

0 0535

Molarity of the solution = - — — x 10 = 0.0535 M. 10 Eq. of FeS2 = 0.0175 Moles of FeS2 =

0 0175

= 1.17 x 10"3

15 Wt of FeS2 = 1.17 x 10"3 x (56 + 64) = 0.1404 gm 0 1404 percentage purity = — x 100 = 14.04%

1

5.

AG° = -RTInK = -2.303 RTIogK o r , 440 = - 2.303 x 2x 298 x log K K = 0.4779 CH3CH2COOH(l)+C2H5OH(l) C 2 H 5 C00C 2 H 5 (l)+H 2 0(l) n(1-a) n(1-a) na na •

{n(1 - a)}

=

_ 5 i l _ =0.4779 (1-a)2

6.

(a)

[\~] to precipitate Agl =

Q

^

[1]

[2]

[2]

1 J

a = 0.4087 if n = 0.5 mole then number of moles of C 2 H s COOC 2 H 5 (l) produced is = na = 0.5 x 0.4087 ^ = 0.2044 O C y IfY"17

[2]

[2]

= 8 5 x 10 1? M

"


RSM12-P1-2-T(M)-CH(S)-4

2 5x10" [ H to precipitate Hg2l2 = J ' — = 5 x 10'13 M

[2]

as [l~] required to precipitate Agl are lesser than that required to precipitate Hg2l2, thus Agl precipitates first. It will continue till the [l~] becomes 5 x 10~13 when Hg2l2 begins to precipitate and thus maximum [l~] for Agl precipitation = 5 x 10~13 M

17 1

•^•^-fri^- - ^"

121

v 0.1M [Ag+] becomes = 1.7 x 10"4 M [Ag+] in solution % [Ag+] in solution =

1 7 x 1 0

x10

0.100

°

= 0 .17%

:.% of [Ag+] precipitated = 100 - 0.17 = 99.83%. 7.

No. of gm equivalents of NaOH required for back titration _ 2.39x0.1 1000 = no. of gm equivalents of H 2 S0 4 remained unreacted .-. no..of gm equivalents of H 2 S0 4 reacted ^ 4 5 x 0 . 2 x 2 2.39x0. 1000 1000 = 17.76 x10" 3 = no. of gm equivalents of MiM2CC>3 = no. of gm equivalents of pure M1M2C4H406, 2H 2 0 Let wt. of the pure sample was x gm moles = — 246 .'. no. of gm equivalents = x

x 246

„

[41

L 1

246 or, x = 2.1845 gm

Pco 2 +

2 1845

10.732

PN 2 + PH 2 O

[4]

x2

x 2 = 17.76x 10

.-. % purity =

[1] 1 . [1]

x 100 = 20.355%

[2]

1 1

= 7 65 mm at 35°C

So, at 0°C the total pressure will be

765 x 273 308

mm = 678.1mm

[2]

1 J

Ph2o = (678.1 - 6 4 5 ) = 33.1mm So pN2 + p C02 = 645 mm at 0°C pN^ = 345 mm at 25°C y 97^ mm = 316mm p N2 at 0°C=

[2]

298

••• Pco2 = ( 6 4 5 - 3 1 6 ) = 329

Since p = mole fraction x total pressure Mole fraction of H 2 0 =

33 1

= 4.9% 678.1 31 © Mole fraction of N2 = = 46.6% 678.1 Mole fraction of C0 2 =

678.1

= 48.5%

[2]

FIITJCCICESHouse (Opp. Vljay Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Ph:6854102, 6865182, Fax: 6513942

RSM12-P1-2-T(M)-CH(S)-5

9.

Energy of photon liberated from He+ during emission of H a line of Lyman series = hc-RHZ2

1

1

2

1

= 6.625 x 10~27 x 3 x 1010 x 109678 x 22

v4y -11 = 6.54 x 10"" erg This energy is used in liberating electron from H atom from ground state, therefore,

[2]

6.54 x 10-11 = Ei of H + - m u 2 2

[2]

= 13.6 x 1.602 x 10" 12 + - mu2 — mu2 = 6.54 x 10~11 •2.178 x 10, - n

2

-11 erg = 4.362 x 10~" 4.362 x 10"11 x 2 u2 = 9.108 x 10"28 u = 3.09 x 108 cm sec -1 10.

Let V ml blood is present in patient a) r 0 ofNa 2 4 = 2 x 103 dps = 2 x 103 x 60 dpm = 120 x 103 dpm for V ml blood 24 r of Na = 1 6 d p m / m l a t t = 5hr = 16 x V dpm/V ml Jo. = No.

r N N0 _ 120x10 3 N ~ 16V , 2.303. N2 t= log- K N _ 2.303x5, 120x10 3 5= log 0.693 16V V = 5.95 x 103 m! b) Activity of blood sample after 5 hours more i.e. t = 10 hours •t = 2.303.log— N0 a K N 2.303x15, 120 x103 10 = _ —log0.693 ~ A A = 75 x 103 dpm per 5.95 x 103 ml 75.6 x10 3 . = r- dpm per mi 5.95x 10 = 12.71 dpm per ml = 0.2118 dps per ml 11.

a)

[2]

[2]

[1] [1]

[1]

[1]

[1]

[1]

p + V42 [ V - b ] = R T RT a or P = ( V - b ) V2

FIITJCC ICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Ph:6854102, 6865182, Fax: 6513942

RSM12-P1-2-T(M)-CH(S)-6

Multiply by [V] then, PV = or PV = RT

_ (V - b)

[1]

VRT_

1 - i f — ! L V J VRT

b V

PV =RT

axV

V

or PV = RT

Now 1

RTV V-b

[2]

= 1 + —+

V

,

1+

b

a

V

VRT

+

fb

—

\\L,

— ) • — + ('— + • or PV = RT 1 + 1 b -RTJ v Ivj

(b) X =

0.693 tn / o

[1]

0.693 = 2.689 x 10-7 sec"1 29.82x24x60x60

Number of atoms in 10 mg of ™1Ce

[1]

_ 6.023x10 23 x0.01g 141g/atom

= 4.2696 x 1019 atoms Rate of disintegration per sec, = decay constant per sec x number of atoms = (2.689 x 10'7) (4.2696 x 1019 atoms) = 1.148 x 1013 or, number of p - particles emitted per sec = 1.148 x 1013 Total energy emitted per sec = (1.148 x 1013 x 0.4132) MeV = 4.745 x 1012 MeV = 4.745 x 1012 x 1.6 x 10^ erg 1 watt = 107 erg 4.74 x 1012 x 1.6 x 10"6 Energy in watts = = 0.7592 watt 12.

[1]

[2]

[2]

107

12

a = 400 x 10" m = 400 x 10"12 x 100 cm = 4 x 10"8cm So, volume of unit cell edge length = a3 = (4 x 10"8)3 As we know that nxM m density p NA x a3

[1]

here, n = no. of atoms per unit cell = 2 Mm = mass of 1 mole atom = 110 gm N A = Avogadro's no. = 6.023 x 1023 = 2x100x10

2x100

6.023 x10 23 x(4x10"-8a X)3

24

= 5.188 gm/cm 3 6.023x10 23 x 4 x 4 x 4 v 100 gm of A contains 6.023 x 1023 atoms of A T5 10 10 gm of A contains 6.023 x1023 x 100 = 6.023 x 1022 atoms of A v 2 atoms remain present in 1-unit cell of bcc type 6.0 23 x1022 remains present in

1x6 023x 10

'

[2]

[2] u n j t c e || S =

3.0115 x 1022 unit cells

[1]


fllTJCC RANKERS STUDY MATERIAL IIT-JEE, 2002 PHASE - IV

CHEMISTRY

SOLUTIONS O 'CH - CH3 H (A)

fc

O - CH3

_

Ii-CHBr?

2Br,

(B)

(C)

OH OH

(C) —

-h;Q „

-CHO

(y

i) OH", IntramoiecularCannizzaro ii) 2H*

?

if"

H

CH - COOH

(D) 2.

a) O

NaOMe

O + MeOH

OMe

C0 2 Me

(A)

(3-ketoester

(B)

(B)-

C0 2 H

mild

+ MeOH

Hydrolysis 3-oxo-acid or (3-keto acid

(C)

-co 2

+ C0 2

b) Diekmann condensation (intramolecular Claisen ester condensation) fllTJ€t

ICES

House, Sarvapriya Vihar (Near Hauz Khas Term.), New Delhi-16

Ph •

^

RSM12-P-IV -T(M)-CH(S) - 2

c) Mechanism of decarboxylation

-co, keto 3.

i)

A B

= =

O

Bu - C = C - L i Bu - C = C - CH - Ph

AH

C

=

&

Bu-C = C - C - P h

BU

o- 0C a ii)

[4 x 1 1/2]

E=

F= iii)

G = Ph - C = C - MgX H = Ph - C =C - CH2Ar

trans 4.

CH3

[ 1 + 1 + 2 = 4] CH,

CH:

CH3-- C — C H = I - •CH3

CH3

CH3-i—C—C H 2—C—CH2

Lh3

CH3 (C) CH3 1

(D)

CH3-- C — O H !:H3 (E) (A) and (E) both reacts to give (C) and (D) only so they must involve same intermediates. 5.

H

A = HOH 2 C—C—C0 2 H I CH3 [2] B=

?

H 2 C=C—C0 2 H CH3

[2]

FIITJCC ICES House, Sarvapriya Vihar (Near Hauz Khas Term.), New Delhi - 16, Ph : 686 5182, 685 4102, Fax : 6513942


c = H2C=(|:—coci CH3

[1]

D = H2C=Cj^—CON(CH3)2 CH3 E = CH 3 CH 2 CHO

[1] [2]

a)

CHO b)

c)

H+

HCO.H

N0 2

O

H3O+

OH HO^J—H

N0 2

ql

N j o2 J

0-5°C

NaN0 2 + HCI

N0 2

NO2 Or Cu/HBr

LvJ^

r d) PhCOCH3

N2+

^ PhCOCH2Br

Ph

^p

cr

> [Ph 3 P C H 2 C O P h ] B r -HBr

Ph 3 P=CHCOPh 7.

PhCOCH3

>

phC(CH 3 ) = C H C O P h

a) Addition on carbon - carbon double bond: £fi^CH—CH 2 —C=CH+ HBr

> CH3CHCH2C=CH 2° carbocation :Br~

<

CH3(j)HCH2C=CH Br (Actual product) Addition on carbon - carbon triple bond:

FIITJCCICESH o u s e ,SarvapriyaVihar (Near Hauz Khas Term.), New Delhi - 16, Ph : 686 5182, 685 4102, Fax :6513942


-> CH2=CH—CH2—C = CH 2 (Vinylic carbocation)

CH2=CH-CH2—C£CTA+ HBR-

:Br CH2—CH—CH2—C—CH2 I

2° - carbocation being relatively far more stable than the vinylic carbocation, addition occurs at ethvlenic linkage rather than on acetylenic linkage. b) h2C-—-CH2 + RNH 2 - — > CH2—CH2OH R—ILH (3-amino alcohol The product p-amino alcohol is less nucleophilic because of stabilisation by intramolecular hydrogen bonding between the lone pair of electron on nitrogen and partially acidic hydrogen of the alcoholic group. H s+ H

^J

5

-

>

2

r / ^ c ^N h 22 R' c) In this case, the mechanism involves the participation of the carboxylate group as given below

Step: 1

Inversion H ;H

+ AgBr

Br. Ag+

An a-lactone (highly strained)

Step 2: COO" -H

CH3

+

H

CH3

OH

The net result of two inversions is an overall retention of configuration. The phenemenon is called neighbouring group participation. 8.

a) The most probable mechanism appears to be elimination addition one involving benzyne intermediate as shown below.


RSM12-P-IV -T(M)-CH(S) - 5 H 2 CH 2 NHMe

:C6H5 -c 6 H, Intramolecular nucleophilic attack

X j l ISk

, / H

J

intramolecular proton transfer Me CI C!

a "f ~ J elimination "

HL +

carbene J^CI ^

J) Eiectrophile

(Remainder amount) C H 3 - C - ChLPh II o (A)

2C(OH)CH,Et

(B)

CD, 2 = — Cw.H i 2 (C)

OPh

a /

NDH

(F) Me =CH

CH=CHNO 2 (H)


FIiTJ€€ RANKERS STUDY MATERIAL IIT-JEE, 2002 PHASE- III-IV CHEMISTRY SOLUTION 1

(a,

o H5C6—C

fcH-

COOC2H5

'H 3 (B)

R"P>R-CH=CH-COOH N

(c)

O

^

CH. U = CH—C—OH

(d)

Me I Me-C-C-Me O Ph

(e)

O

1

2.

Nh

[5x2]

P i

! his ketone is more acidic because the resulting enolate ion Obey's Huckel's rule and is thus more stable.

3-

<3)

*p"h

o/h

1

I

•u

/H

C ^ ^ c h , HO\/0*—CHS

FIITJCC.ICESHouse (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi -110016.

H O

ÔCHS

Ph:6854102, 6865182, Fax: 6513942

RSM12-P3-4-T(M)-CH(S)-2

(b)

/O-Zn-Br C j

(c)

+ Br - Z n - CH2COOC2H5

HCI3 + tBuOK

, f

CH2COOC2H5

H,0 Zn(OH)BrL

COOC2H5

» KCI + t-BuOH

(d)

(e) [5x3] 4.

(A)

(B)

O II CH, - C = C H - C - C H 3 I CH3

Et02-^\o2Et (C) Et0 2 O (D)

HO21 O (E) >1 o 5.

[5 x 2]

OEt

(a)

Br\

^N02

N02

[3]

(b)

/-ci A =

A =

IJJ

CHO

N

H FIITJCC.

ICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi -110016.

[3] Ph:6854102, 6865182, Fax: 6513942

RSM12-P3-4-T(M)-CH(S)-3

(c)

CH3CH 2 CH 2 CH=CH2

[3]

CH 3 MgBr

)

r

^

CH2CH2COOH

7.

[6x2]

Five hydroxy acids have the formula C 4 H 8 0 3 , are OH i)

CH3-CH2-CH-COOH a-hydroxy acid

OH ii) C H 3 - C -

COOH

^Hs a-hydroxy acid

iii)

CH3-CH-CH2-COOH H p-hydroxy

acid

CH3 iv) CH 2 - C H - C O O H P-hydroxy acid

v)

CH2-CH2-CH2COOH

[3]

OH y - hydroxy acid

CH 2 CHI -

C> j

CH

2

O-H 5TQH

-C-OH^-

J

O-H +

o

J

o

A y-lactone y-hydroxy acid

CH3

CH 3

CH, - CH - COOH

> CH 2 = C - - C O O H

OH p-hydroxy acid

o CH, C H - C H , - C - OH—-—>CH,CH = C H - C O O H

I

a, p unsaturated acid

H P-hydroxy

acid

FIITJCC. ICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi -110016.

Ph:6854102, 6865182, Fax: 6513942

RSM12-P1-2-T(M)-CH(S)-150

A

CH2CH2 - C H - C O O H -

> CH3-CH2-CH

/ \

OH

O II c—o

CH-CH2-CH3

/

O —C

a-hydroxy acid

Ii

o

o 9Ha

CH3 - C - C

\

O

O

PM 3 \ CH

H

I OH

c

C H /

/ C \

O

a-hydroxy acid

o \ — C

/

C

lactide

C H

\ c H3

O A lactide

8.

[3] 143.45g of AgCI contains 35.45 g CI" ions 35 45 x 2 87 2.87g of AgCI will contain _ = 0.709 g CI" ions 143.45 266.35 g CrCI 3 -6H 2 0 contains n x 35.45 g of ionisable CP ions (where n = no. of CP ions outside the coordination sphere). Thus, 2.665g CrCI 3 -6H 2 0 will contain

n x 3 5

-45x2-6^g 266.35

nx 35.45x2.665 — = 0.709 => n « 2 266.35 Keeping in view the octahedral geometry of the complex, its structure may be written as [CrCI(H 2 0) 5 ]CI 2 -H 2 0

Also

9.

[5]

Li forms only oxide, Na forms oxide and peroxides and remaining all alkali metal form oxides, peroxides and sulphur oxides. Li + O? -> 2Li,0 (Oxide)

4Na + 0 2 -—•> 2Na 2 0 (Oxide)

2Na + 0 2 •— > Na 2 0 2 (Peroxide)

K + O2

->

KO

2 (superoxides)

Peroxide [ - 0 - 0 - ) 2 has not unpaired electron therefore. It is diamagnetic, on the other hand super oxide [02]~ has an unpaired electron therefore it is paramagnetic super oxides are stronger oxidizing agent then peroxides Stability of the peroxides and super oxides increases as the metal ions become large peroxide and super oxide are larger. Same if both ions are similar in size the coordination number will be high and this gives a high lattice energy. K0 2 is used in space capsules, breathing masks because it both produces dioxygen and remove carbon dioxide. 4KQ2 + 2C0 2 > 2K 2 C0 3 + 30 2 4K0 2 + 4C0 2 + 2H2Q » 4KHC0 3 + 30 2 [5] 10.

i)

The gas which is liberated on heating the mixture with NaOH gives red ppt. With K 2 Hgl 4 so gas is NH3 and mixture contains N H / ion.

ii) The aqueous solution gives white ppt. with BaC!2, so mixture contains SO4"2 ion. iii) Mixture on heating with K 2 Cr 2 0? and H 2 S0 4 gives red vapours (of Cr0 2 CI 2 ) so mixture contains CI" ions. FIITJCC ICES House (Opp. Vljay Mandal Enclave), Sarvapriya Vihar, New Delhi -150.Ph:6854102, 6865182, Fax: 6513942

RSM12-P1-2-T(M)-CH(S)-16

iv) Aqueous solution of mixture gives blue colour with K3[Fe(CN)6] and thus, it contains Fe+2 ion. v) Thus mixture has NH4+, Fe+2, S04~2, CI" Reaction: i)

NH4+ + NaOH

> NH3 + Na+ + H 2 0

ii)

Hg K2[Hgl4] + 3 N a O H + OH3 — - >

^ N H

Brown

iii) S0 4 " 2 + BaCI2

v) 3Fe 11.

I + 4KI + 2HzO + 3Nal

Hg

» BaS0 4 I +2CI" white crystalline ppt.

iv) 4Cr + K 2 Cr 2 0 7 + 3H 2 S0 4 +2

2

» Cr0 2 CI 2 + K 2 S0 4 + 3H 2 0 + 2S0 4 2 (Red) (A)

» Fe3[Fe(CN)6]2 + 6K +

+ 2K3Fe(CN)6]

[10]

(B) Blue

a) Among the halides of lithium LiF, LiCI, LiBr, Lil the covalency is maximum in Lil according to the Fajan's rule and LiF has maximum ionic character. So the melting point of LiF is maximum. [3] b) Bond orders of both N2 and N2 are 2.5 B.O. of N2 = B.O. of N2 =

5-0 - =2.5

2

6 - 1

2

=2.5

But in N 2 , one of the N-atom being positively charged attracts the bonding electron cloud of the other N-atom in higher extent comparing in N 2 . So the bond length in N2 is lower than the bond length in N 2 . And bond energy of N2 > N 2 . 12.

a) (i) sp3

13.

a) Half reactions are MnO 2 '

[4]

(ii) sp3

[11/2 x 2]

>Mn0 4

Mn0 4 ~ >Mn02 Balancing the atoms 1st, then the charges and the e~s for both the half reactions 1st step MnO 2 " > Mn0 4 MnO 2 " + 4 H + — — » Mn0 2 + 2H 2 0 2nd step

MnO 2 "

» Mn0 4 + e"

MnO 2 ' + 4H+ + 2e" rd

2

3 step

[MnO ^ 2

adding 3Mn0 " + 4H b) Half reactions are

+

->Mn0 2 + 2H 2 0

>Mn04+e-]x2 > 2Mn0 4 + Mn0 2 + 2H 2 0

[2]

AS2S3 > H 3 As0 4 + S N0 3 > NO st Balancing 1 the atoms, then the charges and the e~s for both the half reactions 1st step As2S3 » 2H3As04 + 3S N0 3 + 4H" nd 2ind step

As2S3 + 8H 2 0

FIITJCC ICES House (Opp. Vljay Mandal Enclave),

NO + 2H 2 0 > 2H3As04 + 3S + 10H+ + 10e" Sarvapriya


6865182,

Fax:

6513942

RSM12-P1-2-T(M)-CH(S)-150 N 0 3 + 4H+ +

>N0 + 2H20

3rd step

[As2S3 + 8H 2 0 [NO3 +4H + +3e"

> 2H3ASO4 + 3S + 10H+ + 10e" ] x 3 >N0 + 2H 2 0] X 10

adding

3As2S3 + 4H 2 0 + 10N0 3 + 10H+

> 6H3As04 + 9S + IONO

[2]


r

FIITJCC RANKERS STUDY MATERIAL IIT-JEE, 2002 PHASE - 1 to IV

CHEMISTRY

SOLUTIONS fi • 1.

a) Uranium present =

= ^ ^ g atom 100 238 = 2.10 x 10"3 g atom 2.425 g atom D . present. = 2.425 g = Pb 100 M 100x206 y 2 425 x 93 Pb formed from Uranium decay = 100x206x100 = 0.109 x 10~3 g atom Thus, N = 2.10 x 10"3 g atom N 0 = (2.10 + 0.109) x 10~3 g atom = 2.209 x 10~3 g atom 2.303, N t Now t = log— X N 2.303 , 2.209 x10~3 log 1.52 x10"10 2.10 x10-3 t = 3.3x10® yrs

b) E, of H atom = -13.6 eV 6.625 x10"34x 3.0x10 t u t Energy given to H atom = — 1028 x 10~10 = 1.933 x 10~18 J = 12.07 eV Energy of H atom after excitation = -13.6 + 12.07 = -1.5 eV • E =Ii • n n2 2 = - 1 3 . 6 = 9. .-. n^ -1.53 n=3 Thus electron in H atom is excited to 3rd shell; I induced Xi = E ——— 3-E, vE,

= - 1 3 . 6 eV; E3 = - 1 . 5 3 eV


RSM12-P I to IV - T ( M ) - C H ( S ) - 2

6.625 x 10~34 x 3 x 108 (-1.53 + 13.6) x 1.602 x 10" = 1028 x 10"10 m = 1028 A he II induced X2 = (E2-E,) EI = -13.6eV; E2 = - — e V 4 . _ 6.625x10" 3 4 x3x10 8 k2 13 6 - + 1 3 . 6 X 1.602x10" 4 -

= 1216 x 10~10 m = 1216A he III induced X3 = E3-E2 El=-i3.6eV;

E2 = - — eV; E3 = 4 6.625 x10" , 4 x3x10 8 \ 13.6 13.6 X 1.602x109 4

, _ A3 -

• +

- ^ e V 9

-

= 6568 x 10"10 m = 6568A 2.

a) mm added mm left

Na 2 C0 3 7.5 (7.5-a)

+

^ 35.5

= 7.32x10" 5

Given [ C r [ = ^

Also conc. of CP formed

2AgCI ^ ^ 2NaCI + Ag 2 C0 3 excess 0 0 excess 2a a

— = — Volume in ml 5

2a 0.0026 a = 1.83 x 10"4 millimole 5 35.5 .-. m mole of Na 2 C0 3 left in 5 ml = 7 . 5 - 1.83 x 10"4 =7.5 or [ c o r ] = N o w K

.S .A „ CcO P

02

.-.[Ag + ] 2 = +

8

^ =[Ag12[co;-]

3

n

- 2 x 1 ° 1 2 = 5.46 x 10~12

7.5/5

.'. [Ag ] = 2.34

x

10-6

KSP of AgCI = [Ag+] [CI] = 2.34 x 10 - A -y A

An-10

x

0.0026 35.5

K SSP p - 1.71 x 1 0 _ 1 °

b) Meq. of MnO; added = 200 x 0.75 x 5 = 750 Mn+7 + 5e > Mn+2 Meq. of MnO; left unused = Meq. of Fe+2 used = 175 x 1 x 1 = 175 v Fe+2 > Fe+3 + e Now Meq. of MnO; used = 750 - 175 = 575 MnO; is used for Cu2S and CuS to give For Cu2S:

Cu;

- > 2Cu+2 + 2e

S"2

»S +4 + 6 e

FIITJ€«ICES House, Sarvapriya Vihar (Near Hauz Khas Term.), New Delhi - 16, Ph : 686 5182, 685 4102, Fax : 6513942

RSM12-P I to iV - T ( M ) - C H ( S ) - 3

Cu2S

> 2Cu

S"2

ForCuS:

+ S+4 + 8e

> S+4 + 6 e

Let Cu2S and CuS be a and b g respectively :. a + b = 10 ...(1) .•.eq. of MnO; used = Meq. of Cu2S + Meq. of CuS a b x 1000 + x 1000 575 = 159.2/8 95.6/6 .-. Solving Eqs. (1) and (2) a = 4.206 g b = 5.794 g 5 794 .-. % of CuS in mixture = x 100 = 57.94% 10

(i)

sp3d

...(2)

sp d

r (iii)

:e

^

F F

sp3d2

(v)

a) i)

JL *

In trimethylamine carbon atom having no vacant d-orbital of suitable energy cannot undergoes rc-bond formation with the nitrogen atom and hybridisation of trimethylamine remains sp3. But in trisilylamine silicon having vacant d-orbital, undergoes 7t-bond formation with the lone pair of nitrogen atom and hybridisation changes into sp2. So lone pair of nitrogen is not free for donation in trisilylamine and its basicity is lower.

ii) Due to the smaller size of F-atom, when it accept electron into the valence orbital inter electronic repulsion becomes so prominent that acceptance of electron is energtically unfavourable. In CI atom this inter electronic repulsion is not very high due to its larger size and electron affinity of CI is greater than F atom. b) i) K 2 Cr 2 0 7 + 14HCI » 2KCI + 2CrCI3 + 7H 2 0 + 3CI2 ii) 3P + 5HN0 3 > 3HP0 3 + 5NO + H 2 0 200 mm pressure of O

t=47 min

4000 mm pressure of mixture For pure 0 2

) 15Q0

m m

t=74 min -» 1:1 (0 + gas) 2


RSM12-P I to iV - T ( M ) - C H ( S ) - 4 PL=r±

P2 n2 When r\i and n2 are original no. of moles of 0 2 and moles of 0 2 after 47 minutes n, ^ 2000 " n2 ~ 1500 .'. n2 = 3/4n! or moles of 0 2 diffused in 47 minutes = r\i -

4

=— 4

n x 74 or moles of 0 2 diffused in 74 minutes = — 47x4 74 74 .. -n„ = — if ni =1 188 1 188 = 0.3936 Since diffusion of 0 2 in mixture also occurs at partial pressure of 200 mm. (The ratio of gas and 0 2 being 1:1) Now gas and 0 2 both diffusing in form of mixture of through same orifice at the partial pressure of 2000 mm each nO; 14 n

X

74 79^ n" ~ V V 32

g = n02 x .

w

74 188

v79 y

= 0.249

Moles of 0 2 left after 74 minute = 1-0.3936 = 0.6064 Moles of gas after 74 minute = 1-0.249 = 0.7510 0 2 : gas = 0.6064 : 0.7510 1:1.236 6.

ol (B)

}

H2S04

CH3

ICHMBIJ

(j)H

2

'H3

(O

^

CHs —C — CH3

COOH CH3—CH—CH3


RSM12-P I to iV - T ( M ) - C H ( S ) - 5

a)

b) Compound (A) gives red ppt. with Cu(NH 3 )2. Hence it is terminal alkyne C3H7 — C = CH +

C U ( N H 3 )+2

> C3H7 — C = C Cu + NH3 + NH;

C3H7 — C s CH + NaNH2 > C3H7 — C s C Na + NH3 C 3 H 7 — C = C Na + C H 3 — C H 2 — CH 2 Br » C 3 H T — C = C — C H 2 C H 2 C H 3 + NaBr Since (B) gives only one carboxylic acid on ozonolysis followed by hydrolysis, compound (A) is CH 3 —CH 2 —CH 2 —C^CH (i)0 3 •> 2CH 3 CH 2 CH 2 C00H C H 3 C H 2 C H 2 - -C=C—C H 2—C H 2—C H3 |H 2 0 (C)

(B)

a) H3C = C" + CH3CHO

->CH 3 C = C CH-CH 3

i

CH3OH

CH 3 C= C - C H - C H 3

b)

CH

4

"O-

(c) } N +2 He a) i)

i)H C(CH3)3

-> 8 7 o + ; H

1,3-Butadiene is a conjugated diene and is a resonance hybrid: - C = C-C = C-<-*-C-C = C-C-<-*C-C = C - C I I I I I I I I I I I The uncharged structure involves a larger contribution than the charged structures. The latter induces some double bond character in the central C - C bond leading to the shortening of this bond. Besides this, the cental C - C bond involves sp2 -: sp2 carbon atoms which also shortens the bond as compared to the C - C bond in nbutane which involves sp3 - sp3 hybrid orbitals.

ii) The nitro group in nitrobenzene strongly deactivates the benzenes ring. This decreases the reactivity of benzene towards Friedel - Crafts alkylation. b)

200 mL 10"2 M (pH = 2) of HCI 300 mL 10~2 (pH = 12) of NaOH On mixing 100 mL of 10~2 m NaOH will be left unneutralised. Since the total volume of solution would be 500 mL so [OH] would be

100 500

x

10"2M = 2

x

10~3M

Hence pOH = -log [OH1 = 2.7 .-.pH = 1 4 - 2 . 7 = 11.30


RSM12-P I to iV - T ( M ) - C H ( S ) - 6

10.

2P(g) Po

-+4Q(g) + R(g) + S(l) 0 0

P 0 -P'

2p'

y after 30 min

0

2p0

^

after long time

After 30 min, the total pressure = P0 - p' + 2p' + ^ + vapour pressure of S(l) =

Po

+ 3Pl + 32.5 2

p

After a long period of time, total pressure =2P0 + y + V.P. of S(l) 5P Hence, — + 3 2 . 5 = 617

2

P0 = 233.8 mm Hg Also, P0+ — P' = 33.8

+32.5 = 317

2.303 k= ——

/

log

rp

o ^

Po-P' k = 5.21 x 10'3 min-1 A 5 . 2 1 x 1 0 - = 2 ^ 3 log 233.8^ r ' 75 v rPo - P .

P 0 -P' = 158.18

p' = 75.62

.-. Total pressure after 75 min = P0 + -3p' j - + 32.5 = 379.73 mm Hg 11.

Since compound (A) gives four mono chloro derivatives it must be 2-methylbutane.

f

CH3—pH—CHa—CH3 + Cl2 CH3 (A)

CH2CI—(jDH—CH2—CH3

(B)

CH3 CH3—CpOl—CH2—CH3

(C)

CH3 CH3—CH—CHCI—CH3

(D)

CH3

?

CH3—CH—CH2—CH3

(E)

CH2CI

CH 3

(j^CI

CH2

CH 3

Ale. K 0 H )

C H 3

CH3

CH3 (D)

r

Q = C H

_ -CH (

3

CH3 (F)

(C) CH3-(^H-CHCI-CH3

_

Ale. KQH)

CH3

_Q=CH_CH3

r

CH3 (F)


FIITJCC ALL INDIA TEST SERIES IFF-JEE, 2002 PART TEST - 1

CHEMISTRY

SOLUTIONS ,1.

Representing bromophenol blue as HBb HBb + H 2 0 H 3 0 + + Bb~ HOJBbl [HBb] pH = 5.1 .-. [H 3 0 + ] = 7.94 x 1CT6 M . [Bb-J_ 5 . 8 4 x 1 0 - = W 5 [HBb]

7.94 x 1 0 6

% in acidic form = = 2.

7.35 + 1

[t[Bb] [Bb"] + [HBb}

x 1Q0

x100 = 11.97%

1[ 2J1

a) Vapour pressure of the mixture is P = 750 XA + 362 When XA = 1 mole fraction of SnCI4 = 0 .-. Vapour pressure of CCI4 = 750 + 362 = 1112 m of Hg and vapour pressure of SnCI4 = 362 mm of Hg. [2] b) in equimolar mixture 1112 Vapour pressure of CCI4 = Pc cu x cci4 = — — = 556 mm Hg O g2 & vapour pressure of SnCi4 = P°nCU XSnC,4 = — = 181 mm Hg Total vapour presure = 556 + 181 = 737 mm 556 Mole fraction of CCI4 in vapour = = 0.7544 737 and mole fraction of SnCI4 = 0.2456 .-. Weight ratio of CCI4 and SnCI4 in the vapour are 0.7544 x 154 : 0.2456 x 261 = 116.8 : 64.1 Weight % of CCI4 =

116-18

180.28

x 100 = 64.44%

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[4]

1 J

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AITS2002-PT-ll-CH(S)-2

The activity of sample A is 2100 counts per week. After a certain time t, its activity will be reduced to 1400 counts per week i.e. a fraction of the active 14C nuclei will decay in time't'. Sample B must be t times older than sample A. [2] M At = A0 e1400 = 2100 e -xt x 5730 /

t=

= 3352 year

0.693

[4]

Let the weight of polystrene prepared be 100 g. 10.46 No. of mole Br in 100 g of polystyrene = = 0.1309 mole 79.9 From the formula of polystyrene we have No. of mole Br = 3 x mole of Br3C6H2(C8H8)n 0.1309 = 3 x weight _ 3x100 molecular weight 314 + 104n . n = 19

[3]

Balmer series Lymann series — =R K

-1 =R

HET

n t = 2, n2 = 3 n-i = 1, n2= 2

Z' 1 _ 1 4 9

RHe

+

Z2(--—\ = R He " 1 4J

g

4x3

1

= 133.7x10^

9 5R

He

R

HE

[2]

4

22 3RHe

+

15R He

= 1.096 x 107 m"1

[4]

Vapour pressure of pure solvent at 300.15 K P2= 760 torr T2 = 353.15 K Pi = ? T-i= 300.15 K 1 |n _ V vap 1 P, R vT, T.2 / But AH'm, n vapour T 2 A S m , vapour . P2 AS x AT . _ AT •'• log — = where AT = T2 - T P, 2.303RT, 87.03x53 760 or loq = 0.8026 P, 2.303x8.314x300.15 Pi = 119.7 torr Mole po _pfraction of solute -

119.7-100 = 0.1646 119.7 Boiling point of the solution 1 1 RlnX TAH T, TB° 1 i- = , 8.314 x 2.303log(1 -0.1646) 353.15x87.03 T 353.15 "

[4]

X,=

[2]

= Q QQ27Q

^

.. T b = 359.3 K fllTJCf. ICES House (Opp. Vijaxj Mandal Enclave), Sarvapriya Vihar, New Delhi - 110016. Ph:6854102,

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7.

Let a be the edge length of the cube so that 4rA = V2a a = 2V2rA Now since the atom B has occupied the body central octahedral hole. 2rA+ 2rB = a 2rA+ 2rB = 2V2rA ^ = V 2 - 1 =0.414 Volume of cube = a 3 = 16V2r3 4 4 D Volume occupied by A & B = 4 x—7trfal + f —nr

= ^(4r A 3 +r B 3 )

[3]

Volume occupied per unit volume of unit cell 4

-(4r A 3 + r B 3 ) _ 3 <> l 16V2rA3

12V2

4+

fr >

3"

VA J

3.14x4.071

= 0.75 12x1.414 Void space per unit volume of unit cell = 0.25 Formula of the compound = A4B

[2] [2]

MnC>4 / CI' dissolved ->Mn2 Mn 3 0 4 Mn 3+ Normality of KMn0 4 is 0.117 against oxalate where n-factor of KMn0 4 is 5. But in the above reaction n-factor of KMn0 4 is 4.

8.

Normality of KMn0 4 in the above reaction is — x 0.117

[3]

Equivalent of KMn0 4 = | x 0 . 1 1 7 x 3 1 . 1 x 1 0 " 3 Equivalent of Mn2+reacted = 2.9 x 10~3 Moles of Mn2+reacted or produced = 2.91 x 10~3 2 91 x 10~3 Moles of Mn 3 0 4 in the sample = — % of Mn 3 0 4 in the sample =

2.91 x1Q- 3 x 229x100 = 40.75% 3x0.545

[5]

For ZnS not to be pptd. from a solution of Zn2+ & Pb2+ [Zn2+] [S"] < Ksp of ZnS [10-2] [S"] < 1.0 x 1CT21 or the maximum [S 2 ^ = 10~19 at which ZnS will begin to ppt. or upto this concentration, no pptn. will occur. [2] H2S 2H+ + S" [H+]2[S"] = 1.1 x 10-22 [H+]2 [10"19] = 1.1 x 10"22 [H+] = 3.3 x 10' 2 M Thus of [H+] = 3.3 x 10"2 or slightly higher, the precipitation of ZnS will not take place and only PbS will precipitate. [4]

9.

10.

1x1 PV = 0.0409 RT 0.082x298 Thus energy needed to break H - H bonds in 0.0409 mole of H2 = 0.0409 x 436 = 17.83 kJ. Also energy needed to excite one H-atom from 1st to 2nd energy level Moles of H2 present in one lit =

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= 13.6

eV

1 -

= 10.2 eV = 10.2 x 1.6 x 10~19 J .. Energy needed to excite 0.0409 x 2 x 6.023 x 1023 atoms of H = 10.2 x 1.6 x 10~19 x 6.023 x 1023 x 2 x 0.0409 = 80.36 kJ Thus total energy needed for the process = 17.83 + 80.36 = 98.19 kJ 11.

[2] [1]

Cell is Pt, H 2 1 H + (1M) || Ag 2 S0 4 (satd) | Ag Eceii = E° AgT/Ag

H+/1/2H2

n 05Q 0.711 = 0.799 + — - l o g [ A g + ] 2 or [Ag+]2 = 10"3 => [Ag+] = 3.2 x 10"2 Now the solubility equilibrium is Ag 2 S0 4 2Ag+ + S0 4 2 2 KSP = [ A g l [ S O / l

[2]

= (3.2x10-)2r3-2x10"2' [4]

= 1 . 6 x 10~5

12.

a) log— = log

300 K,

AH

i2

i

2.303R• v 1-2

J

— 27.216x10 30 2.303x8.314 523 x 553

:. K, = 421.2 b)

[6]

PCI5 - - PC;3 + ci 2 Before dissociation 0.1 0 0 At equilibrium 0.1 ~ x x x Volume of container = 8 lit. X

Kc

- [Pcyccy [PCI5

X x

8 8 0.1-x)

...(1)

From PV = nRT for the equilibrium mixture we get 1 x 8 =(0.1 + x) x 0.082 x 540 x = 0.08 ...(2) From (1) & (2) 0.08x0.08 = 4 x 10"2 mol IS*.—1 lit Kc =

[3]

Also Kp= Kc(RT)An An = 1 = 4 x 10"2 x (0.082 x 540) = 1.77 atm

[3]

8(0.1-0.08)

13.

2H2(g) + 0 2 (g) initial moles 2a a Final moles 2a - 2x a- x _ 2ax80 ._ Given 2x = = 1.6a 100 x = 0.8a

> 2H 2 0(g) 0 2x

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Thus after reaction H2 left = 2a -1,6a = 0.4a moles 0 2 left = 0.2a moles H 2 0 formed = 1.6 a moles Total moles at 120°C in gas phase = 0.4a + 0.2a + 1.6a = 2.2a Now given at initial conditions P = 0.8 atm T = 293 K PV = nRT 0.8 x V = 3a x R x 293 . v = 3a x R x 293 "

0

.

[3]

8

The volume of container remains constant 3axRx293 _ P X = 2.2a X R X 393 0.8 „ 393x0.8x2.2 . P= — — = 0.787 atm 3x293 14.

[31

Pt, H 2 1 H + 1 Ag+| Ag 1 Anode: -H2 > H+ + e 2 Cathode: Ag+ + e > Ag 1 Net reaction: Ag+ + - H2 „co

Eceii = E u c e i i - 0 . 0 5 9 l o g

> Ag + H+

1 [Ag+]

or 0.503 = 0.79 - 0.059 log

1 [Ag+]

[Ag+] = 9.65x10" 6 M

[3]

350

moles of Ag+ = — — x 9.65 x 10"6 = 3.38 x 10^ 1000 mass of Ag = 3.38 x 10-6 x 108 = 3.65 x 10^g % of Ag = 15.

3 65 x10"4

1.05

X100 = 0.0347%

[31 11

At equilibrium - AG0 = 2.303 RT log KP ...(1) Also AG0 = AH0 - TAS° (Given AH0 for NH3 = - 46kJ) a n d AS 0 R e a ction " 2 x A S ° H 3 - A S ° 2 - 3 x

AS°2

= 2 x 1 9 2 - 1 9 1 - 3 x 130 = - 197 J Also T = 273 + 25 = 298 K Thus, AG0 = - 92 x 103 - 298 x ( - 197) since (AH° for reaction - 46 x 2kJ) = - 92000 + 58706 AG0 = - 33294 J Thus, from equation (1) + 33294 = 2.303 x 298 x 8.3 log KP .. log Kp - - 5.845


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[3]

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FIITJCC ALL INDIA TEST SERIES IIT-JEE, 2002 PART TEST - II

CHEMISTRY SOLUTIONS 1.

CH3

Me

CH3 - C - CH = CH2

HCI

> CH3 - C - CH2 - CH2

CH3

Me Me

HCI

cr

CH3 - c - CH2CH2CI (A)

Me IO,

Me, Me

CH3 - C - CH - CH3

CL

> CH3 - C - CH - CH3

Me

Me CI (B)

alkyl shift

CH3-C-CH-CH3-

CI CH3 - C - C H - C H

Me Me

CH 3 CH3 (C)

2.

[ 1 + 1 + 1=3]

2 bromo-5-nitroacetophenone exists in two forms syn and anti. CH3 ch3 NO x

J V

OH

[2]

Br

(Anti w.r.t. methyl)

(syn w.r.t. methyl)

O O;

C - NHCH, o .

Br

pels cold

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o

CH 3

CH 3 -

N0 2

JQi

PCI5

NO

c.

'OH N-

NaOH

No ring closure

x BK " ^ Br [3] In the anti form due to close proximity of the - OH and Br group ring closure is possible which not possible in the syn form. On reaction with PCI5 both the forms undergoes the Beckmann rearrangement.

[3] 3.

a) i)

[1 + 1 = 2 ] ii)

square pyramidal

[1 + 1 = 2 ]

iii)

c / Io N

[1 + 1 = 2 ]

b)

F^N. In BF3 due to back bonding the boron-fluorine bond assumes some double character. This back bonding is facilitated by the availability of a vacant pz orbital on boron. In (CH3)3 N > BF3, the p z orbital of boron is not vacant as it has received lone pair of electrons from nitrogen, so back bonding is not possible due to which bond multiplicity does not take place and hence B - F bond length is larger in the complex. [3] fllTJCf. ICES House (Opp. Vijaxj Mandal Enclave), Sarvapriya Vihar, New Delhi - 110016. Ph:6854102,

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A1TS2002-PT-I l-CH (S)-3

CICH2-qH-pH2

°J

\ j L

OMe

JN.G.P. CH2 - CH - CH 2 - OMe O [ 1 + 2 = 3] OMe - attacks the less crowded carbon and then O" undergoes neighbouring group participation due to which CP goes out. EtCH - CO x I > 0 CH 2 CH

—

(A)

L

EtCHCO • |

EtCH - CH 2 OH UAIH4 (B)

CH 2 CHOH

CH 2 — CH >

MeC = CH 2

ME-CHOH

MeCO

[2 + 2 = 4]

|H!04

EtCHCH 2 OH + MeCHO CH 2 CHO (C)

6

[2 + 2 = 4]

(D)

a) 2Na 2 S 2 0 3 + 3HgCI2 + 2H 2 0 = HgCI 2 . 2HgS + 4NaCI + 2H 2 S0 4

[2]

b) i) Ammoniumdiamminetetrakisthiocyanato(-N)chromate(lll) ii) Potassiumtrisoxalatochromate(lll) iii) Potassiumdinitrogenbisoxalatodixoygenferrate(lll)

[2]

a)

Me 9

Me 9

Me OH

HPh

PhCHO

[2] [2] H 2 - Ph

LiAIH

4

OH"

(A)

.CHPh hydride shift

[2 + 2 + 2 = 6] b) PhCHCI 2

t-BuOK

^ P h C C I

PhC»CPh

(D)

[2 + 2 + 2 = 6] FIITJCC. ICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi -110016.

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The stannous salts forms a grey ppt. on reaction with H 2 S while the stannic salt forms a yellow ppt. SnS2 with H2S. Now when yellow ammonium sulfide is added to SnS it does not dissolve while SnS2 dissolves on addition of yellow ammonium sulfide. [4] 9.

i)

NO 2

[3]

CH 2 CH 2 OH

ii) > H202/0H"

©\

1 K 2 Cr 2 0 7 /H + V

CH 2 COOH

cq) Phenyl acetic acid

10. A =

11.

X

CHO

B =

C = [2 + 2 + 2 = 6]

a) In SF6 sulfur is in sp d hybridised state and the molecule is of octahedral geometry. All bond angles are equal and no distortion due to any non-bonding electrons. Therefore the structure is symmetrical and SF6 in exceptionally stable and hence electron affinity is low. In SF5 which is actually a free radical has a strong tendency to attract an electron to form the sixth a bond to complete its maximum coordination no. Hence electron affinity of SF5 is high. [3] b)

12.

OH

[3]

2H 3 P0 4

215° C

->

~H20

Reducing property 2Mn0 4 ~ + 5H 2 0 2 + Reducing property 2Fe(CN) 6 3 " + H 2 0 2

H4P2O7

pyrophosphoric acid

-

*300°C -H20

>

2HP0 3

meta phosphoric acid

prolonged heating

[3]

->P2O5

in acid medium 6H+ = 2Mn2+ + 8H 2 0 + 50 2 in alkaline medium + 20H' = 2Fe(CN)64~ + 2H 2 0 + 0 2

[ 2 + 2 = 4]

13.

In this process the sulfide ore is first roasted into the oxide in air and then smellted with exclusion of air, when the oxide reacts with the unchanged sulfide producing the metal. 2CU2S + 3 0 2 = 2CU 2 0 + 2S0 2 2CU 2 0 + Cu2S = 6Cu + S0 2 [ 1 + 2 = 3]

14.

2CrCI3 + lONaOH + 3H 2 0 2 (A)

2Na 2 Cr0 4 + H 2 S0 4 (B)

-> 2Na 2 Cr0 4 + 6NaCI + 8H 2 0 (B)

-> Na 2 Cr 2 0 7 + Na 2 S0 4 + H 2 0 (C)


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4NaCI + Na 2 Cr 2 0 7 + 6H 2 S0 4

> 2CrO,CL + 6NaHS0 4 + 3H 2 0

(D)

(E)

Cr0 2 CI 2 + 2NaOH

2

> Na 2 Cr0 4 + 2NaCI | Pb(OAc)2 PbCr0 4 + 2CH3COONa (F)

4Na 2 Cr 2 0 7 + 2NH4CI

> (NH 4 ) 2 Cr 2 0 7 + 2NaCI (G)

N 2 + H 2 0 + Cr 2 0 3 (H) (I) 15.

[8]

a) i) 2CI0 2 + Sb0 2 " + 2 0 H - + 2H 2 0 > 2CI0 2 " + Sb(OH)e" ii) 258KOH + K4[Fe(CN)6] + 61Ce(N0 3 ) 4 > Fe(OH)3 + 61Ce(OH)3 + 6K 2 C0 3 + 250KN0 3 + 36H 2 0 [2 + 2 = 4] b) P 0 OH

B=

A=

C= [1+1+1=3]

16.

CH3

Et

A = CH3 - C = C H - C - C = C - c h 3 rile

Li/EtNH2

Lindlars

h 15CS\ ;

HISCSN^

c =c H/

\

x

Me

/ CH3

c =c

H/

/

N

H

(B)

Et A•

HOT KMN 4

°

>

\

C = O + CO2H - c - CO2H + CH3COOH Me

(D) Optically inactive

Et 03/Zn

->

^ C = O + CHO - C - COOH + CHGCOOH Me Optically inactive (C)

[5]


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STUM MATERIAL

IIT-JEE, 2002 PRACTICE FULL TEST -1 CHEMISTRY SOLUTION 1.

Clearly 'B' is Na 2 B 4 0 7 which gives the glassy bead of B 2 0 3 on strong heating and Na 2 B 4 0 7 absorbs water to form borax, Na 2 B 4 0 7 , 10H 2 0. Borax is also less soluble in cold water but highly soluble in hot water. So after cooling a precipitate of borax appears also the solution of 'B' is alkaline. [2] So, A = Ca 2 B 6 0n, B = Na 2 B 4 0 7 , C = NaB0 2 D = B 2 0 3 , E = Cr 2 (S0 4 ) 3 F = Na 2 [(0H) 2 B(0—0) 2 B(0H) 2 ].6H 2 0 [1 x6] So the reaction are Ca2B6011 + 2Na 2 C0 3 = 2CaC0 3 i +Na 2 B 4 0 7 +2NaB0 2 [1] (A) (B) (C) 'B' normally exists into hydrated from Na 2 B 4 0 7 .10H 2 0 (borax) Na 2 B 4 0 7

Strong

heating

> 2NaB0 2 +B ? 0, (C)

[1]

(D)

"E" is Cr 2 (S0 4 ) 3 which gives green colour both in oxidising flame and reducing flame C r 2 ( S 0 4 ) 3 — » Cr 2 0 3 Cr 2 0 3 + 2B 2 0 3 (D)

>

2Cr(B0 2 ) 3

(green in both flame)

2NaN0 2 + 6H 2 0 + 2H 2 0 = Na 2 [(0H) 2 B(0 - 0) 2 B(0H) 2 ].6H 2 0 (C) ' " (F)

[1] [1]

Sodiumpervoskite

2.

(a) Initial number of moles of two gases are 1 mole each Let number of moles of N0 2 effused out is 'a', number of moles of other gas effused out is 'b' moles and molecular weight of other gas is M. a + b = 1.5 [2] 1 According to the question a = — . where M2 is average molecular weight of gas left M, and M = 1.5 Mt where M^s average molecular weight of gas effused out a x 46 + b x M ( 1 - a ) x 4 6 + (1-b)M 46-46a + M - b M .-. MT MT = = and M2 = — = [2] 1.5 (1-a) + (1-b) 2 - ( a + b) _ 46 + M - (46a + bM)

,i2

2-1.5 _ 46 + M-1.5M, _ 46 + 1.5M1 -1.5M, _ 2-1.5 0.5 M2 = 92

[2]


RSM12-PT-l(M)-CH(S)-2

So average molecular weight of 2nd gas is 92.

[1] [2]

(b)Ammineaquanitropyridineplatinum(ll)nitrate (c) TeF5 - hybridisation sp3d2 and shape is square pyramidal 3.

[2]

(a) -•^^NVIe :NH—Me

i -ej^Hi >i

•—Me [3] o the product is

| Q

[1] (b) (i)

[1.5 x4]

(ii) C= CH 3 —CH—CH 2 OH 4.

D= CH 3 —ch 2 —CHO

NH2 (a) In pervoksite structure bivalent cations are present in primitive cubic lattice and tetravalent cation is present at body centre and anions are in face centre. If atoms are removed from one of the face diagonals then two Mg2+ ions from two corners and one 0 2 ~ ion from face centred will be removed. 1 3 Effective number of Mg2+ = — x 6 = — [1] 8 4 71 Effective number of O = — x 5 = 5/2 [1] 2 +4 Effective number of Ti =1 [1] 3 5 mass of unit cell = — x 24 + — x 16 + 1 x 48 [1] 4 4 Molar mass .-. density of unit cell = Molar volume 106 6.023 x 10 23 x (2 x 0.72 x10' 8 ) 3 _ 106 23 6.023 x10 x 2.986 x10~24 = 58.94 gm/cc

[2]


RSM12-PT-ll(M)-CH(S)-3

-> Zn2+ + H2

(b) Here eel! reaction is Zn + 2H"

0.059 , [Zn ].pH — r — log — [HI 2 „ 0.059. 0.1x1 Or, 0.701 = 0.76 log +— 2 [H ] + -1 .-. [H ] = 0.0316 mol lit Total H+ is consumed by NaOH [V= 1.5 lit] moles of HCI in 1.5 L solution = 0.0316 x 1.5 Let weight of NaOH required is = W gm pure NaOH = 0.7 W 0 7W

Ft c e l l -

t

c e

n

.'. z^Lzl. = 0.0316 x 1.5

[1]

[1] [1]

[1]

1 1 40 W = 2.7 gm After addition of NaOH to cathode solution [H+] become 10"7 since both acid and base are neutralised. Then e.m.f of cell 0.059 , 0.1 Ecsiu- E° — ' ° 9 [1] [HI 2 0.059 . 0.1 n__ = 0.76 - — log 14

2

"

10~

= 0.3765 V .-. The e.m.f of cell is decreased by (0.701 - 0.3765) = 0.3245 V For pure T 2 0 if pT = 7.62, then pOT = 7.62 pKw for T 2 0 = 7.62 + 7.62 = 15.24 Mili mole of TCI reacted = mili mole of NaOT reacted 15 x 0.25 = 3.75 mili mole of NaOT So excess NaOT = 3.75 - 10 x 0.2 = 1.75 milimoles 1 75 Concentration of "OT = —— = 0.07 M 25 pOT = -log (0.07) = 1.155 pT = 15.24- 1.155= 14.085 (a) ATf = 2 7 8 . 4 - 2 7 6 . 9 = 1.5 K For the acid 3HA C C(1-a)

(HA)3 0 Ca — 3"

[2]

[2]

[1]

a is degree of dissociation

a 2a^ Vant Hoff factor = 1 - a + — = 1 3 3 and equilibrium constant for trimerisation =

[1]

[2] Ca/3 {C(1-a)} 3

[1]

Now ATf = Kfm(1-2a/3) RT2 2a ^ -m 1 [l^ - latent heat of fusion in cal/gm] 1000L 8.31 x (278.4)2 m 10.042 x103 1000x 78 2a .... ) (i = 5m

:. 1.5 =

1 -

2a

[1]

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RSM12-PT-ll(M)-CH(S)-4 .

.

W, x 1 0 0 0

Molality oi riA in benzene = —3 M1i X W 22 U A

W x 1 0 0 0

=

M, x

W2

— - x

M,

A '1 x 1000 n1 x1000 _ \ n,1 + n 2 y n2M2 M, n., +n 2 0,03x1000 0.97x78 = 0.396 m Putting the value of'm' in equation (i)

[13

1.5 = 5 x 0.396 [ l - — a = 0,36

.-. Equilibrium constant for trimerisation =

Ca/3 {C(1-a)} 3

m considering molality = molarity]

0.396x0.36 3 x (0.396)3(1 - 0.36)3 = 2.92

[1]

(b) Cu2+ has electronic configuration 3d94s°. So it seems there is no completely 2 vacant '3d' orbital to take part in dsp hybridisation. But in presence of strong ligand (NH3 is moderately strong field ligand) the unpaired electron from 3d orbital is promoted to vacant 4p orbital and the vacant 3d orbital is available for hybridisation. So Cu2+ can undergo dsp hybridisation after promotion of unpaired electron. [3] 7. ,-COOH A = 0^2 CH2 B -= ^n C H33VC O O H C = CH3COCH3 D XO XDOH OH D = CH3—CH2—CH—CH2—CH =C—CH3 E = CH3—CH2—CH—CH2—CHO CH 3 F — CH3CH2CH—CH2CH2OH

CH 3 CH, G - CH 3 CH 2 C-CHCH 3 (through rearrangement of carbocation)

CH3 H = CH3CH2COCH3, I = CH3CHO

CH2 [1x9]

hc 34 s ,(a), Energy rU . r- = — ofx one photon E = 6.626 X 10" X 3 X 10 J , X 850 x 10 = 2.3386 x 10" 19 J

u ofr photons u * • . = .-. m Number required

[2]

Total energy — Energy of one photon

3.15 x10~14 2.3386 x10~19 = 1.35 x 105

[2]

(b) Here the process is FIITJCC, ICES HOUSE (Opp. VIJAY MANDAL ENCLAVE), SARVAPRIYA VIHAR, NEW DELHI -16. Ph:6854102, 6865182, FAX: 6513942

RSM12-PT-l(M)-CH(S)-5 HCN + I" (aq)

[1]

ICN(aq) + 2l"(aq) + H+(aq)

AS 4 0 6 > H 3 As0 3 (aq) (+3) (+5) H 3 As0 3 (aq)+l 3 (aq) > AsO 3 " + 31"

[1]

1.22 gm of As 4 0 6 = ±===- x 4 = 1.23x 10~2 396 = moles of l3 concentration of I: =

1 23 x i r r 2 10.42

x100Q = 1.1846 M

In 5.21 ml of I, solution moles of l3 =

1.1846x5.21 Tooo

[3]

= 6.172 x 10"3 = moles of HCN concentration of HCN in blood =

[1] 6.172x10'

x 1000

[1]

= 0.4114M 9.

(a)

®

CO; followed

MeCI anhydrous

AICl3

by accidification

(Maintaining proper condition for mono alkylation)

COOH

[3] (b) Here PhS acts mainly as Nu (i.e.,nucleophile). But some eliminated product is also produced here. So two products A and B are PhS—CH 2 —CH=CMe 2

and

CH2 = CH - C = CH2 Me

respectively. (c) A,B,C are ortho, rneta and para chlorotoluene 1e has maximum dipolemoment. So C is !

[2]

Q

In presence of NaNH2 and Liq. NH3 toluidine is produced through benzyne intermediate. Now A gives X,Y,Z ail products in NaNH2 and liq. NH3. So A is m - chlorotoluene. Here reaction are as follows. [2]


RSM12-PT-l(M)-CH(S)-6

NMH,

H

[3]

(A) NH, Path-11 Me NH3

[1]

Me NH,

NH3 only one benzyne

[13

[1]

As 'Y' is common product from (B) and (C), so Y is m - toluidine (C) is giving Z also i.e., Z is p - toluedine and (B) gives X other than Y. So X is o-toluidine. 10.

K , x W x 1000 40x0.0852 A_ AT — = here ATf = 1 7 9 - 1 6 7 = 12 f = ——— [13 MxW Mx2 = Molecular weight of A =142 0.375x22400 Molecular weight of (B) = = 30 [1] 280 Since (A) is alkyl halide, so (B) alkene. Because in presence of Na in dry ether, alkyl halide gives alkane (Wurtz reaction). Molecular weight of CnH2N+2 = 30 or, 12n + 2n + 2 = 30 [2] n=2 So (B) is C2H6 and A is CH3—X Now atomic wt. of X = (molecular wt. of CH 3 -X) — (Wt. of CH3) [13 = 1 4 2 - 1 5 = 127 X = I and (A) is CH 3 — I [23

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FIITJCC RANKERS STUDY MATERIAL

IIT-JEE, 2002 PRACTICE FULL TEST - II CHEMISTRY SOLUTION K.E. = hv — hv0 or v — VQ = KE2 = 2K.E.! KE,

V2 -

v0

=

- ;

V!

-

V

0

=

K.E.

KB

h ' ' " h Dividing these equations yields y2~vo _ K E 2 / h _ yi~yo KE1 / h .'. v 2 - v 0 = 2 v i - 2v 0

[2]

v 0 = 2v-i - v 2 = 2(2 x 1016) - (3.2 x 1016) = 8 x 10

2.

15

Hz

[4]

a) ii) o pyramidal

linear

Square pyramidal

b) i) 2Ca 3 (P0 4 ) 2 + 6Si0 2 + 10C — » 6CaSi0 3 + P 4 + 10CO ii) (NH 4 ) 2 S 2 0 8 + 2H 2 0 + MnS0 4 > Mn0 2 + 2H 2 S0 4 + (NH 4 ) 2 S0 4 iii) 8AI + 3NaOH + 3NaHS0 4 > 4AI 2 0 3 + 3H 2 0 + 3Na 2 S

[ 3 x 2 = 6]

[ 3 x 2 = 6]

Assume complete precipitation of the Pb2+ followed by solution of the equilibrium concentration to be determined. 50 m i x 0.2 = 10 m mol of Pb2+ 5 0 x 1.5 = 75 mmol of CP1 2+ 1 Pb + 2 C r > PbCI2 We have to find out the lead ion concentration which can exist in a solution of CP ion (75 - 20) = 55 mmol in 100 mL. [2] 2+ PbCI2 Pb + 2CP K s p =[Pb 2 + ] [Cn 2 = 1.7x 10"4



.. [Pb2+] = 1

1

- 7 x 1 0 " 4 = 5.6 x 10-4 M (0.55)

[2]

1 1

:. Amount of Pb2+ left in solution = 5.6 x 10"4 x 0.1 x 206 = 11.55 mg 2.303 l o g l o g — Ao

= - U

2.3 log 0.01 = - 4.6 • t =

=

4 6 0 t

^ = 6.64 t1/2 X 0.693 For 131 1 ; t = (6.64) x 8d = 53.1 days For 90Sr, t = (6.64 x 19.9y) = 132 years .\ 90 Sr is likely to be serious; the iodine will soon be gone

5.

[2]

[6]

i)

H 2 0 is a weak nucleophile and CF is not a good leaving group and so hydrolysis is slow. I" from Kl is a powerful nucleophile and it displaces CP to form CH3CH2I. Now r being a good leaving group H 2 0 can displace it rapidly. [3]

ii)

When ethyl alcohol gives iodoform it first gets converted to CH 3 CHO by the oxidising action of sodium hypoiodite (NaOl) which forms by the reaction of l 2 /NaOH 2NaOH + l2 = Nal + NaOl + H 2 0 x NH 4 OH + l 2 > NH4O! + NH4I + H 2 0 (Reaction does not take place) NH4OI does not exist. So C 2 H 5 OH does not get oxidised to CH 3 CHO (with l 2 /NH 4 OH) which can form CHI3. For the iodoform reaction acetone does not require any oxidising agent as it has the required keto methyl group. [3]

6.

RT2 — M [M = mol.wt. Tf = normal freezing pt]

Kf = A H

fUsion

_ 8.314 x(273) 2 x 18 f

1000 x 6 x 10 3

K f « 1.86 K.kg mol"1 Now ATf = Kf x m (m = molality) AT, (273-271) = 1 £ m = — L == = 1.07 moles/Kga Kf 1.86 But m =

moles of solutes weight of solvent (in Kg) /

n = moles of solute = 1.075 x

0.9x18A 1000 j

v

7.

[2]

= 1.74 x 10~2

[2]

Also XS0|Ute = ^ ^ — — - = 0.079 (where X = mole fraction) 760 1 74x10~ 2 Total moles (N) = = 0.22 0.079 Moles of solvent (H 2 0) = 0.2026 Mass of ice separted out = (0.9 - 0.2026) x 18= 12 gm

[3]

i)

[2]

CuSn + 2HCI

(ii) Cu + 4HN0 3

—» Cu+SnCI 2 +H 2 t (B)

(C)

- » CU(N0 3 ) 2 + 2NO s + 2H 2 0

[2]

Blue solution (D)

(iii) SnCI2 + 2HgCI2 Hg2CI2 + SnCI2

>SnCI4 + Hg2CI2 (silky white ppt.) > 2Hg + SnCI4

[2]

(grey)



(iv) Cu(N0 3 ) 2 + 2NH 4 OH

-> CU(OH)2 + 2NH4NO3 Blue ppt

Cu(N0 3 ) 2 + 4NH 4 OH

-> [ C U ( N H 3 ) 4 ( N 0 3 ) 2 ] + 4H 2 0

[2]

In each litre the number of molecules is given by PV 1x1 n = -— = - — : : — = 0.0406 mol RT 0.082x300 3.17g = 78.1 g/mol [2] 0.0406 mol The formula weight of HF is 19g/mol. The large apparent formula weight from the gas density means that the gas is appreciably associated even in the gas phase presumably by hydrogen bonding. The average duster of HF molecules is about 4. [1] i) CrP 3 /pyridine

Ph-,P=CH,

[3] ii) CH=CH2

[3] 10.

a)

.CHO

COOH hot KMn04

COOH

(m.f. = C10H16O) b) millimoles of H2 used up = millimoles of compound =

^COOH (A) Ievulic acid

CH 3 COCH 3 (B) [3 + 3 + 2]

8.40 22.4 10.02

80 mmolesof H2 used up _ 8.4/22.4

= 3 [2] mmolesof compound 10.02/80 The hydrocarbon therefore contains 3 double bonds or one double bond and one triple bond. The ozonolysis product shows only 3 carbons but molecular weight in the range of 80 - 85 indicates 6 C atoms. Evidently 2 moles of each ozonolysis product are obtained per mole of hydrocarbon. Since HCHO can only come from a terminal unsaturation and the dialdehyde from an inner segment of a chain, the hydrocarbon is 1,3, hexatriene. O 3 /H 2 O CH2 = C H CH = C H - C H = CH2 -> 2HCHO + 2CHO—CHO [2] 1,3, hexatriene



11.

2 Kp=-^2_=0.66 PN 2 O 4

PNO2 + P N 2 O 4 P

= 380 torr = 0.5 atm

2

0.5 - p

= 0.66

pN02 = 0.332 atm

ar

>d pN2o4 = 0.168 atm [3] Since each mol of N 2 0 4 which dissociates produces 2 moles of N0 2 , the percent dissociated is given by 0.5pNO2 _ 0.5(0.332) x 100 = 50% [3] )°2o4 0.5x0.332 + 0.168

12.

CH3(CH2)nCOOH + | (n+2) 0 2

» (n+2)C0 2 + (n+2)H 2 0

Volume of HCI used for neutralisation of NaHC0 3 in half portion of solution = 40 - 30 = 10 ml 10x2 5 Equivalents of NaHC0 3 in half of the solution = — — 1000 10x2 5 x 2 Equivalents of NaHC0 3 in total solution = = 0.05 [31 1 J 1000 Moles of NaHC0 3 formed = moles of Na 2 C0 3 formed = moles of C0 2 formed = 0.05 Moles of C0 2 formed = On solving n = 2

l 1 x ( n + 2)

60 +14n

= 0.05 [5]

13. Compound MCI4.2NH3 MCI4.3NH3 MCI4.4NH3 MCI4.6NH3

Exists as [M(NH3)2CI4] [M(NH3)3CI3] [M(NH3)4CI2] [M(NH3)6CI4]

Mode of ionisation [M(NH3)2CI4] [M(NH3)3(CI3f + c r [M(NH 3 ) 4 CI 2 r + 2CI[M(NH3)6]4+ + 4 c r

Number of particles 1 2 3 5

[3] The more the number of particles (here ions) more is the conductance. According to Werner's it is the atoms in the ionising sphere that ionises. The secondary valency i.e., the coordination number for a element is fixed, so to maintain the coordination number (here 6) different number of chlorine atoms are remaining in the inner sphere and the rest of the chlorine atoms present on the outer sphere ionises. The above modes of ionisation is consistent with the above fact that to maintain the coordinator number of 'M' always six ligands are attached with it and the rest is in the outer sphere that actually ionises. [3]

..

14.

PXA 3 XN A

M= -

n solving for M by substituting the data

,

„ 2.32 x (1.221 x10~7)3 x 6.023 x1023 ,_., M= i = 636 g mol 1 [2] 4 Subtracting the molar mass of anhydrous salt and dividing by the molar mass of water 636-423.5 n = = 12 [2] 1 1 18


RSM12-PT-ll(M)-CH(S)-179 15.

j»h COPh

(0 L

II

^ ^

^ ^ C - O M g B r

PhMgBr/CuCI

h H

'

Ph

A

(C19H20O) Two geometrical isomers are possible for this product. Na/liq.NH;

)

[ f ^ j ]

b

()

NBS

(CH3) 2 CuLi

?

l

Br

(C)

)

[2]

( f ^ j ]

l

Me

(D)

[1+1+1]


FIITJCC ALL INDIA TEST SERIES liT-JEE, 2002 FULL TEST - III

CHEMISTRY

SOLUTIONS 1.

CH3S A-

=

B = X

H/

c=

OH

/CH3 H

Br OH

O

/ \ CH3 — CH — CH — CH3

D=

CH3 — CH — CH — CH3 CH3

CI jp —

I

CH 3 - CH - CH - CH 3

CH 3 - C H = C - CH 3

CH3 — CH2 - C - CH3

F=

CH 3

[1 mark each]

CH3

2.

a) Civ A=

ci

,L

/CI

B= [2]

CI

Ph

Ph

[2]

CI

C= PH/ÔH

HO

Ph

[1] CH 2 Br

b) B=

CH 2 Br

D= [1 mark each]

FIITJCC Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Terminal), New Delhi - 16, Ph : 6515949 , 6865182, 6854102, Fax : 6513942

AITS2002-FT-lll-CH(S)-2

Conversion of (B) to (C) is Beckmann rearrangement. Since only one product is obtained, the carbonyl must be a symmetrical carbonyl. (D) on reduction with DIBALH gives (A). Hence (A) has to be an aldehyde. A = CH 2 0 B = H2C = NOH C = HCONH2 D = HC0 2 H E = NH3 [1 mark each]

a)

-CH2OH H*

-OH

H,0

PCI5

-CI

Mg/DE

'MgCI

CH2O/H3O*

-CH2OH [3] b)

CHO

PH3P = CH2

CH = CH2

-CH2CH2OH

HBO

PCC

CH2CHO

[3] c)

dil OH"

H2N - NH2 NaOEt

1 0 3 /Zn/H 2 0

o H [4; no part marking] H 5.

a) CI - CH2 - C * - CI. The proton attached with C* is very acidic due to the-l effect of both

ii the chlorine atoms. [2] b) Vinyl chloride cannot form the carbocation easily whereas allyl chloride can. [2] c) The carbonyl compound formed by the first mole of RMgX, again reacts with the carbonyl to give alcohol as the final product. [2] FIITJCC Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Terminal), New Delhi - 16, Ph : 6515949 , 6865182, 6854102, Fax : 6513942

a)

x

OH

cr

NaOH

•


^ o

Se/A

[3] b)

O

O

ii

ii

(-•OH O h®

s i

ii

CH3- C - C - O H ^ = ^ CH3- C - C - O H

1

OH O I C H 3 - C - C-?OX-H

CHA-'I-OH |H9

CH3-CH-OH |-HE

CH3CHO 7.

A = B = C = D = E = F = G =

FeCr 2 0 4 (or) F e 0 C r 2 0 3 Fe 2 0 3 Na 2 Cr0 4 Fe4[Fe(CN)6]3 Na 2 Cr 2 0 7 K 2 Cr 2 0 7 K 3 [Cr(C 2 0 4 ) 3 ]

[3] [1] [V2]

[%]

[V2] [V>]

[VA [V2]

The reactions involved are i) 4FeCr 2 0 4 + 8Na 2 C0 3 + 70 2 — 2 F e 2 0 3 + 8Na 2 Cr0 4 + 8C0 2 ii) 4FeCI3 + 3K4[Fe(CN)6] * Fe4[Fe(CN)6] + 12KCI iii) 2Na 2 Cr0 4 + H 2 S 0 4 — > Na 2 Cr 2 0 7 + Na 2 S0 4 + H 2 0 iv) Na 2 Cr 2 0 7 + 2KCI > K 2 Cr 2 0 7 + 2NaCI v) K 2 Cr 2 0 7 + 7H 2 C 2 0 4 > K 2 C 2 0 4 + Cr 2 (C 2 0 4 ) 3 + 6C0 2 + 7H 2 0 vi) Cr 2 (C 2 0 4 ) 3 + 3K 2 C 2 0 4 > 2K 3 [Cr(C 2 0 4 ) 3 ]

[1] [1]

[1]

[1]

[1] [1]

a) Manganese is active enough to react with the iron compounds of sulphur and oxygen.

Small quantities of the products of these reactions dissolve in the metal without disruption of the lattice. Large quantities of MnO (or) MnS would form a slag which could be skimmed off the molten metal. Any excess manganese acts as cathodic protection for the iron. [3] 7+ 2+ b) Mn + 5e > Mn 3+ » 4P5+ + 8e" Thus, Meq.ofKMnO.^ Meq.ofP 4 Q 6 =

100X5X1000

158

1 0 0 x 8 x 1 0 0 0

219.9

=3164.56

= 3638.02

[1]



Meq. of P 4 0 6 in excess = 473.46

[1]

wx8x1000

219.9 => w = 13.01g (excess of P 4 0 6 ) 9.

[1]

a) Mass of the balloon = 100 kg = 10 x 104g \ 4 , 4 22 20 Volume of the balloon = — 7tr = — x — x x100 3 3 7 6 3 = 4190 x 10 cm - 4190 x 103 L PVM Mass of the gas in the balloon = RT 1x4190x10x4 = 68.13 x 10 4 g 0.082x300 Total mass of gas and balloon = (68.13 x 104) + 10 x 104 = 78.13 x 104g f • ^ . ^ 1.2x4190x10 s _noQ ,_ 4 Mass of air displaced = — = 502.8 x 10 g 3

Pay load = Mass of air displaced - ( mass of balloon + mass of gas). = 502.8 x 10 4 -78.13 x 104 = 424.67 x 104g PV 1x1 b) Moles of H2 present in one litre = — = — — = 0.0409 RT 0.0821x298 Energy needed to break H - H bonds in 0.0409 moles of H2 = 0.0409 x 436 = 17.83 kJ Also energy needed to excite one H atom from 1 to 2 energy level = 13.6

[1]

[1]

[1]

[1] eV

= 10.2 eV = 10.2 x 1.6 x 10"19J Energy needed to excite 0.0409 x 2 x 6.02 x 1023 atoms of H = 10.2 x 1.6 x 10~19 x 0.0409 x 2 x 6.02 x 1023 = 80.36 kJ Total energy needed = 17.83 + 80.36 KJ = 98.19 kJ Energy required to break H - H bonds = E = hv 436 x10 3 = 6.625 x 10~34 x v 6.02 x1023 v = 10.93 x 1014 Hz

x

6.02 x1023

j

c) No. of a-particles (or) He formed = 2.24 x 10 min No. of He particles formed in 420 days = 2.24x 1013 x 420 x 1440 = 1.355 x 10119 Also at 27°C and 750 mm; He = 0.5 mL Using PV = nRT 750 0.5 . „„„„ „„„ x = n x 0.0821 x 300 760 1000 =>n = 2.0x 10~5 moles 2.0 x 10"5 moles of He = 1.355 x 1019 particles of He 1.355 x 1019 1 mole of He = = 6.775 x 1023 particles 5

[13

[1] [13

[1]

2 . 0 x 10"

d)

. Avogadro's number = 6.775 x 1023 particles/mol O W 105° XH

[13

^H2O = V ^ 0 H + ^ 0 H + V c o s ( 1 0 5 ) FIITJCCLtd.,ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Terminal), New Delhi - 16, Ph : 6515949 , 6865182, 6854102, Fax : 6513942


Since H 2 0 has two vectors of O - H bonds acting at 105°, let dipole moment of O - H bond be 'a'. 1.85 = V2a2(1 + cos105) (or)a, i.e. HOH= 1.52D [1] = 1.52 X 1CT18 esu cm Now, JJ.OH = 8 x d where 8 is charge on either end 1.52 X 10"18 = 8 x 0.94 x 10-8 8 = 1.617 x 10~10 esu [1] Since O acquires 28 charges, one 8 charge from each bond and thus charge on O atom = 2 8 = 2 x 1 . 6 1 7 x 1 0 " 1 0 = 3,23 x 11T10 esu cm [1] e) — = 1.75 K1

[1]

Ti= 298 K T2 = 308 K 2.303 log

K 2 _ Ea T2 - T, K,

R

T,X2

= 2.303 log 1.75 = - ^ - x — M 1.987 308x298 Ea - 10=207 kca! mol -1 10.

[1] [1]

Cu2+ + 4NH3 [Cu(NH3)4]2+ 2+ Y - [CU(NH3 )4] f [Cu2+][NH3]4 The blue colour will be noticed if [Cu(NH3)4]2+ is equal to 1 x 10~5. At this stage [Cu ] = L

' 1.1x10

r = 9.1x10

15

M

x (0.1)

Also, o deposited ^ ,= Cu

Eit 63.5x3.512x1368 = 1.581 . g 96500 = 2x96500 1.581x1000 = 9 9 6 x 1Q _ 2m

L

[2]

M1 [1]

J

63.5x250 [Cu ] present initially = 0.1 (or) 10 x 1CT2 M [Cu2+]left = 10 x 10~2 - 9.96 x 10~2 = 4 x 10^ M [2] 2+ Thus, solution will show blue colour as it will provide appreciable Cu to form complex. 2+

11.

At boiling point Pmix = 736 mm Thus at boiling point, P^20 = 526 mm P; = 210 mm Also, Pîx = Pmix x mole fraction in vapour phase Let 'a' g of liquid and water is collected or this is the amount of vapours at equilibrium 2 5x a Thus, Mass of liquid vapours = " ' " ^ 3.5 a Mass of water vapours = — 3.5

|1]



Now for liquid 210 = 736

x

2.5 x a 3.5 x m 2.5a 3.5x18 3.5x m

...(1)

[1]

...(2)

[1]

Now for H 2 0 3.5x18 2.5a 3.5x18 3.5 x m From (1) and (2) 210 _ 18x2.5

526 = 736 x -

• +

•

526 ~ m => m = 112.7 12.

[1]

a) N 3H2 1 3 (1 - x) (3 - 3x)

2NH3 0 2x

Given mole fraction of NH3 = 0.012 at P = 10 atm 4— 2x = 0 . 0 1 2 => x = 0.0237

[2] 2x

Kp =

NH,

X

Pu

xP 4 -2x 1-X 3-3x xP xP 4-2x 4-2x

4x 2 (4-2x) 2 5 ,-2 3 2 [x = 0.0237; P = 10] = 1.431 x 10~ atm (1-x)(3-3x) P Let mole % of NH3 in equilibrium mixture be increased to 10.4 at pressure P. 2x 10.4 " 4 - 2x ~ 100 => x = 0.1884 4x 2 (4-2x) 2 Now, Kp = => P = 105.41 atm (1-x)(3-3x)3P2

[2]

[1]

b) Suppose 'V' mL of solution contains 0.1 M Mg2+ and 0.8 M NhtCI. Now V mL of 'a' molarity NH3 is added which just gives a precipitate of Mg(OH)a then [Mg2+] [OH-]2 = Ksp Mg(OH)2 0.1V 2V

... millimoles [OH1 2 = 1.4 x 10,-11 [Mg2+1] = Vol. in mL

[OH-] = 1.67 x 10"5 M [2] Now, if [OHT = 1.67 x 10~5, an addition of NH3 in NH4CI, then Mg(OH)2 will precipitate. Now, for a buffer solution of NH3and NH4CI [NH4CI] - log [OH~] = - log Kb + log [NH3] 0.8x V / 2 log 1.67 x 1 0 = - log 1.8 x 10"b + log a x V / 2V > a = 0.7421 M [2] . .. 0.7421 x V r M U r [NH3] in solution = — = 0.3710 M [1] 2V FIITJCC Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Terminal), New Delhi - 16, Ph : 6515949 , 6865182, 6854102, Fax : 6513942


13.

3

a) For complex [Mn(CN)6] ", the number of unpaired electrons is calculated as, 2.8= Vn(n + 2) => n = 2 [1] [Mn(CN)6]3_ has two unpaired electrons. Hence the geometry is octahedral with d2sp3 hybridisation. [1] 2_ For complex [MnBr4] , the number of unpaired electrons is calculated as 5.9 = Vn(n + 2) => n = 5 [1] [MnBr4]2~, has 5 unpaired electrons. Hence the geometry is tetrahedral with sp3 hybridisation. [1] b) HN0 2 + 2H 2 S0 3 + H 2 0 A = HO - N = O C=

H-N-OH 1

H

> NH2OH + 2H 2 S0 4 B= HO-S >0 I OH D= yrO

[1]

HO - S. 0 H

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[V.2 mark each]

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FIITJCC ALL INDIA TEST SERIES IIT-JEE, 2002 FULL TEST - IV

CHEMISTRY

SOLUTIONS 1.

The energy of the incident photon is ._ he 4.14x10 - 1 5 x3x10 8 „„ w Ei = — = =3.1 eV X 400x10 The maximum kinetic energy of the emitted electrons is Emax = ET - W = 3.1 - 1.9 eV = 1.2 eV It is given that emitted electrons of maximum energy + 2He2+ > He+

in 4th excited state

[2] + Photon

The 4th excited state implies that the electron enters in the n = 5 electronic state. In this state the energy is uz 5 The energy of the emitted photon in the above combination reaction is E = Emax + (-E 5 ) = 1.2 + 2.1 = 3.3 eV [3] After recombination reaction, the electron may undergo transitions from a higher level to a lower level thereby emitting photons. The energies in the lower electronic levels of He+ are _ -13.6-22 .... E4 = ——22 = - 3.4eV 4 -13.6 • 22 E 3 = ~ 1 3 2t 2 2 = - 6.04 eV 3 -13.6 • 2 2 I " = - 13.6 eV 2 The possible transitions are n=5 >n = 4 AE = E4 - E5 = - 3 . 4 - ( - 2 . 1 ) = -1.3 eV n=5 >n =3 AE = E 3 - E5 = -6.04-(-2.1) = -3.94 eV n =4 »n=3 AE = E 3 - E 4 = -6.04 - (-3.4)= -2.64 eV Hence the photons that are likely to be emitted in the range of 2eV to 4 eV are 3.3 eV, 3.94 eV and 2.64 eV. [3] E2=

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Vihar (Near Hauz Khas Bus Terminal), New Delhi - 16, Ph : 6515949

, 6865182,

6854102,

Fax : 6513942

AITS2002-FT-IV-CH(S)-2

2.

a)

l / C 0

E t

CH3

b)

N02

3.

2

j

CH3

N02

NHCOCH3

[3]

Let solubility of X2Y3 be S moles/litre. X2Y3 2X3+ + 3Y"2 2S

3S

Ksp = (2S) (3S) = 108S P°-P n 5S Now N " 55.56 5S 31.9-31.8 31.8 55.56 S = 0.035 moles/litre Ksp (at40°C) = 5 . 6 7 * 10^ AH (T'2 '1 Now log spV ' K sp (30) 2.303R 2 J 6 r 5.67 x10~ AH 10 log 5 3.5 x10" 2.303x8.314 303x313 4.

a)

[3]

AH = 401.13 KJ/mole R'\ CH - C0 2 H

[4] =A

O

R _ I R—r.

O ^ Wf\ S -CH - C - O - H

\ chn CsHsN

R'

1

> R - C, = CH - O "

C o R'v C- C-H

R'

' a

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b) i)

The starting material in the first case will be CH 3 COCH 3 & CH 3 C0 2 Et. When we add base the mechanism that will follow is

CH2 — c — CH3

-> CH3 — C—CH2— C — CH3

&

o

A diketone

The carbanion will not be formed at the ester because of its low acidity and hence a, (3 unsaturated ester will not be obtained. [3] ii) The starting materials are acetone and benzaldehyde

&

OH"

CH3 — C — CH3

CH3COCH = CH - Ph <-

- > C H 3 - C - CH

A

C H 3 - C - CH 2 CH - Ph

A

0H7PhCH0/A

L

o II

[2]

PhCH = CH - C - CH = CH ~ Ph This reaction is feasible. a) The equilibrium constant is given by AG0 -118.08 x10 3 Jmol" 1 InK = = =-6.175 RT 8.3145x2300 \ K = 2.08 x 10,-3 H20 Initial

5=

+

H2

n

0

(1 - a ) n

an

[2]

1 -02 2 0 1 . — an

Equilibrium

Partial pressure .

K

_

PH 2

' Po? _

PH 2 O a

1+

^

—a 2 a

1 r, - a P 2

aP

(1 - a ) P

a

1+ — 2

a

1+ — 2

3/2p1/2

,1/2

(1-a)(2 + a)

3/2.p1/2

K=

j = — [neglecting in comparison] V2 v P = 1 atm .".a = (V2k)T 3 = 0.0205

i.e. 2 percent of water has decomposed

[4]

b) PV = ZnRT PV or n = — — ZRT FIITJCC Ltd., ICES House, Sarvapriya


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w and n = — gm moles m PVM .-.Weight of 0 2 (w) = ZRT

100x5x32

= 693.7g

0.96x0.082x293

. . . . P,V, 10x1000 a) The gas taken is n = ' 1 RT 0.082x273

[5]

446 moles

7 = - = 1.66 3 fn A

• V, = (1 O f 5 x 1000 = 3981 lit

Final volume Vf =

[2]

vP f y i i— - p— f v f- = 10x1000-1x3981 = 9119 litre atm y-1 1.66-1 b) 1 mole HCHO, 1 mol C0 2 , 2 moles of HCOOH Adiabatic reversible work. =

7.

p v

0/P\X0 / ON O=P O -f P=O

/ 0 = p\

o

o

[2]

V0H

a)

\

[3]

o

N -P = O / O

V

o

"OH

o

not isolated H20

o

o

II

II

HO-P-O-P-OH 0 O 1 I OH-P-O-P-OH ft fl Tetrametaphosphoric acid H20

O

O

O

2 HO-P-O-P-OH (

Hz

°

O

O

OH-P - O - P - O - P - O - P-OH

OH

OH

O

OH

Pyrophosphoric acid

OH

OH

OH

Tetrapolyphosphoric acid

H20

O II

Overall reaction is P4O10 + 6 HzO -—> 4 H3P04

4 HO - P - OH OH

[1x4]

Phosphoric acid

b) XeF 2 + AsF5

[1]

> [XeFf [AsFe]" complex

8.

a) A = Al D = NH3

FIITJCC Ltd.. ICES House, Sarvapriya

B = AIN E = AlCIs

C = AI(OH)3 D = KAI0 2


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b) The compound that is used is Lead tetraethyi Pb(C2H5)4 and is used as an 'antiknock' additive to increase the octane number of petrol. [2] It can be prepared by reacting vapours of ethyl chloride with sodium lead alloy in presence of a catalyst. 4C2H5CI + 4Na + Pb = Pb(C2H5)4 + 4NaCI [2] 9.

a) M = Na 2 S0 4 R = Sulfur

N = Na2S S = S0 2

Na 2 S0 4 + 4C — N a Na2S + 2HCI H2S + Ci2

2

P = H2S [5]

S + 4COt

» H2S + 2NaCI >2HCI+

Si

Yellow Ppt.

S + 02 > so2 S0 2 + K 2 Cr 2 0 7 + H 2 S0 4 = K 2 S0 4 + Cr 2 (S0 4 ) 3 + H 2 0

[3]

Green

b)

The other option of migration will lead to

10.

But this product will experience more steric crowding and hence less stable. So the above compound (A) will be the major product. [4]

OMe

a) OMe

MeO

OMe

MeO (CH3)2SO4 KOH

CH2 — CH — CH2

CH2 — CH = CH2

(B)

(A)

CH3O

OMe + HCHO Radioactive compound

CH2CHO Non radioactive compound

j-^j

b) The redox potential of KMn04 depends on the pH of the solution. In acid solution the standard reduction potential i.e. E° = 1.51 V FIITJCC Ltd., ICES House, Sarvapriya

Vihar (Near Hauz Khas Bus Terminal), New Delhi - 16, Ph : 6515949,

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The reaction is MnO'4 + 8H+ + 5e

Mn2+ + 4H 2 0 [Mn ] 0.059, log - + 8 [H ] -[MnO^]

. P • • teen - t Ce||

= 1.51 - 0 . 0 9 3 p H

+

^

l 5

o

g ^ [Mn ]

[2]

At pH = 1, Eceii = 1.417 which is above the standard reduction potential of Cl 2 /Cr, Br2/Br" and l 2 /l" system and hence oxidizes the halides to respective halogen. But at pH = 6, Eceii - 0.952 which is above the l2/l only and so it can only oxidize I to l2i but not bromide and chloride. [3] 11.

I he mechanism is aromatic SN2 type where Br will be displaced by the lone pair on nitrogen. Now the —N02 group at the para position due to its - R effect enhances the 5+ on carbon bearing bromine and hence it is easier for the nucleophile to attack that carbon. But presence of electron pushing (R) i.e. alkyl groups reduces the electron deficient character of that carbon resulting in a lower reaction rate. [4]

12.

i)

2KI0 3 + 5S0 2 + 4H 2 0 = l 2 + 3H 2 S0 4 + 2KHS0 4 l 2 + S0 2 + 2H 2 0 = 2HI + H 2 S0 4 ii) (CH3COO)2Pb + Ca(OCI)CI + H , 0 =- Pb0 2 + CaCI2 + 2CH3COOH Ph

13.

\

Q

_

Q

C =N

NH2OH

Ph/

Ph'

[2 + 2]

\

(A)

0 H

(B) PCI5

Ph - C - N H — P h (C) H 3 0*

PhCOCI <

soc 2

'

PhCOOH + NH2Ph (D)

NH2Ph

o II

Ph - C - N H — P h (C)

Phs C =O Ph' (A)

Mg/H"1"

Ph

OH

I , 'Ph

\l

C - C

'Ph (F)

ONa Phh C

Ph'

FIITJCC

Phx jjjo^

C=O

2

Ph/

(A) Na

Ph,

(E)

N

Ph/

Zn/OIT

CH-OH

OH

Ph/

ONa Ph

/

C

\

p h

[6x1]

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14.

Composition of electrolyte used in the electrolytic reduction is i) Pure alumina 20% ii) Cryolite (AIF3, 3NaF 60%) iii) Fluorspar (CaF2, 20%) Nature of electrodes - Electrolysis is carried out in the iron tank lined inside with gas carbon which serves as cathode. The anode consists of a series of graphite rods suspended vertically which dip into the molten electrolyte. [2]



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FIITJCC ALL INDIA TEST SERIES IIT-JEE, 2002 FULL TEST - V

CHEMISTRY

SOLUTIONS 1.

i)

The solution is a buffer and so salt pH = pKa + log —— acid = 4.74 + 0.176 pH = 4.92 Now if we add 0.02 moles of HCI it will combine with the salt CH3COONa + HCI > CH3COOH + NaCI So CH3COONa left over is (0.03 - 0.02) = 0.01

[2]

and CH3COOH formed = 0.02

Total C H 3 C O O H = CH3COOH (formed) + CH3COOH (left over) = 0.04 So the medium now contains 0.01 mole CH3COONa and 0.04 moles CH3COOH. So it is again buffer. pH — pKa + log 9

0.04

= 4.74 - 0.6

pH = 4.14 [3] ii) CaC2 + 2H 2 0 = Ca(OH)2 + C2H2 AI4C3 + 12H 2 0 =4AI(OH)3 + 3CH4 Mg2C3 + 4H 2 0 = 2Mg(OH)2 + CH3C = CH The difference in reaction is due to the difference is the carbanion that is present in the above carbides. [ 3 x 2 = 6] 2.

3.

i)

A = ZnS0 4 .6H 2 0 B = ZnS0 4 , H 2 0 ii) F = NH3 G = NaP0 3

i)

C = ZnS0 4 D = ZnO, E = S0 2 [7x1-7]

5Sn+ 2OHNO3 = H 2 Sn 5 0 11 .4H 2 0 + 20N0 2 + 5H 2 0 Metastannic acid

5Sn0 2 + 5H 2 0 ii) FIITJCC

NaBr0 3 + XeF2 + H 2 0 = NaBr0 4 + 2HF + Xe

[2]

Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Terminal), New Delhi - 16, Ph : 6515949 , 6865182, 6854102, Fax : 6513942


Cu is very slowly oxidised on the surface in moist air giving a green coating, which is basic copper carborate CuC0 3 .Cu(0H) 2 . 2Cu + H 2 0 + C 0 2 + 0 2

> CUC03.CU(0H)2

[3]

basic copper carbonate

Reaction at cathode Reaction at anode

Hg2CI2 + 2 e " — > 2Hg + 2CP 2Ag + 2CI" » 2AgCI + 2e"

Hg2CI2(S) + 2Ag(s) Now Eceii = 0.0455 0.0455 = 0.334 - E°AgCI rwA iri. / Ag / CI ••• E°aAgCI n / /AAgin/ CI = 0.2885 V AgCI Ag++ CP + Ag + e > Ag AgCI + e" > Ag + CP /.AG 0 ! = AG° 3 -AG° 2 = -27840.5 + 77103.5 = 49263 .-.Ksp = 2.32 x 1CT9

> 2AgCI(s) + 2Hg(l)

[2]

AG0! = - 2.303 RT log K•sp AG2° = - 96500 x 0.799 = - 77103.5 AG°3 = - 96500 x 0.2885 = - 27840.25 J

[2]

Ksp(AgCI) = [Ag + ] [CP]

••• [Ag+] =

2.32x10 "

= 2.32 x 1Q-

10' 1

[2]

a) Kinetic energy of electron = hv - hvcrit he he ^crit = he

1 X

X,crit

6.6 x 10~34 x 3 x 108 10" =1.65 x 10 J

1

1

400

600

= 1.03 e V

[5]

b) V ^ ! = V2N2 Vi x 10 -3 = 30 x 0.001 x 4 [Pyrophosphoric is a tetrabasic acid so during reaction with NaOH n-factor = 4] Vt =

= 120 cc

10"3

Volume of NaOH = 120 cc 2.303 log ^

v

1o

or 2.303 log

-

p

[3]

V-T, .

t , T

4.5x10 1.5x10

[1]

2

Ea

7

373 - 323 8.314 373 x 323

E a = 2.2 x 104 J moP1 _ AQ-Ea/RT Now K = Ae

[2]

2.2x10"

. 4.5 x 107= Ae • A = 5.42 x 1010

8 3x373


[3]


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AITS2002-FT-V-CH(S)-3

8.

a) w-, = 96.4g w 2 = 3.6g 0.002 x (392.25)*

=

1000 lf

[2]

9

- 6 x 1 0 0 ° = 255.36 96.4x5 . . . 255.36 A4 Atomicity m 08 32 m 2 =34.19*

3

[2]

[2]

Formula S 8

b)

21 2 At equilibrium 1 - a 2a Total moles 1 + a Since pressure is proportional to no. of moles of gas P . — = 1+ a Where P = observed pressure P 0 = calculated pressure (having no. dissocation) P-Pn a = o 2a 1-a Now P p - = = - P & P,2 = — P 1+a 1+ a P-Pq ••• KP =

4a2 1-a2

Po J

V

P =

\2

p =

4X0.1151xQ112

=5

_82xl0-2atm

1-0.1151

1

v When a = 0.9 we have

rP

o

[3]

y

Kp(1-a2) P =- "pv 2 = 5.82x10 -2 (1-0.81) _= 3.41 x 10 _J atm 4x0.81 4a

[3]

9. BrCH 2 CH 2 CH 2 - C - H

(CH 2 OH )2 /HA

CH v^n22 -• ) B r C H 2 CH 22 2v^n

CH

(A)

/°1 \o-

Mg/ether

/

o —

MgBr(CH 2 ) 3 CH (B)

Ô

—

CH3CHO/H2O

CH3 OH < (D) Acetal

OCH 3

CH3 H

°

CH3 - CH - (CH 2 ) 3 CHO (C)

[6; no part marking]



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AITS2002-FT-V-CH(S)-4 ii)

? ch coch

CH = CLi

-

H

3 > CH 3 - C - C - C H "

^

(F)

)H o

O II CH2 = C - C - C H = CH CH3 (F)

10-

i)

OHCH2CH2 - CH2Br

PhCHO/OhT

- CH3 — C — C — CH3

Ph

CH3 (G)

[ 1 x 4 = 4]

CH

2=C(CH3)2 > (CH3)3CO - CH 2 -CH 2 CH 2 Br H+

CH = CLi 1r

HOCH 2 CH 2 CH 2 C = CH < ii)

CH3

(CH3)3C - O - CH 2 CH 2 CH 2 - C = CH

CH3

[3]

CH3 CH3COCI

Sn/HCI

NO2

NHCOCH3 KMnc>4

COOH H,0+

NHCOCH3 11.

[3]

The optically active compound is 4 methyl 2 phenyl-2hexene CH3 CH3 = CH - CH - CH2 - CH3

[2]

It has one asymmetric centre and hence optically active and gives acetophenone as one of the products. Geometrical isomers P H

\

/CH3

H-

/CH S

^CH-CH3 C2H5 (A)

F!ITJ£€ Ltd., ICES House, Sarvapriya

CH3-CH'

H

C2H5 (B) Vihar (Near Hauz Khas Bus Terminal), New Delhi - 16, Ph : 6515949

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(CD 2 OH) 3 C - CDH

12.

A =

o

-CO2H

N

B =

H (C)

NH D =

O 0 II

18

c—o

o = c/ \ o-

E =

CH2

/

CH2

18

13.

\

[ 5 x 2 = 10]

Since B reacts with HI0 4 to give benzaldehyde so it should contain one benzene ring and two adjacent - O H groups. Moreover since (A) gives iodoform test so it must contain - C H - CH3 groups. The structures of compound and reactions are 'OH Br CH - CH - CH 3 OH Iodoform ii) KOH ii) H +

OH-CH-CH-CHs OH

HIO,

[5]

14.

a) 2PbS + 30 2 = 2PbO + 2S0 2 2PbO + PbS » 3Pb + S 0 2

[2]

b) [3Ca 3 (P0 4 ) 2 CaF 2 ] + 14H 3 P0 4 -

-> 10Ca(H 2 P0 4 ) 2 + 2 H F

[2]

triple superphosphate

15.

- N 0 2 group is normally identified by reducing it to - N H 2 group and then converting to diazo group and then coupling with alkaline naphthol. Now here this is not possible as - N H 2 is already present. So the nitro group is reduced by Zn / NH4CI solution where it gets converted to phenyl hydroxylamine and this product when treated with Tollen's reagent gives a ppt, of metallic silver. - N H 2 can be identified through diazo test. NHOH Zn /NH4CI

[Ag(W3)2r

Ag4 [4]



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FIITJCC ALL INDIA TEST SERIES IIT-JEE, 2002

OPEN TEST (Mains) CHEMISTRY

SOLUTION 1.

One form does not undergo reaction with AgN0 3 indicating absence of free CI" i.e. CP in the outer sphere and with ethylene diamine indicating that the CP and N0 2 " are anti to each other. Had they been cis, they should have been replaced by ethylene diamine which is a bidentate ligand. Second form reacts with AgN0 3 but not with ethylene diamine which indicates availability of free CP. The third compound reacts with AgN0 3 and ethylene diamine indicating CP in the outer sphere and presence of N02~ in the cis-position. [3] A = [Co(en) 2 (N0 2 ) CI]N0 2 B = [Co(en) 2 (N0 2 )]CI C = [Co(en) 2 (N0 2 ) 2 ]CI

I

r

o2

Co

I

02

en

N0 2

Co

en

CI

N0 2

Trans-chloro-bis(ethylene diammine)nitrocobalt(lll)nitrite

7rans-bis(ethylene diammine)dinitrocobalt(lll) chloride

CI

02 NO:

Co

CI en

B cis -bis(ethylenediammine)dinitro coablt(lll) chloride

[33 We have — = J — d, or

100 51.5 d = 60.3

[2]



PCI3

PCI5

+

Cl2

Initially 1 0 0 At eqm 1- x x x Total moles at equilibrium = 1 + x d

1

Now — = — [D = vapour density of undissociated PCI5 = D 1+x M

x=

Molecular weight

208.5

[d = vapour density at equilibrium]

104.25-60.3

= 0.729

60.3 Thus PCI5 is 72.9% dissociated

[4]

a) The strength of a base depends on its ability to donate an electron pair. The more easily it can be donatable the stronger is the basicity. The nitrogen lone pair in the compound cannot get delocalised into the benzene ring as the lone pair does not remain parallel to the Tt-electrons in benzene. This occurs because due to bulky groups on nitrogen as well as in the ortho positions of the ring there occurs appreciable steric crowding. To avoid the steric crowding C - N bond with its lone pair rotates and becomes perpendicular and thereby does not undergo delocalisation and hence becomes a strong base. [3] b) X=

[2] a) Hypochlorous acid may be differentiated from chlorine by shaking the substances with mercury. Hypochlorous acid gives a light brown precipitate of basic mercuric chloride soluble in HCI and chlorine gives white mercurous chloride insoluble in HCI. [1] 2Hg + Cl2 =Hg2CI2 2Hg + 2HOCI = HgCI2-HgO + H 2 0 [2] b) An aqueous solution of Na 2 C0 3 will hydrolyse to furnish NaOH and thereby making the medium alkaline. Now if we try to perform the brown ring test then there will be a problem because the FeS0 4 added to perform the brown ring test will react with NaOH resulting in the precipitation of Fe(OH)3. So brown ring test cannot be performed to identify N03~ radical in the mixture. To identify the N03~ radical the mixture is treated with sulphanilic acid and a-napthylamine and a red colour of the solution indicates the presence of N03~ radical in the mixture. [4] a) 1 , 2 dibromoethane spends most of its time in the anti-staggered confirmation whereas 1, 2 ethanediol will exist mostly in the gauche form due to intramolecular hydrogen bonding.

Br In the antiform the dipoles cancel each other thereby resulting in a low dipole moment.

H In the gauche form the dipoles do not cancel each other and hence net dipole moment is high.

[3]



b) P4O10 is a more powerful dehydrating agent than Cl 2 0 7 since the latter is the product of dehydration of HCI0 4 by P4O10. The reverse reaction cannot be expected to proceed so well. [3] Flask A:

Flask B:

Cone, of OH" = 1CT3 mole/100 m Cone, of OH" / L = 10~2 moles/Lt. pOH = - log [OH"] = - log [10"2] pOH = 2 pH = 1 4 - 2 pH = 12 Change in pH = 5 Before addition of NaOH [Salt] = 0.1 [Acid] = 0.1 [Salt] pH = pKa + log [Acid] pH = 7 +log

OA

0.1 pH = 7 After addition of NaOH [Acid] = 0.09 [Salt] = 0.11 [Salt] p.H = pKa + log [Acid] 0.11 pH = 7 + log-—— = 7 + 0.08715 0.09 pH = 7.08715 Change in pH = 0.08715 7.

[2]

[3]

02N

CHS



CHS

rr

h

O CH3 8.

[2]

(j)"Mg+EBr

i)

CH3 -
(1,4 addition)

r

H)

CH3 - CH = CH - y - CH3 Et a/2

(1, 2 addition)

[3]

a A/3 = 4rh rh + rf =

a

+

v2y

V2

V

Dividing by r h 1+rf

_ a x1 _ a ^ 4 rh V2 rh V2 aV3

,

r

4V6

h

6

f 1 + — = r

, 1

a/2 l

A

TP

4V6

= ^ - 6 = 0 . 6 3 3

aV3 = 4rh a

2

/

a

\

i uJ uj

rh + rf = J

—

+

—

aV5

Solving 1 = 0.29

10.

[3]

[4]

a) Solubility of any given compound depends on its lattice and hydration energy. U oc

1 r, + r



AH

oc -

1

r+

+

1 r_

-

where U & AH are lattice and hydration energies and r+ & r_ the radius of the cation and anion respectively. 1 In case of Ag halides Ag+ ion is a large ion. So — is quite low. Now when r_ changes 1 1 from fluorine to iodine, the sum of —+ — changes significantly resulting in a larger drop r+ r_ of hydration energy as compared to lattice energy and hence solubility decreases with lithium halides Li+ being small in size, the case is just the opposite. [3]

b)

H3

I

f

CH2 - CH - CI t EtS?

H3

N GP

y?

V'

- > CH2 - CH

>

^

|

f

H3

?

H2 - CH - SEt

OCH3

(A)

(major product)

CH3 H

°

>(pH2-(pH2

OCH3 (B) [3] In the first case sulfur of the neighbouring group is stabilising the carbocation which is not possible with oxygen because of higher electronegativity. So, CH3OH attacks the carbon via ordinary S N 2 attacks and product (B) is obtained. But in the first case the CH3OH attacks that carbon in the bridged sulfonium ion which is less crowded and therefore product (A) is obtained which is bound to be much slower than the internal nucleophilic attack which is better known as neighbouring group participation. 11.

OEt

At T = 298 K E = + 0.0713 V, so AG = - nFE = - (1) x 95.485 kC mol"1 x (0.07131V) = - 6.809 kJ mol - 1

[1]

The temperature coefficient of the cell potential is HP — = - 4.99 x 10^ VK~1 - 2(3.45 x 10"6) (T/K - 298)VK"1 dT At 298 K this expression evaluates to dT

= - 4.99

x

10"4 VK_1

'dE x AS — vdT,Vp nF 4 1 AS = 1 X (9.64 X 10 C mor ) x (-A.99 x 10^ VIC1) = - 48.2 JK"1 mol -1 AH = AG + TAS = - 6.809 kJ mor 1 + 298 x (-0.0482 kJ K" 1 mor 1 ) = - 21.17 kJ mol -1 The reaction entropy is calculated as


[3] [2]

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12.

CH3\

/COzEt

COsEt

y C H - CH. X

C0 2 Et

C02Et

\

CH

/CH3

C - (C0 2 Et) 2

(A)

I

CH2-CH2-CN (B) C h 3

\

/

CH3

C0 2 Et

(C0 2 Et) 2 Dieckmann Cyclisation

CO2E1

CH2 - CH2 - C02Et

(D)

(C)

-C0 2 Et v

C0 2 Et

CH3

(E)

13.

a) Equivalent weight of KMnQ 4 =

M

°'

eCU ar weight

Equivalent weight of Fe2+= 56

'

5

CO2H

[8]

=31.6

Normality of KMn0 4 =

= 0.11 (N) 31.6 As 20 ml of Ferrous salt requires 27.25 ml of 0.11 (N) KMn0 4 solution m 1+ ofr tferrous ion = 0.11x27.25 = 0.15N Normality 20

[2]

Strength of ferrous ion = 0.15 x 56 = 8.4 g/lit But actual strength of ferrous ion = % of ferrous ion =

8.4 58.8

x

100

= 58.8 g/lit

100 = 14.29

[2]

b) CsBr3 and CsBrCI2 are compounds containing interhalogen ions, Br3~ & BrCI2~ respectively. The alkali metal has a + 1 oxidation state in the above compounds. So the existence of these compound is not a contradiction of the statement that alkali metals have only one positive oxidation state. [3] 14.

a) "

H

H30

CH3OH CH 3 O - C = O

In the given compound H + does not attack double bond oxygen due to the fact that the tetrahedral intermediate that will be formed will be sterically conjested. So the reaction was via acylium ion formation [3] FIITJCC Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Terminal), New Delhi - 16, Ph : 6515949 , 6865182, 6854102, Fax : 6513942

AITS2002-OT-CH(S)-7

b) The nucleophilicity is defined as the rate of attack of a nucleophile to an electron deficient centre. CP being small in size compared to Br~, it is easier for CP to attack. In dimethysulfoxide which is a aprotic solvent the anions do not get solvated and so the smaller ion attacks more rapidly. Therefore CP is a stronger nucleophile than B r . [2] 15.

H2(g)+^02(g)—>H20(g) AH = AE + AnRT AE = - 5 7 . 7 9 - A n R T = - 57.79 - (-0.5) x 2 x 298 x 10~3 = - 5 7 . 4 9 2 x 103 cal/mol For 0.1 mol AE = -57.49 x 103 x 0.1 cal/mol = -5.749 x 103 cal C v for water (g) = 5.92 - 2 = 3.92 AE = nC v AT 5.749 x 103 = 0.1 x 3.92 x AT AT = 14665.8 K T 2 = (14665.8 + 298) = 14963.8 K

16.

[2]

[4]

a) The more stable the conjugate base is, the stronger is the acidity. Now in the

halophenoxide ions CP atom can act as d-rc acceptors and thereby the negative charge on oxygen atom gets delocalised to a larger extent.

9'

9~

d7t-p7l

o In fluorine this is not possible due to new availability of d-orbitals and hence it is less stable than p-chlorophenoxide and thereby the corresponding acid i.e. p-fluorophenol is a weaker acid than p-chlorophenol. [3] b) The carbon monoxide collected can be estimated by reacting with l 2 0 5 & then titrating the iodine liberated with sodium thiosufate solution. From the equivalents of Na 2 S 2 0 3 required, we can estimate the equivalents of l2 and hence l 2 0 5 . I 2 0 5 + 5CO = 5C0 2 + l 2 l 2 + 2Na 2 S 2 0 3 = 2Nal + Na 2 S 4 0 6 [3]

17.

The number of photon is E Pt XPt N = N

[P = watts, t = time] hv hC/X he _ (5.6x 10" 7 )nx(100Js" 1 )x 1,0sx0.5 = 1 4 x 1 Q 2 o (6.626 x 1034 JsX2.99 x 10 8 ms- 1 )

[3]

CH3 18.

a) CH2 = C - CH = CH2 b) H3BO3 4HB0 2 metaboric acid

100 C

"

140 C

°

(isoprene)

[2]

> HB0 2 >H 2 B 4 Q 7 pyroboric acid

[2]

heating

boron trioxide

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FIITJCC Rankers Study Materiel IIT - JEE 2002 PHASE TEST - 1

MATHEMATICS SOLUTIONS 1.

Let (at2, 2at) and (at-i2, 2ati) be two points on the parabola y2 = 4ax. Tangents at these points intersect at [att^ a(t+ t-i) ] . Since this lies on the hyperbola xy =c2. a 2 tt 1 (t+t 1 ) = c2 i.e a2 t-it2 + a2 t-,2t - c 2 = 0 This is a quadratic in t. Let t2, t3 be its roots. It means that the points [at,t2, a(ti+ t2) ] and [atit3, a(ti+ t3) ], which are the points of intersection of tangents at ti and t2 and at ti and t3 lie on the hyperbola. [2] c2 Also t2 + t3 = -U and t2t3 = — — [2] a\ => at2t3 a(t2 + t3) = - 4 - ( " t 1 )a2 = c 2 a t1 Hence the point of intersection of the tangents at t2 and t3 also lies on the hyperbola.

2.

The pair of lines given by y2 - 4xy + 3x2 =0 . . .. (1) intersect at the origin. The equation of the circles through the origin is x2 + y2 +2gx + 2fy = 0 We homogenise this equation with the heip of x +2y = 1, i.e. x2 + y2 +(2gx + 2fy)(x +2y) = 0 (2) should be identical with (1). Hence = ~4 = — — = > f = -4/21, g = -1/7. 1 + 2g 4g + 2f 1 + 4f Hence the equation of the required circle is 21x2 + 21y2 -6x - 8y = 0.

3.

[4]

[2] [2] [2] [2]

ax2 + bxy +cy2 = bx2 + cxy +ay2 => (a - b ) ~z + ( b - c ) - + (c -a) = 0

y

y

=

y

[2]

a -b q

g

Q

Since b, a, c are in HP, — — = — a -b b

/ 2 \ — =1, c/b . Taking — = - , we find that —— + 2 cy2=d y y b ^L c 22 fr -t1 1i ^ ay 2 — + c —+ — = d

db2 _ j db(c + b) => y =\ a(b2 + cb + c 2 ) ~ -y 2c(b2 + be + c 2 ) I

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[2]

[2]

RSM12-P3-T(M)-MA(S)-2

x=

I

dc(b + c)

[4]

]] 2b(b2 + bc + c 2 )

The equation of the tangent to the ellipse at any point (a cose, b sine) is x cos 0 y s i n f =1 ~a b It passes through the focus (0, c) of the parabola => sin6 =b/c Hence the equation of the tangent becomes

[2]

x-fK+l-1

all c c If the point (2ct, ct2) lies on the tangent, then at2 +2tVc 2 - b 2 - a = 0 [2] If ti, t2 are the roots of this equation, then tit 2 = -1. Hence points t1 and t2 represent the extremities of the focal chord, => length of the chord = yc 2 (t 2 - t 2 ) 2 +4c 2 (t, - t 2 ) 2

[2]

= W ( t 1 - t 2 ) 2 f c + t 2 )2+4}= cVi(t1+t2)2-4T1t2Ji(tl+t2)2+4} = c[ (U +t2)2 +4] = c

4(c2-b2)

+4

4c 1 •

c2 - b 2

[2]

ax2 + bx + c is always non - negative => b2 < 4ac bx2 + cx + a is is always non - negative => c2 < 4ab cx2 + ax + b is always non - negative => a2 < 4bc But equality sign is not valid for all the three simultaneously Therefore, a2 + b2 + c2 < 4(ab + be + ca) a2 + b2 + c 2 < 4 • (1) ab + bc + ca Also, we know that a2 + b2 +c2 - ab - be - ca = => a" + b* + c > ab + be + ca

[(a - b)2 + (b - c)2 + (c - a)2 ] > 0

ab + be + ca

> 1.

[2]

[2]

[2]

(2)

From (1) and (2), we get 1< 6.

ab + bc + ca

< 4 => given expression can never lie in (-oo, 1) ^ [ 4 , oo)

[2]

Clearly the point (1, 2) lies on 7 x - y - 5 = 0 Also centre of the circle must lie on the bisectors of the lines x + y + 1 3 = 0 and 7x-y-5=0, given by x + y + 13 _ + 7 x - y - 5 V2 ^ ~ V50 => x - 3y = 35 and 3x + y = -15 [3] If (h,k) be the centre of the circle, then h - 3 k = 35 (1) and 3h + k = -15 -...(2) k—2 Clearly, CB is perpendicular to BP - x 7 = -1 h-1 =>h + 7 k - 1 5 = 0 . . (3) Solving (1), (3) and (2), (3). we get the centres as Ci = (-6,3) and C2 s (29, -2) r,2 = 50 and r^ = 800 F I I T J C C , ICES HOUSE (Opp. VIJAY MANDAL ENCLAVE), SARVAPRIYA VIHAR, NEW DELHI -16. Ph:6854102, 6865182, FAX: 6513942

[2]

RSM12-P1 -T(M)-M A(S)-3 => Smaller circle has the radius = V50

[2]

Therefore, area of the quadrilateral ACBP = 2 — x V 5 0 x -v/200 2

= 100 sq unit

[1]

Any point N on the parabola x2 = 4y is (2t, t2) Tangent to the parabola on any point T is given by tx = y +12 . . . . (1) [2] Since tangents are drawn from x + 4 = 0, coordinates of the point M(on the line) = (-4, -4t-t2) Let P(h, k) be any point, which divides MN in the ratio 1:2 Therefore,

^

M

^

I

. 2t - 8 t2 - 8t - 2t2 h = —-—,k = 3 ' 3 Eliminating't' we get

Q

i 3 - 8t -1 2

W t=

[2] .. — 8t — t2 and k = 3

3h + 8 2

N

p h + sY

J 3h + 8 f 8 + 3k = 0 J I 2 J

[2]

I 2 2 => 9h + 96h + 12k + 192 = 0 Generalizing, we get the required locus as 9x2 + 96x + 12y + 192 = 0 or 3x2 + 32x + 4y + 64 = 0.

[2]

x" v2 The given ellipse is — + = 1 . . . (1) a b Let P(h, k) be the point of intersection of tangents. Kv t-—2 = 1 . . . (2) [2] a" b The equation to the lines joining the points of contact to the centre will be given by making Equation of the chord of contact of P(h,k) w.r.t. (1) is given by x v 2 ' h x ky (1) homogenous with (2). So we, get — + —2 = a" b a2 " b2

1

hx

J

\

V 1 1y 2 + 2 -hk |x2 + 4 ^ j x y = 0 . . . (3) I b2 J a2b~ ,b The lines given by (3) subtend an angle 45° at the centre

-K

h2 2 I™ . V a4b4 a2 tan 45° = - v 2 2 h . k _ 1 a2 a4 + b 4

a

rn la 4

1

b4 1^ b2

+±Sf 2

( 1

+

b4

[2]

a4b2 a2b4 a2b2 b Generalizing, we get the required locus as

[2]

=0

-\2

1 x y = 0 2 2 b a b a b ab , 4 . ,2 ,4u2N2 4 u 4 / u 2 „ 2 , _ 2 . ,2 z => (b4x2 + a Y - a V - a V ) z - 4 a V ( b V + a Y - a,2u2x b^) _= 0 . x"

y

1

2

-4

Let e and e' be the eccentricities of the hyperbolas 2

2

2

2

2

2

X 2

a

[2]

V x —2 = 1 and 2 b a

V —2 = - 1 respectively. b

Hence , b = a (e -1) and a = b (e' -1) e:

e'2

...(1)

x v Equation of the line making intercepts e and e ' is —+ —? = 1 e e

[3] [2]

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RSM12-P1 -T(M)-M A(S)-4 Its distance from the origin =

= 1. (using equation (1) 1 — + e' e' Hence it always touches the circle x2 + y2 = 1. 1

i(n2 + l)(n2 + 2 - 2 n ) 10.

Let 9 = sec"'

2

(n

30 that

>

sec20 =

(n2 - n +1 ^

or sec20 = 1 +

n • 1)

"

n2 - n + 1

or 0 = tan"1n -tan~1(n-1), so that Tn = tan"1n - tan"1(n-1) T1 = tan"11 - tan"10 T2 = tan"12 -tan" 1 1

[2]

Tn = tan"1n - tan"1(n-1) => S n = tan"1n.

[4]

Given equation x2 +y2 +8x - 10y +40 = 0 represents a circle centered at C(-4, 5) and of radius 1. Let P be the point (-2, 3) 5-3 Slope of CP = = - 1 [2] -4+2 -2 => Inclination of CP with the positive direction of the x-axis is 135°. Equation of CP is

C(-4\5) ^

^Xp(-2, 3)

x+4 cos 135°

sin 135°

1 4

2-

„

1

1

= 2 4+

V2

2

9 _ _4_ = 9-4V2 . 7= = 2 V2_ 2~V2

a = max[(x +2)2 +(y -3) 2 ] = QR2 = | 2 + = 2

9

4

)

\

V2

If Q and R are the points where CP intersects the circle, then A 1 ( 1 R . 1 - 1 - 4 , - ^ + 5 '• v 2 v2 Now, b = min.[(x +2)2 +(y -3) 2 ] = PR2 =

•w,

O

[2]

' ' ~V2

12.

[2]

(n2 - n +1)2

n - (n - 1 ) 0 = tan"' 1 + n(n - 1 )

1

or tan0 =

[1]

(n2 + l)(n2 + 2 - 2n

(n2 + i f + (n2 +1)~ 2n(n2 +1)

=

or sec 9 =

11.

(n2 - n +1 f

'

[2]

= 9 + 4V2 => a + b = 18.

2+

[2]

[2]

V2

[2]

The given equation can be written as I x ' + 4x + 7 —r

I x + 4x + 6 Let t =

-(a-2

x 2 + 4x + 7 2

x +4x + 6

x2 + 4x + 7 ^ x2 + 4 x + 6

= 1+-

1

(x + 2 ) 2 + 2

(a - 3) = 0

[1] [2]

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RSM12-P1 -T(M)-M A(S)-5 Therefore x + 4x +6 > 2

left ""

[2]

I 2. Now the given equation reduces to t2 - (a - 2 ) t +(a - 3 ) =0 ^ t . (a-2)±V(a-2)2-4(a-3) or t = ±

i—i 2

or t = a - 3, 1

[2]

At least one of these must lie in => 1 < a - 3 < 3/2 , 4 a

9

6

|(A4, -"

[3]

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FIITJCC Rankers Study Materiel NT - JEE 2002 PHASE TEST - II

MATHEMATICS SOLUTIONS Let y = cos0(sin0 + Vsin2 0 +sin 2 a ) => y - cos0 sin9 = cos9 Vsin2 9 +sin 2 a Squaring both sides we get y2 + cos29sin29 - 2y sin9cos0 = cos20sin20 + sin2acos20 => y2tan20 - 2ytan0 + y2 - sin 2 a = 0 (where cos20 * 0) Since tan0 is real, we have 4y2 - 4y2(y2 - sin2a) > 0 => 4y2 [y2 - (1 + sin2a)] < 0 => —Vl + sin2 a - a < cos0| sin 0 + Vsin2 0 +sin 2 a ] < a -H sin2 a)

where a =

For cos© = 0, the result is trivial. Domain of the increasing function f(x) is [a.b] => f '(x)>0 in [a , b]. Domain of the decreasing function g(x) is [a,b => g '(x)<0 in [a , b]. Now h(x) = fog(x)+gof(x) => h'(x) = g'(x)f 'og(x) + f '(x)g'of (x) < 0 Hence h(x) is a decreasing function . hence the range of h(x) is [h(a), h(b)] Also range of g(x) is a subset of [a, b] => Domain of fog(x) is [a, b]. Similarly, Domain of gof (x) is [a, bj. Hence domain of h(x) is [a, b]. _ 1 1 1 9 + —X + —X + ... + x 2 3 10 Lt x-»0

_ 9

1 1 1 — + —X + ...+ 10x Lt 1 + 2" 3 x-»0

Lt

= ex"o

JJM

x-*0

k k l ^ k l X

1 1 1 —+—+...+ 9„ X 2" 3 10x

. 1 1 9 2" 3*

— + — + . . . +

ufell.tl], x X

= e x"o

(1 + X) 1/X + e ( x - 1 ) _ sin - 1 X

9/x

|JM

1

10x

+

i H X

-

e

JiV-nfa, UJ UJ

+lnf—1 IioJ

-

c

Inf—1 11 l 101 J _ J_ 10!

(1 + x\1/X )'/x +e(x-1) x

Vsin

1

x;

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RSM12-P2-T(M)-MA(S)-3 f. lim

( ln(1+x) ^ e x -e

.XX 1—+

^ +e

x-»0

V

lim x-»o

lim —— *->° sin x

2

e

...

3

-e

•+ e

lim — x-ô sin xi

V

f r

X

X

2

\

lim x->0

lim

+ e

f

r

sin""1 x ;

+ e x1 =

V

Let A = B, then 2A + C = 180° and 2tanA + tanC = 100 Now 2A + C = 180° => tan2A = -tanC ......(1) Also 2tanA + tanC = 100 2tanA - 1 0 0 = -tanC (2) From (1) & (2) 2 tanA - 1 0 0 = Let tanA = x, then

2x 1—x

2

2 tan A

1 - tan A

= 2x - 100 => x3 - 50x 2 + 50 = 0

Let f(x) = x3 - 50x + 50. Then f '(x) = 3x 2 - 100x. Thus f '(x) = 0 has roots 0, f(0).f

100

. Also

100

< 0. Thus f(x) = 0 has exactly three distinct real roots. Therefore tanA and hence v 3 , A has three distinct values. Thus there exists exactly three non similar isosceles triangles. Let I = I e*3"6 (sece - sine) d9 tane = t sec 2 6 d6 = dt de =

dt

dt

I =Je< V T + 7 -

1 + t"

1+t

Vi+" Integrating first part by parts we have,

2

-

Iel

——— . e d t - 1J .. e 4 dt + c = ~ e + 'f .. 0.1/0 o 2 3 2 (1 + t2}3/2 (1 + t ) ' t

/

r

1

t j i

f

o+t2)3'2

VTTt^ e

t

£

Given that f(x) + f(x + 4) = f(x + 2) + f(x + 6)

•

+

dt

c, = e 1 a n e cose + c

(1)

Replacing x by x + 2, we get f(x + 2) + f(x + 6) = f(x + 4) + f(x + 8)

(2)

From (1) and (2) we get f(x) = f(x + 8) x+8 Now let g(x) = Jf(t)dt

(3)

X

=> g '(x) = f(x + 8) - f(x) = 0, by (3) => g is a constant function.

Let t = — => dt = y y So

[ A , f _ J 1 + t 2 1/x J l1+ y '

X

dy- -

1/x dt f r 2 1+y ~ ! 1+t 2

J

dy

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The region enclosed by the parabola x 2 = 4y and the line 2x = y is shown in the figure as OACBO. We require the ratio of the area of regions OABO and CABC. The point A is (1,2), B is (2,1), and C is (8,16). 8 2 2V Now Area OACBO = f v( 2 x - — ) dx= x 2 - — J a' 19 = 64-

128

64

= |2xdx + j V 5 - x 2 d x - J x 2 / 4 dx 0 1 0

Area OABO

5 . . , 2 . 5 . . , 1 1 2 5 . . = 1 + 1 + —sin — r = - 1 — s i n — j = — = —+ —sin 2 V5 2 V5 3 3 2

UJ

=> Area of CABC = — - - s i n " 1 ^ 3 3 2 4 + 15sin" 1 Required ratio =

15

1 2 4 - 1 5 s i n -1 10.

The given differential equation can be written as y 4 dx + 2xy3dy +

xy

(xdy~ydx) x

xy 7 dx + 2x 2 y 6 dy + d(y/x) = 0

=> — ,xy7dx + —,2x 2 y 6 dy + - d W = 0 => x 2 y 6 dx + 2x3y5dy + ^ ^ y y y IxJ y/x

=0

=> - d(x3y6) + d(logy/x) = 0 => x3y6 + 3logy/x = constant 3 11.

f(X) = 1 + 1 X X4 - X. 2 +. —+ 4

=

o2

1 + -

f

^^ 2J

v

X -

V2.

1

+— 4

1

if x = —=, f(x) will attain maximum value => max. f(x) = 5 V2 And if x oo f(x) - » 1 => range is (1, 5] 12.

Since g is the inverse of f, f(g(x)) = x. Differentiating both sides w.r.t. 'x' we get f (g(x)). g'(x) = 1 => g'(x) =

Now

1'

f'(g(x))

=

1' —

1+(g(x)) 3

J i—T77u3" = J W W 1 i+(g(t)) 3 i

=

9< x )-9( 1 ) = 9( x )- as f (0) = 1 => g (1) = 0 •k

rk

"k

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HITJCC RANKERS STUDY MATERIAL IIT - JEE, 2002 PRACTICE TEST - PHASE-I, II MATHEMATICS SOLUTIONS 1.

Let f(x)= v a 2 cos 2 x + b 2 sin 2 x + Va 2 sin 2 x + b 2 cos 2 = Va 2 cos 2 x + b 2 sin 2 x + -Ja2 + b 2 - [a 2 cos 2 x + b 2 sin 2 x] Let P = a2 cos2 x + b2 sin2x 2 2 2 P= 1 2 2

Let f(x) y = #

a

+b

+(a

-b )cos2x

+ \/a2+b2-P

Now y2 = a 2 + b 2 +2 A /p(a 2 + b 2 ) - P 2 = a2 +b2 + 2 j l ( a 2 + b 2 ) - | l ( a 2 + b 2 ) - p j 2 y2 will be greatest when P =

2.

3

ymax =

=> => =>

y2 will be least when P is least i.e. cos2x = -1 Least value of P = b2 Least value of y2 = (a2 + b2) ymin = a + b

=>

(a + b) < f(x) < V2(a2 + b 2 )

a2 + b 2 )

sin x is continuous for all x < 0 so f(x) is continuous for all x < 0. Now g(x) = [x] + V x - [ x ] Case I: Continuity at +ve integers x = a (a e N) g(a") = a-1 + V a - a + 1 = a g(a) = a + Vo = a g(a+) = a + Va - a = a so continuous for all integers Case II: If x is not an integer x = b, in this case [b] = [b"] = [b+] FIITJCC.LtdICES House (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi -214.PH: 6865182, 6854102, Fax: 6513942

RSM12-PT-PH-I, II (M)-MA(S)-2

g(b") = [b-] + yfb-[b~] g(b) = [b] + V b ^ t b ] g(b + ) = [b+] + V b - [ b + ] So it is continuous at b so continuous for all + ve x and at 0 it is continuous. So continuous for all real x. 3.

F'(x) = 4 sinx + 3 cosx = 5 sin

( x + tan" 1 3I

n 3n —< x < — 4 4 , 3 n . tan 1 - + - < x + tan 4 4

1

4,

3 371 13 - < — + tan" 1 4 4 4

3 - < % => F' (x) > 0 => F (x) is an increasing function in

0< x +tan

n 37t 4'T

Least value of F(x) is at x = 7C F

( f )= 1 (4sint ^ ' 7t

+ 3cost

) 1 d t = I - 4 cost+ 3 sin t|

6 r

4

V 42

4(A).

3 ^ f-4.V3 +

42

/

v

3.0_

1 42

3

7t cosec 2x - 3 J -y^dx Tt 1 sin 2x

4V3

3

+•

4V3-3-V2

1 - Jf - f / \ dx, J sin ' 2x - — sin 2 x - I 3; I 6J

put 2x = t

N

f

1

sin^ t dx =

71

f t-

3,

V

"® 11 Jj ej

J - x i tJ dt \ ( .it 2 7 1 71 • f t sin t — sin f t - ^ sin t - — sin t -sin — 3 6 I 6J V rsinacosp-cosasinp , , ,_ . = - J dt, where a = t - 7 i / 3 and p = t — 7t/6 sinasinp

J

CO

= — [fcot(t — 7r /6) dt — [cot(t — 7t/3)f dt] it \ sin 2x V f \ TC sin 2x In sin(t-ir/3) - I n sin(t - 7i/6)+ c = In ej < (b).

l = - j

+c

(tanx - 1)se.c2 x dx

(tan x + 1)Vtan3 x + tan 2 x + tanx Put tan x = t (t-1) (t + i)Vt 3 + t 2 + t

dt=-J

t2-1 (t 2 + 2t + 1)Vt3 +1 2 +1

dt

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RSM12-PT-PH-I, II (M)-MA(S)-3

1 -

,+2

,+1+

nw l

I=- J

dt

Put 1 + t + - = u 2

= - 2 tan

1

u + c, where u = J l + tan x +

1 + u'

^ tanx

The pair of lines given by y 2 - 4xy + 3x2 =0 . . . . (1) intersect at the origin. The equation of the circles through the origin is x 2 + y 2 +2gx + 2fy = 0 We homogenise this equation with the help of x +2y = 1, i.e. + y +(2gx + 2fy)(x +2y) = 0 (2) -4 1 should be identical with (1). Hence f = -4/21, g = -1/7. 1 + 2g 4g + 2f 1 + 4f Hence the equation of the required circle is 21 x2 + 21 y 2 - 6 x - 8y = 0. 6.

ax 2 + bxy +cy2 = bx2 + cxy +ay2 => (a - b ) ^ - + (b - c ) - + ( c - a ) = 0 => - = 1, ^ y y y a-b Since b, a, c are in HP,

c-a

c

a^b

b

X X c 'ac2 + 2c => - =1, c/b . Taking - = - , we find that y y b

f ay' x=

c ( 1 1 —2 + c - + - = d b c b J)

y=

db 2

I

2

2

]la(b +cb + c )

db(c + b) 2c(b2 + bc + c 2 )

dc(b + c) V 2b(b2 + bc + c 2 )

ax + bx + c is always non - negative => b 2 < 4ac bx2 + cx + a is is always non - negative => c2 < 4ab cx2 + ax + b is always non - negative => a2 < 4bc But equality sign is not valid for all the three simultaneously 2 u2 i. - . Therefore, a 2 + b + c2 < 4(ab +. be + ca) •%2

.

a2 + b 2 + c 2 <4 ab + be + ca •j

Also, we know that a 2 + b 2 +c2 - ab - be - ca = => a + b + c

> ab + be + ca =>

a2 + b2 + c2

[(a - b)2 + (b - c)2 + (c - a) 2 ] > 0

>1.... ab + be + ca a2 +b 2 + c2 From (1) and (2), we get 1 < <4 ab + be + ca => given expression can never lie in (-oo, 1) u[4, oo). 8.

(2)

Clearly the point (1, 2) lies on 7 x - y - 5 = 0 Also centre of the circle must lie on the bisectors of the lines x + y + 1 3 = 0 and 7x-y-5=0, given by x + y + 13 _ + 7 x - y - 5 42 450 => x - 3y = 35 and 3x + y = -15 If (h,k) be the centre of the circle, FIITJ€€, ICES House, (Opp. Vijay Mandal Enclave), Sarv;

....(1)

x+y+13=0

P(-1,-12) 7x-y-5=0 B(1, 2)

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RSM12-PT-PH-I, II (M)-MA(S)-2 then h - 3k = 35 and 3h + k = -15

(1) . . . . (2) k-2

Clearly, CB is perpendicular to BP

h-1

x

7 = -1

=> h + 7k - 15 = 0 . . . (3) Solving (1), (3) and (2), (3), we get the centres as C-| = (-6,3) and C2 = (29, -2) => r,2 = 50 and r22 = 800 => Smaller circle has the radius = V50 Therefore, area of the quadrilateral ACBP = 2 9.

"1

v 50 x V200

100 sq unit

Any point N on the parabola x 2 = 4y is (2t, t2) Tangent to the parabola on any point't' is given by tx = y +12 . . . . (1) Since tangents are drawn from x + 4 = 0, coordinates of the point M(on the line) 2 s (-4, -4t-t ) Let P(h, k) be any point, which divides MN in the ratio 1:2 Therefore, h .

3 8t -1 2

+

. 2t-8 . t — 8t — 2t => h = ,k = 3 3 Eliminating T we get

'3h + 8

+8

, 3h + 8 .. - 8t -1 2 => t = — : — and k = — — 2

3h + 8

+ 3k = 0 V j y => 9h + 96h + 12k + 192 = 0 Generalizing, we get the required locus as 9x2 + 96x + 12y + 192 = 0 or 3x 2 + 32x + 4y + 64 = 0. 10.

The given equation can be written as ' x 2 + 4x + 7 x + 4 x + 6j Let t

V

-(a-2)

x2 + 4 x + 7 2

x2 + 4x + 7

+ (a-3) = 0

x2 + 4x + 6

= 1+

1

x + 4x + 6 (x + 2 / + 2 2 Therefore x + 4x +6 > 2 t G

1

' 2 Now the given equation reduces to t 2 - (a -2)t +(a - 3 ) =0

^

t

_ (a-2)+V(a-2)2-4(a-3)

At least one of these must lie in

or t

1

_ (a-2)±(a-4)

o r t = a - 3, 1

'2

=> 1 < a - 3 < 3/2 , 4 a e

4

-|

* * * *

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FIITJCC Rankers Study Material NT - JEE 2002 PHASE - III TEST

MATHEMATICS SOLUTIONS 1.

0 < cos"1 x < 71, 0 < sin"1 y < n/2, v x , y e R+ 1 1 0 < 2cos" x < 2tc & 0 < 4(sin" y f < n2, :. 0 < 2cos"1 x + (sin"1 y)2 < 2 k + "n2, or, 0 < ptt2 < 7t2 + 2it or, 0 < p < 1 + 2/ti [p] = 0 or 1 [4] !f [p] = 0; tan"1 x + tan"1 y = [p] TX/4 reduces to tan"1 x + tan"1 y = 0 which is not possible Since x, y e R+ if [p] = 1; tan"1 x + tan"1 y = [p] TC/4 reduces to tan"1 x + tan"1 y = 7t/4 [4] x+y 71 Or tan"1 v x y < 1 & x, y > 0 1 - xy 4 Ôr x + y = 1. 1 - xy

y = 1 - x Proved. 1+x

[2]

Let AD = d & BE = I, AC = b

AE = 3 in A ABC; b2 + c2 = a2

•(1)

2

b , in A ABE; — + c2 9 in A ABC; b 2 + c2 = 2

2

3^ ^

..(2)

•(3)

[3]

[Appolonious Theorem] Or Or

+ c2 = 2dz +• b2 + c2 = 2d 2 + ~(b 2 + c 2 )

Or b 2 + c2 = 4d2 Dividing (2) by (4), we get,

[from (1)]

(4)

[3]

b 2 + 9c 2 _ 9I2 _ 9 f B E > [ 2 _ 9 4 b 2 + c 2 _ 4d 2 ~ 41 AD J or b2 = 3c2 From(1); 4c2 = a2 cos Z ABC =

a2 + c 2 - b 2 2ca

_

4^3

4c 2 + c 2 - 3 c ' 2.2.C.C

2

Z ABC = 60°.

[4]

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Let

sin 3 6

cos 3 G

sin(29 + a)

cos(29 + a)

cos0sin 3 0

...(1)

sin0cos 3 0

_

cos0sin(20 + a)

= k _k_

sin0cos(20 + a)

sin0cos0 _ sin(30 + a)

cosGsin 3 0 + sinGcos 3 0 cos 0 sin(20 + a) + sin 0 cos(20 + a)

sin20

[4] [2]

2sin(30+a)

comparing the values of k we get, cos 20

_

sin 20

cos(30 + a) ~ 2 sin(30 + a) or tan 0 = 2tan (30 + a)

Proved.

V l 6 - 4sin 3 0 - 4sin 2 0 + sin0 + 3

[4]

- sin0 + 4 C O S 2 ( 1 + sin0) = 7

...(1)

2

16 - 4 sin 0 - 4 sin 0 + 3 sin 0 = 16 - 2 ( l - c o s 2 0 ) + sin 30 = 5 = =

14 + 2 cos 20 + sin 30 - sin0 + 4 cos2 0 (1 + sin 0) = 5 - sin 0 (1 -4cos 2 0) + 4 cos 2 0 5 - s i n 0 (4 sin2 0 -3) + 2 (1 + 2cos 0) 7 + sin 30 + 2 cos 20 equation (1) reduces to,

Vsin 30 + 2 cos 20 +14 + Vsin30 + 2cos20 + 7 = 7 Let sin 30 + 2 cos 20 = x .'. Vx + 14 + V x + 7 = 7 (x + 14) - ( x + 7) = 7

••(2) ..(3)

.". Vx + 14 - V x + 7 = 1 By (2) + (4); we get,

..(4)

[Dividing (3) by (2)]

2 Vx + 14 = 8 => x = 2 sin 30 + 2 cos 20 = 2 or 3 sin 0 -4sin 3 0 + 2 -4sin 2 0 = 2 or sin 0 (sin2 0 + 4 sin 0 -3) = 0 or sin e (2 sin 0 + 3) (2 sin 0 -1) = 0 1

sin 0 = 0 or sin 0 = 2

[3]

[4]

-3 [• sin0 = — is not possible] 1

= nn o r 0 = nn + V(-1)" - V n e I ' 6

[3]

AC = h cot 45° = h BC = h cot 60° = ^ L V3 DC = h cot 60° = hV3 [2] In cyclic quadrilateral ABCD, Or, AB.CD + AD.BC = AC.BD [5] [Ptotemy's Theorem] or, AB(CD+BC) = AC.BD [ v A B = AD] Or, 10V3 hV3 +

V

= h . (2r) [ v AB = 10 V3 & BD is the diameter where r is the radius]

v3 J

Or, 2r = 40, or r = 20 Radius of the swimming pool = 20 m. FIITJCC, ICES HOUSE (Opp. VIJAY MANDAL ENCLAVE), SARVAPRIYA VIHAR, NEW DELHI - 1 6 . Ph:6854102, 6865182, FAX: 6513942

[3]

RSM12-P1 -T(M)-M A(S)-220

S(-rc2r_2 r=1

= (c0 - c 2 +C4 -

f

= (C, - c 3 + c 5

)2

2r-2

K - r c

. r-1 (1 + x)n = Co - C1X + C 2 x2 +

Cn xn

Let x = ix (1 + ix)n = C 0 - I(C1X) - (C2 x 2 ) - (C3 x 3 ) + = ( C 0 - C 2 + C4 ) + ix(C, - C 3 + C 5 Putting x = 1, we get (1 +i) n = ( C 0 - C 2 + C 4 ) + ( C , - C 3 + C5similarly putting x = -ix, we get, (1 - ix)n = ( C 0 - C 2 + C 4 ) - ix(C1 - C 3 + C 5 Putting x = 1, we get (1 -i) n = ( C 0 - C 2 + C 4 ) + ( 0 , - 0 3 + 05By (2) x (3) we get,

[(1 + i) (1 -i)]n = (Co - C 2 + C 4 n

or 2 = ( C o - C 2 + C 4 or (Co + C, + C 2 + Cn)2 = Hence the result follows. 7.

...(1)

) )

...(2) )

)

...(3)

f + (C, - C 3 + C 5 -

2

)2

) + (0,-03 + C5 -

(Co + C 2 + C 4 -

[4]

f (0, - C 3 + C 5 -

f

[4]

f [2]

1 1 7 S = 1 + — + —- + —— ...(1) 6 18 324 since this is an infinite series it will be of the form ( 1 + x f V n e I" or f (1 + x)n = 1 + nx + - i

A x 2! 3! comparing Tr from (1) and (2), we get,

n(n-l)

1 6' n(n - 1 ) X 2 2! 2 n2 x

x

2!

x2 =

..(2)

[2]

1 18 [3]

18 => n = - — 3

•. X = —-r

2

6

.-. S = (1 + x)n =

= 3V2

2

1

1

sin 39

sin 3 9

2cos9

sin69

sin 3 29

4cos2-1

sin99

sin 3 39

sin9

.00

36

or, n - 1 = 4x

1

^x 3 + .

sine

sin39

cos 29 . „„ sin 39

sin 69 . „„ sin 99

[5]

sin3 9 2 2 sin 3 29 [ y sin 9 (4 cos 9 -1) = sin 9 (3 -sin 9 = sin 36] . 3 sin 39

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[4]

RSM12-P3-T(M)-MA(S)-4 3sin0-4sin3 0

3 sin 29 - 4 sin 3 2G sin 66

sinG

3sin36-4sin

sine

9.

sinSG

sin36

sin30

sin 60

sin 60

sin96

3

36

sin96

sin3 9 sin 3 26

[3]

(Applying Ci = 3CT -4C 3 )

3

sin 30

sin 3 0 sin 3 20 = 0 [ v Two columns are equal]

[3]

3

sin96

sin 36

The system of equation will have a non-trivial solution if A = 0 1

2

[p + 3'

2

[p + 1]

3

[p + 2]

3

1

i.e., if

or,

1

2

jp]+3

2

[p] + 1

3

[p]+2

3

1

1

2

o or ([p]+6) 0

3

3 [P]-i 2 - [p]

1 3

=0

\p] + 3

or ([p] + 6) 1 [p]+1 1

= 0

=0

[4]

1 -[P; - 2 =0 (Applying ^ = ^ + 0 2 + 0 3 and taking [p] + 6 common from Ci) 1

or ([p] + 6 ) (-2[p] + 2 + 2[p] -[p] 2 ) = 0 or, ([p] = 6 ) ( [ p ] 2 - 2 ) =0 ••• [P] = -6 => p e [-6, -5) or [p]2 = 2 => [p] = ± V2 which is not possible Required range of values of p are given by - 6 < p < -5. 10.

[3]

[3]

Prime numbers in the number system greater than 2 & 3 can be either represented by 6n + 1 or 6 n - 1 V n e N Let P (n): (2n - 1 ) 2 n " 1 = has its last digit as 6 or 8 V n is prime be true for n = 6k + 1 For n = 2 & 3; (2n - 1 ) 2 n _ 1 = 6 & 28 respectively .". P (2) & P (3) are true Case ! P (6k +1): (26k + 1 - 1 ) 26k has its last digit as 6 or 8 is true on assumption. =>212k + 1 _2 k has its last digit 6 or 8 according as k is even or odd 4k + 1

4k + 2

4k + 3

...(1)

4k + 4

[ v Last digit of 2 , 2 ,2 &2 is 2, 4, 8, 6 respectively] We have to prove P (n) is true for n = 6 (k + 1) + 1 i.e., for n = 6k + 7 [2] P (6k + 7) = (2fik + 7 —1) 26k + 6 ~ 212k + 1 3 — 26k + 8 = 212 (2 1 2 k + 1 —26k) + 26k 212 —26k + s = 4096 (212k + 1 -2 6 k ) + 26k (4096 - 6 4 ) ..(2) = 4096 (last digit 6) + 2 6 k . 4032 if k is even [from (1)] = (last digit 6) + (last digit 6) .4032 = (last digit 6) + (last digit 2) = last digit 8 Also from (2); P (6k + 7) = 4096 (last digit 8) + 2 6k .4032 if k is odd. [from (1)] = (last digit 8) + (last digit 4) .4032 = (last digit 8) + (last digit 8) = last digit 6 [4] FISTJ€€, !CES HOUSE (Opp. VIJAY MANDAL ENCLAVE), SARVAPK'YA VIHAR, NEW DELHI - 1 6 . Ph:685410l\ 3865182, FAX: 3513942

RSM12-P1 -T(M)-MA(S)-5 Case II P (6k - 1 ) : (2 Sk " 1 - 1 ) 26k~2 has its last digit as 6 or 8 is true on assumption. => 2 1 2 k " 3 -2 6 k ~ 2 has its last digit 6 or 8 according as k is odd or even 4k + 1

4k + 2

4k + 3

...(3)

4k + 4

| v Last digit of 2 ,2 ,2 &2 is 2, 4, 8, 6 respectively] We have to prove P (n) is true for n = 6 (k + 1) - 1 i.e., for n = 6k + 5 P (6k + 5 ) ' (2ek + 5 —1) 2ek + 5 = 212k + 9 —2ek + 4 = 212 (2 12k_3 —26k~2) + 2 6k_z 212 -2 6 k + 4 (4) 6k 10 4 = 4096 (last digit 6) + 2 (2 - 2 ) if k is odd [from (3)] = (last digit 6) + (last digit 4) (last digit 8) = (last digit 6) + (last digit 2) = last digit 8 Also from (4); P (6k + 5) = 4096 (last digit 8) + 2ek (210 - 2 4 ) if k is even [from (3)] = (last digit 8) + (last digit 6) (last digit 8) = (last digit 8) + (last digit 8) = last digit 6 The number 2 n _ 1 (2n - 1 ) has last digit 6 or 8 for prime n. [4]

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FIIIJCC Rankers Study Material IIT - JEE 2002 PHASE - IV MATHEMATICS SOLUTIONS Let z= x + iy then jzj = yjx 2 + y 2 Therefore the given equation becomes 2(x2 + y 2 ) + ( x + iy)2 = 5 - i V3 or 2(x2 + y2) + (x2- y2) + i.2xy = 5 -i V3 or (3x2 + y2) + 2xy = 5 -i V3 Equating real and imaginary part , we get 3x2 + y 2 = 5 : . y2 _

2xy = -V3

4x 2

3x2 + — = 5 or 12 x4 - 20x2 + 3 = 0 4x 2 or (2x - 3 ) (6x2 - 1 ) = 0 x2 = 3/2 i.e. x = ± A/3

y=

2x

V2

or x2 = -

6

i.e. x = ±

. ^ [3 when x = ± J - , y - + — = V2 2v3

.ft V6

_ 1 +-==• V2

. 1 _a/3V6 _ 3 when x = ± - 7 = r , y = + = +-?= v6 2x1 V2 The required complex numbers are [3 Z

2

i_

V2"V2'

_ [3

_1

3i_

V6 " 7 2 '

1_ _3i_ "V6+V2'

Let z8 = -1 = cos(2k +1)TX + i sin(2k +1)rc, where k = 0, 1, .., 7. z8 +1=(z -a 0 )(z - a , ) . . .(z -a 7 ) where a = c o s - + isin - (when k = 0) 8 8 Since a 0 = a7, a, = a 6 . . . . , a 3 = a 4 , hence z8 +1 = (z - ao) (z - ct0) (z - a,) (z - a, )(z - a 2 )(z - a 2 )(z -a 3 ) (z - a 3 ) = (z 2 -(a 0 + a 0 )z+|a 0 | 2 )(z 2 -(ct 1 +a 1 )z+| ai | 2 )(z 2 -(a 2 + a 2 )z+|a 2 | 2 )(z 2 - (a 3 + a 3 )z+|a 3 | 2 )

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RSM12-P-IV-T(M)-MA(S)-2

5n n z - 2 c o s —z + 1 z 2 - 2 c o s — z + 1 z 2 - 2 c o s — z + 1^

8

4 1 =>Z + —7

8

8

= 2 cos49.

If z = cosG + i sine => z + And the expression becomes 2 cos46 = 2 4 cos 6 - cos 8-

cos e - cos

N

v

cos e - cos

The four points z1f z2, z 3 and z4 ZADB = ZACB (or n - ZACB) /Z Z [f Z2 3 arg arg 2 Z Z Z 3 4 l 1

_Z4

Jl

i1

371

cos 0 - cos

~8~

cos46 = 8 cos A6 - c o s 8

Z

8

1 371 1 5tiv 1 0 7ti n z + — 2cos z + — 2cos — z + — 2cos — z 8 z 8 z 8

1 0 it z+ --2cosz 8

Z

771 o z -2cos — z +1

2

371

8

571 Y

7n cos 6 - cos 8 V

5jiY _ c o s e - c o s — cose-COS

8A

8y C(z3)

are concyclic if D(Z4)

\ | /

Z

- 3 - n arg ( 2 Z (Zl-Z4X 2-Z3)

B(z2)

A(zJ

(z2 ~ z 4 X z i - 2 s ) = _ purely real z 2 z3) ( i -^X Using the relation Zi = -z 2 and z ^ 2 + z3z4 = 0, (z2 - z j z , - z 3 ) _ z 1 z 2 - z 4 z 1 - z 2 z 3 + z 3 z 4 (Zl-Z4XZ2-Z3)

Z Z

1 2

Z Z

2 4

Z Z

13

Z Z

3 4

=

— = -1 a real number. Hence the result. Z 1\ 4 - 3) Note: We have considered the case of Z\, z 2 , z 3 , Z4 in cyclic order and leave the other case for you to prove. Z

Z

We split the question in following cases (i) No objects are alike > There are 2n+1 distinct objects. So number of ways _ 2n+1/-N (ii) Two objects are alike — > n-2 objects have to be selected from 2n distinct objects So number of ways = 2nCn.2 (iii) Three alike objects. No of ways = 2nCn_3 and so on Total = 2n+1Cn + ( 2n C n _ 2 + 2n Cn_3 + +2nC0) Now (1+1)2n = 2n C 0 + 2n C 1 + 2n C 2 + ...+2nCn-2 + ( 2n C n -i+ 2n C n + 2n C n+1 )+ 2nCn+2 +...+ But 2nCn+2 + + 2nC2n = 2nC0 + +2nCn.2 o2n o 2nr* 2n r* 2rv _ Z ~Z n-1 ~ n + 2n Q => 2nC0 + zn C! + _ o2n-1 2nr* - 2 - Un.-| -

2n

C2n

2

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Total =

2n+1

Cn + 22n"1- 2nCn.i

2n r *

= 22n"1 +

^

2

"C - -

(2n)!

_ 22n-1.

= 22 1 , ( 2 n - l ) n! n! 2 n!(n -1)!

Let x0 be the number of empty seats to the left of the first person and X| (1 < i < n - 1 ) be the number of empty seats between ith and (i + 1)th persons and xn the number of empty seats to the right of the nth person. Then x0> xn > 0 and Xj > 3 for 1 < i < n - 1 . So number of ways in which n persons can sit in 6n seats such that at least 3 seats are vacant between any two persons is the same as number of solutions of the equation x0 + x, + ... + n = 6n - n - 5n which is equal to coefficient of x5n in (1 + x + x 2 + ,...)2 (x3 + x4 + coefficient of x&n in x3n + x + x2 + )\n -1

) n " 1 or

= coefficient of x2n + 3 in (1 - x)" n " 1 _ 2n + 3 + n + 1 — U2n + 3 _ 3n + 3/-> — u2n + 3 Now if n is even say n = 2m. Similarly, we can show the no. of required ways is ( 2m Pn)( 1 ° m + 1 P m ). Here AB = (i + 2 j + k)-(2? + } + k) = -? + j 1

/>*

A

A \

/

A

^

A \

A

*

AC = (i + j + 2 k j - ( 2 i + j + k ) = - i + k .-. AB x AC = plane n .

( - T + j)x ( - T + k) = }

+ k + T j . to

=> unit vector perpendicular to the plane 71 =±

i +j + k V3

Let PQ = X; then OP = ± ^ ( i + j + k) now the PA = (2? + ] + k)-(? + ] + k ) = i V3 And X = PQ = | projection of PA on PQ j 1 . unit vector along PQ | = + V3

J_ V3

Clearly OQ = OP + PQ = i + j + k + - ^ ( i + ] + k) = (i + ] + k ) ± - ( i + j + k) v3 3 = | ( i + ] + k) o r | ( i + ] + k) But A, B, C and Q are non-coplanar because [ AB BC AQ] * 0 Position vector of the foot of the perpendicular =

+ j + k).

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7.

ABC and PQR are the given triangles. Let the perpendiculars from A, B, C to the sides QR, PR and PQ intersect at O. Take O as the initial point. Let a, b, c , p, q, r be the position vector of A, B, C, P, Q and R respectively. Since OA, OB and OC are perpendicular to QR, RP and PQ. a • (r - q) = 0 , b • (p - r) = 0 and c • (q - p) = 0 Let the perpendiculars from P and Q on BC and CA respectively intersect at the point X whose position vector is taken as x . It implies that (p - x)- (c - b ) = 0 and (q - x)- (a - c ) = 0 => p • (c - b ) = x • (c - b ) and q • (a - c) = x • (a - c) Adding, we have x • (a - b) = p • c - p • b + q.a - q.c = c • (p - q) - p • b + q • a = - p - b + q-a = - b • r + a • r = r • (a - b) => (r - x)- (a - b) = 0 XR is perpendicular to AB . Hence perpendicular from R to AB passes through X.

Alternative

solution:

Let the vertices of triangle ABC be A f o , y-,), B(x2, y 2 ) and C(x3, y3). Let the vertices of triangle PQR be P(p1t q,), Q(p 2 , q 2 ) and R(p3, q 3 ). Now the slope of QR is (q3 - q 2 )/ (p3 - p 2 ), so the slope of the line perpendicular to it is - (p3 - p 2 )/ (q3 -

q 2 ). Therefore the equation of the perpendicular from A to QR is

y - y , - ^ ( x - x , ) q3-$2 => (p2 - p 3 )x + (q2 - q 3 )y = (p2 - p 3 )xi + (q2 - q 3 )yi.. .(1) Similarly, the equations of the perpendiculars from B to RP and C to PQ are (p3 - Pi )x + (q3 - q,)y = (p3 - Pi )x2 + (q3 - q, )y2

... (2)

and (p, - p2)x + (qi - q 2 )y = (p, - p2)x3 + (q, - q 2 )y 3

...(3)

The lines (1), (2) and (3) are stated to be concurrent, i.e., they meet at a point. The x- and y-values on the LHS will then be the same at this point. The left hand sides of (1), (2) and (3) will add up to zero, giving (P2 - Ps)Xt + (p3 - Pi)x2 + (Pi - p 2 )x 3 + (q 2 - q 3 )yi + (q3 - qi)y 2 + (qi - q 2 )y 3 = 0 ...(4) In other words, equation (4) is equivalent to saying that the perpendiculars from the vertices of ABC to the sides of PQR meet at a point. But (4) can be rewritten in the form. (x2 - x 3 )pi + (x3 - xi)p 2 + (Xi -x 2 )p 3 + (y2 - y 3 )qi + (y3 - yi)q 2 + (y, - y 2 )q 3 = 0 ...(5) Equation (5) is the mirror image of (4), which is equivalent to saying that the perpendiculars from the vertices of PQR to the sides of ABC meet at a point. 8.

The four students can be arranged in 24 orders.

g Now the chance that the first selects IIT Kanpur = — 39

g Similarly the chance that second chooses IIT Delhi = — 39

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And so for the third's chance for IIT Bombay = And the fourth to choose IIT Kharagpur = Hence the required probability =

10 39

12

39 24.8.9.10.12

256

39.39.39.39

28561

There are pq seats which had to he occupied by Watson, Holmes and (pq - m -1) other passengers; and m - 1 vaccant. All selection of (q - 1 ) passengers or vaccat seats to complete Watson's coach are equally likely. But there are C^; 1 such possible selections of which C ^ ; 2 exclude Watson.

i-i Therefore the chance of Holmes not traveling with Watson is C ^ ; 2 + C ^ = (pq - q )

+

(pq - 1 )

And the complementary chance is 10.

Q 1 pq-1

For roots of x2 + px + q = 0 to be real and distinct we must have, p2 > 4q => p > 3 as q > 1 There are following Dossibilities; P Total cases q q 1 4 3,4,5,6,7,8,9,10 8 2 8 3,4,5,6,7,8,9,10 8 1 3 4,5,6,7,8,9,10 7 4 1 5,6,7,8,9,10 6 5 2 5,6,7,8,9,10 6 2 6 6,7,8,9,10 6 7 2 6,7,8,9,10 5 8 3 6,7,8,9,10 5 9 3 7,8,9,10 4 1 4 7,8,9,10 4 Total favorable cases 59 Total number of ways in which p ^nd q can be chosen with replacement = 10. 10 = 100 59 Required probability = 100

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RANKERS STUDY MATERIAL IIT - JEE, 2002

PRACTICE TEST - PHASE-III, IV MATHEMATICS SOLUTIONS 1.

0 < cos"1 x < n, 0 < sin"1 y < n/2, v x , y e R+ 0 < 2cos"1 x<2iz & 0 < 4(sin"1 y)2 < %2, :. 0 < 2cos"1 x + (sin"1 y)2 < 2n + TI2, Or, 0 < P7T2 < 7I2 + 271 or, 0 < p < 1 + 2/TC [p] = 0 or 1 If [p] = 0; tan"1 x + tan"1 y = [p] n/4 reduces to tan"1 x + tan"1 y = 0 which is not possible Since x, y e R+ If [p] = 1; tan"1 x + tan"1 y = [p] 7t/4 reduces to tan"1 x + tan"1 y = 7r/4 /

\

Or tan"1 x + y 1-xy Or A i X 1-xy 2.

=

!

7T 4

vxy < 1 & x, y > 0

1-x y = — - Proved. 1+x

Let AD = d & BE = I, AC = b

AE = 3 In A ABC; b2 + c2 = a2

•(1)

b^ In A ABE; — + c2 = I2 9 In A ABC; b2 + c2 = 2 d 2 +

4

..(2)

[3]

•(3)

[Appolonious Theorem] Or

b2 + c2 = 2d 2 +

Or

b2 + c2 = 2d 2 + -^(b2 + c 2 )

Or b2 + c2 = 4d2 Dividing (2) by (4), we get,

[from (1)]

(4)

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RSM12-PT-PH-III, IV(M)-MA-4 9I2

b 2 + 9c 2

9fBE

9 4

41 AD b +c 4d or b 2 = 3c2 /. From(1); 4c 2 = a 2

4 3

2

2

2

cos Z ABC =

3.

a2 + c2 - b 2

4c 2 + c 2 - 3 c '

2ca

2.2.c.c

=-

Z ABC = 60°.

Vl6-4sin 3 9-4sin 2 e + sine + V 5 - s i n 9 + 4cos 2 (l + sin9) = 7 3

...(1)

2

16 - 4 sin 6 - 4 sin 0 + 3 sin 6 = 16 - 2(1 - cos 26) + sin 30 = 5 = =

14 + 2 cos 26 + sin 30 - sin0 + 4 cos 2 0 (1 + sin 0) = 5 - sin 0 (1 -4cos 2 0) + 4 cos2 0 5 -sin 0 (4 sin2 0 -3) + 2 (1 + 2cos 0) 7 + sin 36 + 2 cos 26 equation (1) reduces to,

Vsin36 + 2cos26 + 14 +v / sin36 + 2cos20 + 7 = 7 Let sin 30 + 2 cos 26 = x Vx + 14 W x + 7 = 7 ( x + 1 4 ) - ( x + 7) = 7

..(2) ..(3)

Vx + 14 - Vx + 7 = 1 By (2) + (4); we get,

..(4)

[Dividing (3) by (2)]

2 Vx + 14 = 8 => x = 2 sin 36 + 2 cos 26 = 2 or 3 sin 6 -4sin 3 6 + 2 -4sin 2 0 = 2 or sin 0 (sin2 6 + 4 sin 0 -3) = 0 or sin 6 (2 sin 6 + 3) (2 sin 6 -1 ) = 0 1 -3 => sin 6 = 0 o r s i n 0 = [ v sin6 = — is not possible] ^ 9 = nit or 0 = nn + (-1)° — V n e I 2

4.

1 r_1C

X(- ) r=1

2-2

=(C0-C2 + C4-

f

\2

i2 z(-rc L r=1

2 r

_

\2 f

=(^-03+05.

2

J

(1 + x)n = C o - C,x + C2 x2 + Let x = ix

Cn xn

(1 + ix)n = Co - l(CiX) - (C2 x 2 ) - (C3 x 3 ) + = (Co- C 2 + C 4 ) + ix(C1 - C 3 + C 5 ) Putting x = 1, we get (1 +i) n = ( C 0 - C 2 + C 4 ) + ( 0 , - 0 3 + 05similarly putting x = -ix, we get, (1 - ix)n = (Co- C 2 + C4 ) - ix(C1 - C 3 + C 5 Putting x = 1, we get (1 -i) n = ( C 0 - C 2 + C 4 ) + ( C ^ C s + CsBy (2) x (3) we get, [(1 + i) (1 -i)] n = (Co - C2 + C 4 )2 + (C, - C 3 + C 5 -

...(1) )

...(2) )

)

...(3) f


RSM12-PT-PH-III, IV(M)-MA-4 or 2n = (Co- C2 + C4 f + ( 0 , - 0 3 + C5 2 or (C0 + Ci + C2 + Cn) = (Co + C 2 + C 4 f (Ci - C3 + C 5 Hence the result follows.

f )2

1 1 7 S = 1 + — + — + —— + oo ...(1) 6 18 324 since this is an infinite series it will be of the form (1 + x)n V n e I" or f Nn n(n-l) 2 n(n-lVn-2) (1 + x)n = 1 + nx + ^ 2! 3! comparing Tr from (1) and (2), we get, 1

nx =

n(n-l) 2!

6'

3

+

oo

..(2)

2_ 1 18

n M ) ,

36

2! n2 x 2

18

or, n - 1 = 4x

S = (1 + x) n = 1 - 1

^3 =

2

1

sin30

sin 3 6

2cos0

sin 60

sin 3 20

4COS 2 -1

sin90

sin 3 30

1 sin0

1 ( X = —+ 6

=> n = — 3

2

$12

f[

sin30

sin 3 0

cos 20

sin 60

3

sin30

sin90

sin0

^

beJtl&i k oi^'M^t 5 7 >1 3(9 - 2,2 i v \ t & - ^ - S / V ^ SiVtC35<\l® - LfUrfit® Sine,

S/Vi**^ *

sin 20 [ v sin 0 (4 [4 cos2 0 -1) = sin 0 (3 -sin 2 0 = sin 30] sin 3 30 t.Op

„ ^j,cos Si'n©
(Multiplying C, by sin 0) 1 sinG

1 sin0

3sin0-4sin3 0

sin30

sin3 0

3sin20-4sin320

sin60

sin 3 20

sin90

3

3sin30-4sin

3

30

(Applying 0 , = 30, -4C 3 )

sin 30

sin 3 0

sin30

sin30

sin 60

sin 69

sin 3 20 = 0 [ v Two columns are equal]

sin90

sin90

sin 3 30

Let z= x + iy then \z\ =

+ y2

Therefore the given equation becomes

2(x2 + y2) + ( x + iy)2 = 5 - i V3

or 2(x2 + y2) + (x2- y2) + i.2xy = 5 -i V3 or (3x2 + y2) + 2xy = 5 -i V3 Equating real and imaginary part, we get 3x 2 + y2 = 5 y2 =

2xy = -V3

4x'

+^- =5 or 12 x4 - 20x2 + 3 = 0 4x / „2 or (2x.2 - 3 ) (6x -1) = 0

.-. 3 x

2

c


tô^svyv^

RSM12-PT-PH-III, IV(M)-MA-4 3 1 11 x 2 = 3/2 i.e. x = ± , - or x2 = - i.e. x = ± J— 73 — 2x

y=

. ^ [3 _ V3V2 _ 1 when x =± J - , y = + — ^ = + - 7 = \2 2V3 V2

^ 1 _ V3 V6 _ 3 when x = ± ^ = , y = + = +—=• V6 2x1 V2 The required complex numbers are [3 Z

i_

" V2

_ [3

J~2 '

V2

J_ +

J

V 2 ' V6

3i_

1_ J3i_

V2 '

V6+V2"

We split the question in following cases (i) No objects are alike » There are 2n+1 distinct objects. So number of ways -

wn (ii) Two objects are alike — » n-2 objects have to be selected from 2n distinct objects So number of ways = 2nCn.2 (iii) Three alike objects. No of ways = 2nCn_3 and so on Total = 2 n + 1 C n + ( 2 n C n _ 2 + 2 n C n _ 3 + + 2n C 0 ) Now (1+1 )2n = 2n C 0 + 2n C 1 + 2n C 2 + ..+ 2n C n . 2 + (2nCn.1+2nCn+2nCn+1)+ But 2nCn+2 + + 2nC2n = 2nC0 + +2nCn.2 _ - 2 2"Cn_, 2n, => 2nC0 + " C i + n-2 ~ o

2n

Cn+2 +...+

2n

C2n

+

_

0

2nn

2n-1

-2

-

Total =

Un.-|

——-

2„c

2

n _

22n'1

J-

2n

C

-

2n+1

Cn + =222n2"n1.- 12n+0n_i (2n)! (2n-l)

n! n! 2

n!(n -1)1

= 22n"1+

Here AB = (i + 2 ] + k)-(2? + ] + k ) = - i + ] AC = (i + j + 2 k ) - (2? + ] + k) = - I + k .-. ABx AC = ( - i + j ) x ( - i + k) = j + k + i 1 to plane n => unit vector perpendicular to the plane n =+

i + 3j + k V3

Let PQ = A,; then OP =

v3

+ j + k) now the PA = (2? + j + k ) - ( i + j + k ) = i

And X = PQ = | projection of PA on PQ | unit vector along PQ| = + V3

V3

Clearly OQ = OP + PQ = i + j + k + A ( i + j + k) = (i + j + k ) ± - ( i + j + k) V3 3 = g ( i + ] + k) o r - ( ? + j + k) FIITJCC, ICES HOUSE (Opp. VIJAY MANDAL ENCLAVE), SARVAPRIYA VIHAR, NEW DELHI - 1 6 . Ph:6854102, 6865182, FAX: 6513942

RSM12-PT-PH-III, IV(M)-MA-4 But A, B, C and Q are non-coplanar because [ AB BC AQ] * 0 Position vector of the foot of the perpendicular = ^ (i + j + k), 10.

There are pq seats which had to he occupied by Watson, Holmes and (pq - m -1) other passengers; and m - 1 vaccant. All selection of (q - 1 ) passengers or vaccat seats to complete Watson's coach are equally likely. But there are C ^ ; 1 such possible selections of which C ^ 2 exclude Watson. Therefore the chance of Holmes not traveling with Watson is C ^ 2 + Cj^j 1 = ( p q - q ) + (pq - 1 ) And the complementary chance is

q-1 pq-1

.

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I

FIITJCC Rankers Study Material IIT- JEE 2002 PHASE - 1 - IV MATHEMATICS SOLUTIONS

OA = h cot30° = V3 h, OB = h cot45° = h Area of AOAB = - OA.OB sin60° = > 7 5 = 1 x V3 h.h — => h = 10 m. 2 2 2 OA24-OB2-AB2 on„ cos60 = 2.0A.0B

LJ • r«. AA Height of the +tower = 10 m

AB = VOA 2 +OB 2 -2OA.OBcos60° = J ( l 0 ^ f + ( 1 0 ) 2 - 2 . ( l 0 V 3 ) l 0 . 1 = V300 +100-100V3 Given f

= V400-100V3 = 1 0 V 4 - V 3 m

V

= f ( x ) - f ( y) j Putting x = y = 1, we get f(1) = 0 Now,

f'(x)

=Lim^X h->o

/

+

h

h

) -

...(1)

/

^ = L i m - ^ h

(From (1))

1 + —

= Lim h-»0

vXy f'(x) =

, • /(1 + x) L

since Lx->0 im^

x

=

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RSM12-PI-IV-T(M)-MA(S)-2

f(x) = 3 in X + c Putting x = 1 => c=0 => f(x) = 3 In x = y (say) 3

3

Required area = J x d y = j e y / 3 d y -co

-co

=

= 3(e - 0) = 3e sq. units

Given

f(x) =

Now

f( 1 - X ) =

• ••(1)

ax + ,1-x

ra

1 x

a + Va From (1) and (2) f(x) + f(1 - x) = 1 2n-1

i \

2n)

2n-1 ^n-i

r

M

I

_ \

—

1

-

1

2n-1

/

i

2n

+

...(2)

Va + ax

...(3) /

„ \

2n-1

/"2n-rA

2n

54s.

= 2n - 1

,

= 2n - 1

(putting 2n - r = t)

0 if a is even 1 if a is odd

Each time two balls are taken out and one ball is replaced, the number of white balls in the box either decreases by two or else remains the same. Therefore, if 'a' is even, the last ball cannot be white and the probability is zero. If 'a' is odd, the last ball must be white and the probability is one. x = (6) 1/3 v x = [x] + f where 0 < f < 1 given equation becomes x3 - (x - f) = 5 i.e. x3 - x = 5 - f =>4 2 we have x3 - x = x(x 2 - 1) > 2(4 - 1) = 6 > 5; Therefore 1 < x < 2, => [x] = 1 Now the original equation can be written as x 3 - 1 = 5 whence x3 = 6, i.e. x = (6)\1/3

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RSM12-PI-IV-T(M)-MA(S)-3 Without any loss of generality, the equation of the two circles can be taken as x2 + y2 = a 2 and (x - X)2 + y2 = b 2 Radical axis of the two circles is given by (x - Xf - x 2 - b2 + a 2 = 0 ...(1) -2Xx + X2 + a 2 - b 2 = 0 x=

A2

+ a2 - b2 2X PA = 7 (a cos Q - X f + a 2 sin 2 9 - b 2

Now,

2 2 2 PA = Va -b + X -2a?.COS0

•v/

P(a cosQ, a sin6) M

C1 (0, 0)

\

|

I

J

V

(X, 0)

y2 =

(x-X) Radical axis

and, PM

X2 + a2 - b 2

=

2X

and

C-|C2 = X

=>

2(PM).(C 1 C 2 ) =

- a cosG =

X2 + a2 - b 2 - 2aX cosG

X2 + a 2 - b 2 - 2 a X c o s 0

2X .X = X2 + a2 -b2-

2aX cos0 = (PA)*

Hence, (PA)2 = 2(PM).(C 1 C 2 ) |z-i - 11 = |z2 - 11 = |z3 - 11

=> The point corresponding to 1(say P) is equidistant from the points A, B and C. => P is the circumcentre of the AABC. Now if z-, + z2 + z3 = 3 then the point corresponding to centroid of the AABC is z1 + z 2 + z 3 _

_ g 0 c j r c u m c e n t r e and centroid coincide

=> AABC is equilateral. Conversely if AABC is equilateral, then centroid is the same as the circumcentre i.e. P. Hence centroid

Zl +

+ z

3

3 =1

=> z, + z 2 + z3 = 3. 1 1 Let tan"1x = 0, then -n/6 < 0 < n/6, as - ~
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RSM12-PMV-T(M)-MA(S)-4

Now cos"1

2x 1+x

2

.1 _2x_ + tan = cos 1-x2 "

2tan0 1 + tan 0

= cos"1(sin20) + tan"1(tan20) = cos" cos v \ n n n . ft = - - 20 + 20, as — < 20 < - and - < 2 3 3 6 1 2x ' • -1 f 2x V3 COS + tan 1 + x2 1 1 - x 25 Therefore J e x +1 J_ "V3

j vf r dx ax x J e +1

+

^ r dx ax x J e~ +1

2 tan 6

+ tan"

2

71

1 - tan 2 0

+ tan"1(tan20)

-20

y jt „ „ 5TC - - 26 < — = n/2 2 6 ^

7i dx = — 2

"1 V3 71 71 1 fdx = —X J ~ 2 0 2 73

dx ex+l S 7t

2^3

(a, b, c) lies on the plane 3x + 2y + z = 7 => 3a + 2b + c = 7

(1)

We have (a? + b] + ck). (3? + 2] + k) = 3a + 2b + c = 7

(2)

Also (a i + b] + ck)(3i + 2] + k) = 7 a 2 + b 2 + c 2 , 7 3 2 + 2 2 + 12 cos0

(3)

where 0 is the angle between the vectors ai + bj + ck and 3? + 2j + k. From (2) & (3), we get p o 3 49 7 a + b + c = — sec0 > - as L.H.S. is positive and |sec0| > 1. Equality holds if sec0 = 1, which is the case, when the vectors a i + b j + ck

and

3i + 2 j + k are parallel. 7

Hence least value of a 2 + b 2 + c 2 is - .

2

10.

X v Since there are exactly two points on the ellipse — + ~ a

2

= 1, whose distance from

b

centre is same, the points would be either end points of the major axis or of the minor axis. But

a2 + 2b2

> b, so the points are the vertices of major axis.

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RSM12-PI-IV-T(M)-MA(S)-5

Hence a =

V

a 2 + 2b2

Therefore e = , 1

11.

a 2 = 2b2

A/2 '

Given that f(x) + f(x + 4) - f(x + 2) + f(x + 6)

(1)

Replacing x by x + 2, we get f(x + 2) + f(x + 6) = f(x + 4) + f(x + 8)

(2)

From (1) and (2) we get f(x) = f(x + 8)

(3)

X+8

Now let g(x) =

Jf(t)dt X

=> g '(x) = f(x + 8) - f(x) = 0, by (3) g is a constant function. 12.

Here A(0) = 0 as first two rows of A(0) are identical. Thus lim

A(x)_„_A(x)-A(0) _A> - lim = A'(0) x->0 X - 0 x-»0

x-»0 X

f'(x + a) f'(x + 2a) f'(x + 3a) Now A'(x) =

f(a)

f(2a)

f(3a)

f'(a)

f'(2a)

f'(3a)

=> A'(0) = 0 as first and third rows of A'(0) are identical. Hence lim

A(x)_ = 0.

xÔ X

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FIITJCC ALL INDIA TEST SERIES IIT-JEE, 2002 MATHEMATICS (PART TEST -1)

SOLUTIONS 0 1.

0 0

For n = 1, A(n) = 1 4

0 0 = 0 = RHS. 1 0

[2]

Hence the result is true for n = 1. Let the result be true for n = m i.e. m m-1G _ (m - 1 ) m (m +1) m+1 A(m) = m+2C3 c3 m c 3 6~ ' m+3 /•» m + 2 m+1G ~ c3 °3 For n = m + 1 m+2

A(m + 1) =

M + 2 ^

m+4

mt3p

m+1

m

m+2

c3 m+1 c3

m+1/-.

ID+3Q

m 1

~ c3

c3 c3

c3 m+3Q

m

3

c3

m+2

c3

m

3 m+1^ + m+3 q 3 m+2 m+4 c3 c3 -

m+1,^ 3 m+1 c3

m+2i-> 3 m+3/-. U 3

m+1/-> m+2

c3

Operate Ci - C2 and C2 - C3 on the second determinant so that m+1/-^ m m f^ m+1 m 1 c 2 - m~1c3 c2 3 - c3 m+2 A(m + 1) = rm-2 r*3 m+1 + C 2 - mn3 m + 1 c 2 m + 1 c 3 m+3 /-> m + 2 m+1 m+3 Q m+2 C2 m + 2 c 3 c3 °3 3 3 Now operate C, - C 2 on the second determinant. We have m

A(m + 1) = A(m) +

C1 m+1/-> m+2Q

m 1

~ C3 m/->

~ 3

m

c2 m+1 c2 m+2

C2

m

c3

m+1

c3

m+2

c3

[2]

Now operate Ci + C2. We get C1 m+lQ m

A(m + 1) = A(m) +

m+2 q

m 1

~ c2 mQ

m+

^C

m

m

c2 m+1 c2

m+1

m+2

m+2

C2

o3

C3

Again operate C-i + C 2 so that

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AITS2002-PT1 -MA(S)-2 C 1 - m-1Ci m+1Ci m - C1 m

A(m + 1) = A(m) +

m+1

-

m

m

= A(m) +

m

Cl

2

m+1

m+2

0

C2

m+2

C2

m

c2 C2

m c3 m+1 f\

C2

m+1

m+2

c3

c3

m

1

m+1

c3 m+2/->

m

C2

= A(m) + 1

m+1

1

m+2

c3

c2

C2

m+2

c3

Operate R 3 - R 2 and R 2 - R-i.Weget m

1

c 2 = A(m) + m Ci. m+1C2 m+lA> • ^2

C1 m+1Ci

0

( m - l ) m (1m + l) — -

6

+

c3

m

m

A(m + 1) = A(m) + 0

=

m

C2

m(m + l) m.— -

m

C2. " ^ C ,

[4]

m(m-l) , A i.(m + 1)

2

2

m(m + l ) r , _ . m(m + lXm + 2) = — i [ m - 1 + 3 m - 3 m + 3 j = —^ ^ '-. 6 6 Hence the result is true for n = m + 1. Hence by mathematical induction the result is true for all n > 1.

[2]

The given result can be written as 2a cosA + 2b cosB + 2c cosC = a + b + c Using sine rule we get 2 sinA cosA + 2 sinB cosB + 2 sinC cosC = sinA + sinB + sinC => sin2A + sin2B + sin2C = sinA + sinB + sinC => 4 sinA sinB sinC = 4cos—cos—cos— 2 2 2

[4]

A . B . C . 8sin—sin—sin— = 1 2 2 2 => 4 cos

A-B

A+B

COS

4 sin2 — - 4 c o s

A

B

-

2

•

c

,

sin— - 1 2 c

• ,1 = n0 sin—+ 2

[3]

This is a quadratic equation in sin— which must have real roots. Hence 16cos2 But cos

A-B 2

-16>0

cos2

A-B 2

-1

A-B A-B 2 ^ 1 => cos 2 = 1 => A = B

Similarly it can be prove that B = C = > A = B = C.

[3]

. . sec46-3tan29 . Let r— = k, so that 4 2 sec 9 - t a n 9 (1+ tan29)2 - 3tan29 = [(1+ tan29)2 - tan29].k => 1+ tan49 - tan29 = [1 + tan49 + tan29] k => tan49(1 - k) - tan29(1 + k) + (1 - k) = 0

....(1)

[3]

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AITS2002-PT1 -MA(S)-3 Since tan2e is real, (1 + k)2 - 4(1 - k)2 > 0 =>[1 + k +2(1 - k)] [1+ k - 2(1 - k)] > 0 or ( 3 - k ) ( 3 k - 1 ) > 0

l 0 which is true, and - — - > 0 1-k 1-k 1

[3]

=> k < 1

[4]

Let AB be the tower of height b. Let BC be the flagstaff of length a.

S

The shadow of the tower and the flagstaff on the

/

C'

ground is AP = (a + b) cota When the flagstaff is tilted to BC', let the shadow be AQ. In the tilted position of the

/

,

flagstaff, let L be the projection of C' on the

/

/

/

t

/

/\

0\

/

/<

/

• \i

/

/

/

a

,

/

/

/ /

/ '

/

/

/

/

/

then C'M = a cosG.

/

M

/

ground. If BM is perpendicular from B to C'L,

/

/

/

s

/

/ /

A A«

Also AL = BM = a sine.

Q

Hence AQ = AL + LQ = a sine + ( a cos6+ b)cota

[4]

Hence the required distance = AQ - AP

[3]

= a sine + ( a cos6+ b) cota - (a + b) cota = a [ sine + ( cose - 1) cota] = a[ sina sine + cose cosa - cosa ] coseca = a [ cos(0 - a) - cosa] coseca

. e .

0

= 2a sin— sin a -

5(a)

e

2

coseca.

[3]

The given equation is j r ( x + k-l)(x + k)-10 = 0 k=1

or ^ [ x 2 + x ( 2 k - l ) + k 2 - k ] - 1 0 = 0 k=1

=>nx 2 + x £ ( 2 k - l ) + J V - £ k - 1 0 = 0 k=1

k=1

k=1

22 r / - n]-i + n(n + lY2n ^ + l) i.e . nx + x[n(n+1) 6

n(n + l)^ - 1 0 = 0. 2

10 = 0. ....(1) or x2 + nx + n - 1 — 3 n Since a and a + 1 are the roots of the equation,

[2]


AITS2002-PT1 -MA(S)-4 n -1

2a + 1 = - n and or + a = or a = (n + 1)2

n

+ 1 so that*

n + 1 _ ( n - l X n + 1)

4

10

2~ "

10

3

10 / Jn-1 1 => — = vin +11 — - + n 1 3 2

n~ n+1

=

4

(" + l X n - l ) 12

120 = (n - 1) n (n+ 1) or 4.5.6 = (n - 1) n (n + 1) => n = 5. (b).

Let d be the common difference of the AP. Then £

n

n

Ck Sk = £

k=0

C *[2a1+(k-l)d]

k=0

k=0

a = = = = = 6.

[3]

' k=0

- | ) , 2 - + ^[n.2-+n(n-l)2-]

[3]

3! n2n~1 + dn(n- 1)2n~3 a 1 n 2 " - 1 + n ( a n - a 1 )2 n " 3 n.2 n - 3 [ 4a! + a n - a-,] = n.2 n " 3 [ 3a, + an] 2"- 3 [2na, + n(a, + a„)] 2n~2 [na, + Sn],

[2]

1

Take the centre of the semi- circle as the origin, and the bounding diameter as the x - axis. Let (a, n) and (c, r2) be the coordinates of the centres of the two circles of radii r-i and r 2 respectively. Since the circles touch the semi- circle, 2 2 1 - r, = I a + r

=> 1 - a 2 = 2r,

....(1)

and 1 - r 2 = -jc 2 + r; => 1 - c2 = 2r2

....(2) [3]

Also the circles touch each other 2 => r, + r 2 = (a - c) + (r, - r j => (a - c)2 = 4r,r 2

-•(3)

[3]

From (1) and (2), we have n + r2= 1 -

a2 + c 2

< 1 - |a| |c|

Hence max^r, + r2) = 1 - |a| |c|


AITS2002-PT1 -MA(S)-5 when |a| = |c| => r-, = r2 = |a|

from (3)

Hence 1 - r, 2 = 2r, => r, = V2 Hence max.(n + r2) = 2 ( 7 2 - l ) . 7.

[4]

Let the incircle touch AB at F. Let O be the origin and let zu z 2 , z3 be the complex numbers represented by D, E and F respectively. Since C is the point of intersection of the tangents to 2ZiZ2 . Similarly A and B z, + z ,

the circles at D and E, C represents the complex number 2Z2Z3

represent the complex numbers and

Z2 + Z3

. Let r be the radius of the incircle 1+Z3 => |z-i| = |z2| = |z3| = r. Z

Equation of the line BO is z 2z,z 3

2 Z , Z

z

ZI+Z3

z,+z3

or z =

p ^ Z Z 13

V

z 2Z1Z1Z3Z3

3

r ' f c + z,)

D(Z1)

z ....(1)

Equation of line DE is

Z

z-z.

Zl

z2-z1

Z2-Z1 zr 2 z

~z1 . Z 2 Z1

V Z1Z3 z2-z1

[4]

. Where it meets (1), we have

r2

. Z1Z3 Zi r2 r2 z2

z

[3]

i

=> z _ = ( z i + z j z s which is represented by G. z, + z . 8.

[3]

Let the coordinate system be so chosen that the given straight line is x = p and the equations of the circles are x 2 + y2 = a2, x2 + y 2 = b 2 , x 2 + y 2 = c 2 . The line x = p cuts these circles at A, B, C respectively. The coordinates of these points are A [ p , -Ja2 — p 2 J , B(p t V b 2 - P 2 ) , c ( p , V c 2 - P 2 Equations of the tangents at these points are px + yja2 - p 2 y = a 2 2

2

px + Vb - p 2 y = b

px + Vc 2 - p 2 y = c 2 These tangents intersect at V - V a ^ V b ^

,

7a

2

-p

2

+

Vb

r r

?

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AITS2002-PT1 -MA(S)-6 P2-Vb2-P2 Vc2-P2

2 2 P2 - Vc - PJ— V g 2 - P—2 , Vc I 2 - p 2 +Va n "

- P2

[4]

Area of the triangle formed by the tangents at A, B, C is p 2 - v / a r - p 2 "Vb2

~2

A =

[3]

P2 - a / c 2 - p 2 Va 2 - p 2 -—— —

P ^ V ^ V V ^ P

/I vc

/ i — j P2 + v a ~P

2

V a ^ + V b

1,

2

^

2

1

2p

2

-p2 -

V ^ n [3]

2p CA . BC . AB 2p

Let the two points be P(at! 2 , 28^), Q(at22, 2at2). Since the normal at P and Q meet the parabola at R, we have _

2 -_ =- t ti

t l

2

2 - - => t-|t2 = 2 t,

Also T is [ a t ^ , a(ti +12)]. Hence the vertices of ATPQ becomes P(at12, 2ati), Q

f 4a A

A > 4a

t2 V l1

tl 1 y

r 2^

f , T

2a, a t , + t 1J j V v Let (a, (3) be the centroid of ATPQ. t2 + 7 T + 2 a =

=> a v

v

' rH y,

V

[4]

- 2 /

_

V

I

1

= 3a + 2a

4 2 and p = a 2 t + — +1, + — = a V r t1 t1v v 2 Hence p = a(3a + 2a). => locus of (a, p) is y 2 = a(3x + 2a).

[4]

[2]


AITS2002-PT1 -MA(S)-7 10.

The equation of the tangent to the parabola y 2 = 4ax is y = mx + — . It passes through the m focus ( - ae, 0) of the ellipse => 0 = - aem + — m

m =

Hence the equation of the tangent becomes y

x + ae

1

VT + e

= r

[3]

VT + e

Any point on it

rVe +e

-ae,

vt+ e

Since it lies on the given ellipse, we have C r~ \2 1 rVe ae =1 b 2 (l + e) 7 T+~e v or r*(e + 1 - e3) - 2raeVe(l + e) (1 - e2)2 - a 2 (1 - e 2 ) 2 (1 + e) = 0

[3]

If r, and r 2 are the roots of this equation, then the length of the focal chord = (n - r2) =

A /(r 1 +r 2 )

2

-4r 1 r 2

-J4a 2 e 3 (l + e)M J

2

+4a 2 l

! 3 e 2 F (l + e)(l + e - e ) (1 + e -e 3 ) I

2 a ( l - e 2 ] j ^ + e ) y l e 3 +1 + e - e ' 1+e-e3 _ 2 a ( l - e 2 ) ( l + e) 1 + e-e3

[4]


FIITJCC ALL

INDIA TEST SERIES

IIT - JEE, 2002 MATHEMATICS (PART TEST - il)

SOLUTIONS Each of the friends was present for seven dinners and each one missed seven dinners. Hence the total number of dinners = 14.

[3]

(a) Number of dinner when all of the friends are present = 1 (b) Number of dinners with every five of them = 6C5 = 6. From (a) and (b) we find that with every five of the friends I had two dinners. Now if I select a group of four, then this group is present once in (a) and twice in (b). Similarly any group of 3 is present once in (a) and thrice in (b). Any group of 2 is present once in (a) and four times in (b). In the above manner each one had dinner from (a) and 5 dinner from (b) making a total of six dinners for each one.

[4]

(c) Hence each one had dinner on one day with me, making it six more dinners. The total of (a), (b) and (c) is 13 dinners. Hence I had 1 4 - 1 3 = 1 dinner alone.

[3]

Let O be the initial point. Let

a, p, y be the position

vectors

respectively

of

A,

B

and

C

so

that

jaj = a, |p] = p, jyj = y .

Since OP is the bisector of ZBOO,

OP = X

v p

y

where A. is a scalar to be determined. Equation of the /

line OP is r = X

\P

y/

Equation of the line BO is r = p + t(p - y) /

For the point P, X Pp

+

7

y

[3]

= P+ t(p-y)

=> — = 1 +1 and — = - t y

P

PY

=> X= -^-t—, so that P+y

OP =

Py P+ y

P +1 P

y

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AITS2002-PT1 -MA(S)-2

Similarly OQ =

ya

y

a

y + a ^y

and OR

a

ot(3

a —

a +p a

B +

[3]

—

p

Area of APQR = area of POQ + area of A QOR + area of A ROP Vector area of APQR = | (OP x OQ)+ | (OQ x QR)+ 1 (OR x OQ) . Since OP, OQ and OR are coplanar, vector area of APQR = 1(0PX0Q + 0QxQR + 0RX0Q) apy'

^pxy^pxa^yxa^

(p + yXct + y)

ap

PY

J

py

a 2 py

v x a + -—yx p + a x p (y + aXa + p) ya yp ap ap 2 y

f axpâxy^pxy^

(p + yXa + p) 2a Py (a + pXp + yXy + a ) ' 2 2a Py (a + pXp + yXr + a) Area of APQR

ay

Py JJ

(axp + pxy + yxa)

vector area of AABC. 2a Py

[4]

(a + pXp + y\y + a)

Area of AABC

3(a).

ap

2!

1 We have F(x) = - + x 2x(x +1)

•

+

3x(x + iXx + 2)

+ .... to

QO

.

Hence F ( x ) - F ( x + 1) I

1

x

1

+-

1

1

1! 1

1 2

x (x +1)

2!

x(x + iXx + 2 )

F(x) - F(x+ 1 ) -

_

+

1

J_ x

1

+

2

x(x + iXx + 2)

x (x + iXx + 2 )

1!

2!

x(x +1) ' x(x + lXx + 2) ' x(x + iXx + 2Xx + 3)

-

x(x + iXx + 2Xx + 3)

2! +

....to oo

+ .... to 00 .

2! • +

-2!

= lim n—

x(x + iXx + 2\x + 3)

1!

2

= |im

•

2 x ( x T l ) " 2(x + iXx + 2 ) " 3x(x + ^ x + 2) ~ 3(x + ^ x + 2Xx + 3)

x +1

x(x + 1)

2!

2! • +

x(x + iXx + 2Xx + 3)

+ .... tO oo

+ .... to CO

+.... tO co

z M x 2 (x + iXx + 2)... .(x + n - l)(x + n)

[3]

- 1

x2|%1

+1

n-1

+1

+1

Since x > 0 and all the factors in the denominator are greater > 1, the limit on the RHS ->0. Hence F(x) - F(x - 1) - \ x

= 0.

[2]



(b).

ZXCY = - - A . Let ZCXY = a. 2 Applying sine rule to AXYC, we have R XY YC = 2r' sina n sin| a + - A sin| - A 2 y where r' is the circum radius of AXYC. As Y

> C, XY -> R and a R R

2r' =

cos A

[2]

0. Hence in the limit

[3]

2 cos A

Let the curve be y = f(x). Equation of the normal at any point (x, y) is Y-y = - _ - L - ( X - x ) dy/dx .e.

~ ( Y - y)+ ( X - x ) =0. dx (a-x)+£-(b-y)

Its distance from (a, b) is

1+

dy

>2

dx

The distance of the normal from the origin (0, 0) is x+y

L'

j 1+

\

dy

(a-x)+£(b-y)

dx d

y'

=k 2

1+

]

Idx (a-x)+|(b-y)

dy

[4]

\2

dx

dy x+y — V dx or [k(a - x) ± x] dx + [k(b - y) ± y]dy = 0 => - | ( a - x )

2

=

+

±y-|(b-y)

2

±y

= constant.

[4]

Since the coefficient of x2 is the same as that of y2 and there is no term of xy, the above solution represent a family of circles. For k = 1, a special case is a straight line.

5(a).

I= J

[2]

secx dx 2tanx + s e c x - 1

= Jf ^ = 2 s i n x +1 - c o s x cosec - 2 w z J

2

J

dx f — X X . . x 4 s i n " c o s +2sin 2 2 2 2

dx

[3]

x 2 cot +1 2

Let 2 c o t - = t 2

cosec — dx = dt 2



Hence

— f - ^ L = —L| n |l +1|1 + constant ? Jt + 1 ? ! 2 x = - - I n 1 + 2cot + constant.

[4]

2

(b).

We have 2x

I=

1 + x2

o =

=

V -r 1

= (tan -1 x f f

6.

1 + x2

1 2 tan 1x f ( " 2 )dx J 1+X 0

n2

f _2

16

v36

71

+1

/

V3

2x

^Sin

dx

dx + f t

A

2

dx

1

1+

x y V2

2

( C ° r 1 x2) dx J 1+x

-(cot"1xf^ 7t 2

7%

16,

72

[2]

For x = 0, y = 0, f2(0) + 2 = 3f(0), and since f(0) * 2, f(0) = 1. Also for y = 0, f(x) f(0) +2 = f(x) + 2f(0) => f(0) = 1 For y = 1, f(1) f(x) + 2 = 2f(x) + f(1) => f(1) = 2. f r f X 1+ •f(x) f X + h f V v J) Nowf(x)=lim ( ) - ^ -= nlim f(x)f = lim h->0

[3]

j + 2-f(x)X, h

I

[2]

(i1 + hl x

\

= lim h-»0

= lim h->0

(f(x)-1> f t f n ^ - 2 I V X; J

f(x)-1 X

= f(x)~1.f(l) x f(x)

2

I

h/x =

<( \

X

•f(1)

X

!feLl.2

[4]

2

X

f(x) W = f _ _ £ dx + c •+c x2 J x3 Since f(1) = 2, c = 1, so that f(x) = 1 + x2 f 3 > x3 2 Hence 3 jf(x)dx - x(f(x)+ 2) = 3 x + — + c'1 - x ( 3 + x ) V 3 / = c-1, (a constant).

[4]


AITS2002-PT1 -MA(S)-6 Let the equation of the circle be x2 + y2 = a 2 If L is the latus rectum of the parabola, we take the equation of the parabola as y2 = L(x + a)

(- a, 0)

Any point on the circle is (acos0, a sinG) and since this point also lies on the parabola a2 sin20 = L (a + a cos9) => L = 2a sin2 - .

,3/2 j T U x + a]dx = 2 X - ( x + a)

Area of the given segment = 2 . 16a2 . 9 2a 3 9 i.e A = sin-cos — = 3 2 2 3

[3]

v

. . . (2sin9 + sin29) '

acosu

-a

[2]

dA 2a2 r_ . _ _ol 8a2 9 39 , Hence — = — - 2cos9 +2cos29 J = c o s - c o s — = 0 for max. or min. d9 3 3 2 2 71 9 = n or 9 = [3] and

= — f - s i n 9 - 2 s i n 2 9 Jl < 0 for 9 = 3 3

d9

Hence area is maximum at 9 = — 3 And L = 2a sin2— = 2 a . - =

8(a).

[2]

Since t2 + 2xt + 4 = 0 does not possess distinct real roots, 4x 2 — 16 < 0 => x2 < 4

- 2 < x < 2.

Slope of the tangent at any point (x, y) is dy dx

= 3x - 4 x + 1

u which has a max. or min for dx 2y _ 0 = 6x - 4 i.e x = — 3

Hence

— = 21, dx x=2 dx x=2

At x = - 2, y = - 8 -

5

' dx X-2/3

1 3' °

8 -- 2 = - 1 8 .

Hence the required equation is y + 18 = 21 (x + 2). (b).

R.H.S. =

[3]

[2]

bx[((pxc)-a)q-((pxc)-q)a]+c(qxb](pxq)-(c-(pxa))(qxb)

= b x l(a x p)- c)q - ((c x q). p)a]+ [b • (c x q)](p x a ) - (c • (a x p))(b x q)

[3]

= a x (c x q)x (b x p)] = a x (q x c)x (p x b) = L.H.S

[2]

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6513942

AITS2002-PT1 -MA(S)-6 Let E be the event that two balls from the bag are white. Let A| be the event that exactly i white balls are in the bag so that 2 < i < 5. j , i = 2, 3, 4, 5 4 Hence the probability that all the balls in the bag are white

=> P ( A j ) =

5

=P

_

' E

A

p

J

P(A 5 )

z r -KM

[3]

vAi y

i=2 i=: 5

[3]

Cn

2(~* r\ Uo 3Uo 4/^ Uo 5Wo + + + 5 2 5 2 5f-v2 5 2 10 20

1+3+6+10

[4]

2

10(a). We have to find the area of the region bounded by ABCDOA 0 4 2 = |(l6-x )dx+J(x-4)2dx (

[2] y =(x-4)

y 3A°

(*-4f

16x- —

6 4 - ™ + —[2] 3 3

= 64 sq. units. a

(b).

n

y =16 - x

[1]

n

k

4

n

n

k+l

i i m J L y n c Ji_ = iim_Ly"ck^— n — n k + 1 n-»-nnj^ k +1 1

n

n

1

n

= lim — Y n C k f x k d x = lim — y nCkx n n_>00 n n i~o o 0 Vkô

,.

1V

y, .

dx

[3]

(1 + x r

= l i m —n J (1 + x r d x = n lim v n , ' , n-*» n 0 -»« n (l + n) = Mm <1 +n ~ 1 = lim 1 + - I n->» n n (l + n)

= e.

[2]

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6513942

f l l T J € € RANKERS STUDY MATERIAL IIT - JEE, 2 0 0 2 PRACTICE FULL TEST • II MATHEMATICS SOLUTIONS 1.

| Z 1 - 1 | = |Z2-1| = |Z3-1| => The point corresponding to 1(say P) is equidistant from the points A, B and C. => P is the circumcentre of the AABC. Now if Z l + z2 + z3 = 3 then the point corresponding to centroid of the AABC is z i

+ z

2 + z 3 = 1. go circumcentre and centroid coincide 3 => AABC is equilateral. Conversely if AABC is equilateral, then centroid is the same as the circumcentre i.e. P. Hence centroid

Zl + Z l +

3

=1

=> Z l + z2 + z3 = 3. 2.

Let (at2, 2at) be a general point on the parabola y2 =

Ay

4ax, where t is any real number. Equation of the line containing a chord of the circle x2 + y2 = a2, bisected by (at2, 2at) is at2x + 2aty = a2t4 + 4a2t2 or

at [at3 + (4a - x)t - 2y] = 0

(1)

since (1) passes through a point of x - axis, say(h, 0), so

at[at3 + (4a - h)t] = 0

(2)

(2) will have exactly three distinct real roots, provided 4a - h < 0 i.e., h > 4a. Distinct roots of (2) are t = 0, ±

43

Solving the equations of the given circle and the given parabola, we find that Points A and B are corresponding to at2 = a(V5 - 2) or t2 = (V5 - 2) Thus for points of parabola inside the circle, t2 < V5 - 2 fllTJCC. ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Ph: 6865182, 6854102. Fax: 6513942

RSM12-PT-II (M)-MA(S)-2 _ h-4a _ But from (2), t 2 = — — . So — — < V5 - 2 or, h < a(V5 - 2) + 4a Thus required points are (h, 0), where 4a < h < a(V5 + 2). f f x ) = 3x2 + 2ax + b y = f(x) is strictly increasing f '(x) > 0, V x => (2a)2 ~ 4 x 3 x b < 0 => a 2 - 3b < 0 this is true for exactly 15 ordered pairs (a,b), 1 < a, b < 6 namely (1, 1), (1, 2), (2, 2), (1, 3), (2, 3), (1,4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5), (1, 6), (2, 6), (3, 6) and (4, 6). Thus, required probability =

15

5

36

12

Let the sides of AABC be respectively ar2, ar and a. Since r > 1, the biggest angle is A and smallest angle is C. Given that A = 2C. Now using sine rule sinA

sinC

ar

a

sin 2C

= sinC

cosC =

(as sinC * 0)

Now A + B + C = 180° => 3C = 1 8 0 ° - B

(1)

But C < B < A

180° - A < 180° - B < 180° - C

=> 180° - A < 3C < 180° - C

(Using (1))

=> 36° < C < 45°

(as A = 2C)

=> cos45° < cosC < cos36° => _

1

42

r e

r2

< —

2

M -

<

V5 +1 4 N/5+1

(a, b, c) lies on the plane 3x + 2y + z = 7 3a + 2b + c = 7 We have ( a i + b j +ck). (3i + 2 ] + k) = 3a + 2b + c = 7

(1) (2)

Also (ai + b] + ck)(3i + 2\ + k) = Va 2 + b 2 + c 2 .a/3 2 + 2 2 +12 cose

(3)

where 8 is the angle between the vectors ai + bj + ck and 3i + 2j + k. From (2) & (3), we get !FIITJ€€, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 1 6 . Ph: 6865182, 6854102. Fax: 6513942

RSM12-PT-II (M)-MA(S)-3 49 a + b + & = -— sec© > -

as L.H.S. is positive and |sec0| > 1.

Equality holds if sec0 = 1, which is the case, when the vectors a i + b j + ck and 3 i + 2 j + k are parallel. 7 Hence least value of a 2 + b 2 + c 2 is —. 6.

Since there are exactly two points on the ellipse

x2 a

+

y2 b

= 1, whose distance from centre is

same, the points would be either end points of the major axis or of the minor axis. a 2 + 2b2 But J — - — - > b, so the points are the vertices of major axis. Hence a =

a 2 + 2b2

a 2 = 2b2 Therefore e = J1 -

V2

Let x^ x2 and x 3 be respectively the number of coins of Rs. 1, Rs. 2 and Rs. 5. Then required number of ways is the number of integral solutions of X! + x2 + x3 = 5, where 0 < x1f x2, x 3 < 3. Now number of solutions = coefficient of x5 in (1 + x + x2 + x3)3 = coefficient of x5 in (1 + x)3 (1 + x2)3 = coefficient of x5 in (1 + 3x + 3x2 + x3) (1 + 3x2 + 3x4 + x6) = 12

8.

Let m be the slope of the common

y

tangent AB of the given circle and ellipse

in

the

first

B

quadrant,

intersecting x and y axes at A and B

(T

respectively. Then cVl + m 2 = yla2m2 + b 2 2

m =-

c -b

\

T> P A

K

2

a2 - c 2

•I x

'

Thus equation of the line passing through A and B is y = mx + c V l + m 2 . Hence

( 0,c, y

a a

-b 2

-c2

Due to symmetry the quadrilateral formed by the common tangents will be a rhombus having a2-b2 diagonals along axes of coordinates of length 2 c J — — — and 2c c -b a -b2 Hence required area = — x 2c 2 2 Vc -b

?r X

a

-b 2

V a -c

2

_ "

a2-b2 a2 - c 2

"

2c 2 (a2 - b 2 ) V a ^ V ^ b ^ '

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RSM12-PT-II (M)-MA(S)-4 Here A(0) = 0 as first two rows of A(0) are identical. Thus | i m ^ ) x-»0 x

=

,imA(x)-A(0)=A,(0) x-»0 X- 0

f'(x + a)

f'(x + 2a)

f'(x + 3a)

f(a)

f(2a)

f(3a)

f'(a)

f'(2a)

f'(3a)

Now A'(x) =

=> A'(0) = 0 as first and third rows of A'(0) are identical. Hence lim ^ ^ X 10.

= 0.

Let A P be the tower of height h. We have AB = h cota and CA = h cotp. Therefore BC = AB = h cota In AABC, using cosine rule, we get cos108° = -sin18°

*

2h2 cot 2 a

2cot2a-cot2p 2 cot 2 a

-2sin18°.cot 2 a = 2cot 2 a - cot2p 2cot 2 a (1 + sin18°) = cot2p 2cot a 1 +

cot 2 a

V5 - 1 "

V5+3

= cot p

= cot p

(3 + V5)cot 2 a = 2cot2p.

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FIITJCC

RANKERS STUDY MATERIAL I I T - JEE, 2002

PRACTICE FULL TEST I MATHEMATICS SOLUTIONS 1.

Equation of tangent to the curve, y = f(x) is ; (Y - y) = f'(x) (X -x) X

Equation of tangent to the curve g(x) = yi = [f(t)dt is ; ( Y - y i ) = g'(x) (X -x) = f(x) ( X - x ) Given that tangent with equal abscissas intersect on the X-axis, y y! f(x) y, - =x ^ —= _ f'(x) f(x) f ( x ) f(x) f x ( ) - f '(x) ^ g'(x) _ f'(x) y, ~ f(x) g(x) f(x)

—s x — ^ =>

g(x)

=>

g(x) = Ce kx

=> g'(x) = kCe k x =>f(x) = kCe kx y = f(x) passes through (0, 1) => kC = 1 yi - g(x) passes through (0, 1/n) C = 1/n = > k = n => f(x) = enx 2.

x2 Slope of tangent at (X!, yi) = ~ V? The tangent cuts the curve again at (x2, y2) y 2— - y—i => Slope of the tangent = — x2 -x-i Also, x^ + y^ = a3 and x\+y\=

x1

_y2-y1

y2

a3

x?+y? = x

i-x2

y2+y2+yiy2

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RSM12-PT-II (M)-MA(S)-2

X^2 X.j2 + 2 + ~2 ~~ 2 2 y? y r + y2 + y i y 2 22 22 2 x 2 2 + x 2 2 + x 2 i yf i V2 i V1V2 = y f x f + x^y^ + y^x-|X 2 x

f y f - y l x ? = xiVi[xiy2 - x 2 y i ]

x 2 y-i + y 2 X i = - x 1 y 1 : 3.

yi

At x = x0 => y = Xq Equation of tangent at (x0, x 2 ) is 1 / 2\ - (y + Xq ) = x0x Now coordinate of point A from figure= (x0, x 2 ) AP = x0 For coordinate of point Q; x = 0 J:(y + x 2 ) = 0 y =

-Xq

PQ = OP + OQ = Xq

+

Xg = 2Xg

APAQ = - AP x PQ 2 A = — X02 Xq = Xg .', x 0 e [1, 2]

Maximum area = 23, at x0

The equation of the tangent to the ellipse at any point (a cos9, b sin9) is

xcos

^ +

=1

It passes through the focus (0, c) of the parabola => sine =b/c x I b^ v Hence the equation of the tangent becomes —. 1 — - + — = 1 a v c c If the point (2ct, ct2) lies on the tangent, then at2 +2tVc 2 - b 2 - a = 0 If t-i, t 2 are the roots of this equation, then t ^ = -1. Hence points t-i and t 2 represent the extremities of the focal chord. => length of the chord = Jc 2 (t* - t 2 ) 2 + 4c 2 (t 1 - t 2 ) 2 = c A

- t j j f t , + t 2 ) 2 + 4 j = c V f e + t 2 ) 2 - 4 t , t 2 jfct, + t 2 ) 2 + 4

= c[ (t, +t2)2 +4] = c

~4(c2-b2)H

= 4c 1 +

c2-b2

Given equation x2 +y2 +8x - 10y +40 = 0 represents a circle centered at C(-4, 5) and of radius 1. Let P be the point (-2, 3) ~3 =— --1 -4+2 -2 => Inclination of CP with the positive direction of the x-axis is 135°. Equation of CP is Slope of CP =

5



x+4

y - 5 ,= sin 135°

cos 135°

±1=>

*±4 1

y^5 J^

= ± 1

V2 V2 If Q and R are the points where CP intersects the circle, then

5,

vr

1

CU

1

1

R=

1 4

V2~ ' Now, b = min.[(x +2)2 +(y -3) 2 ] = PR: \2 2 -

_1_

V2

\2

9 _ _4_ = 2 4 + 1_J_ = 2 2 V2 2 V2

72

\ 2

2

a = max[(x +2) +(y -3) ] = PQ = 2

Let

9

4

2

V2

2

= I2+

= 9-472. \2

2

2+

J~2

= 9 + 4V2 => a + b = 18.

sin 3 6

cos 3 0

sin(20 + a )

cos(20 + a )

...(1)

=k

cos 0 sin 3 0

sin 0 cos 3 0

cos 0 sin 3 0 + sin0cos 3 0

cos0sin(20 + a )

sin0cos(20 + a )

cos 0 sin(20 + a ) + sin 0 cos(20 + a )

sin 0 cos 0 k

''

+ 5

~V2

sin 20

sin(30 + a ) ~ 2sin(30 + a )

comparing the values of k we get,

cos 20

sin 20

cos(30 + a )

2sin(30 + a )

or tan 0 = 2tan (30 + a) The four points zu z 2 , z3 and z4 are concyclic if ZADB = ZACB (or n - ZACB) arg

Z

V 1

_Z

z

= arg

Z

2

Z

Z Z

z

D(Z 4 )

3

V 1~ 3

4

z

C(z3)

J

z

arg ( 2 ~ 4 X i - 3 ) ' = 0 (zi-z4Xz2-z3), Z Z Z 4 X 1 Z 3 ) purely real i.e ( 2 Z (Z1 Z 4 2 Z 3 ) " Using the relation z^ = -z 2 and Z1Z2 + z3z4 = 0,

B(z 2 )

A(zJ

X

(Z2

Z

3 ) _ Z 1 Z 2 Z4 Z 1 ~ Z 2 Z 3 + Z 3 Z 4 Z Z z , z 2 - z 2 z 4 - z , z 3 + z3z4 Z1 4 X 2 - z 3 )

( =

— = -1 a real number. Hence the result. 1 VZ 4 Z 3 / Note: We have considered the case of zu z 2 , z 3 , z4 in cyclic order and leave the other case for you to prove. Z

OJ

1 Letl(rn) = mC1 - fl1 + — m C 2 + f l1 + —+ — 3J I 2J _

m+1c

1

1 1 f 11 + - m+1C2 + (< 1 + —+ 1 I 2 3 l 2;

J

m

C3+..

m+1

\

2

3

m

c3+....


RSM12-PT-! {M)-MA{S)-4

+ n

_

(

l ) r

C0 + m Ci - 1 +

'

i

1 1 1 1 + — + — + ... + —m + 1 ,C m + ( - l Y f , ( l + — + ...+—-— m+V'm+1 my 2 3 V 2 m+1

1 11 (mC1+mC2)+^1 + | + l j ( m C 2 + m C 3 ) + .

(-1r -1

= I (m) + m C 0 -

1+

1 + - + .... + - - ( - C m . 1 + m c m ) + ( - i r f i 4 + . . . + m +lj 2 m

n

-

2

C1

' +

1 1 1+ —+ —

2

"C 2 +....(- i j r 1 f i + l + . . . + l ] m

3

(-1 r^

1

Given function f (x) can be rewritten as f (x) =

f(x-1) =

0,

x - 1 e ( - co, - 1 )

1 + (x-1)

x - 1 e [-1, 0]

1-(X-1)

x - 1 e (0,1]

0,

x - 1 e (1, oo)

aiso, f (x + 1) =

o r f (x + 1) =

0,

x e ( - oo, - 1 )

1 + x,

x e [-1, 0]

1-x,

x g (0, l ]

0,

x e (1, oo)

or f ( x - 1 ) =

x +1 e ( - oo, - 1 )

1 + (x + 1)

X

+ 1 G [ - 1, 0]

1 - ( x + 1)

X

+ 1 G (o, 1]

0,

X

+1

0,

x < -2

2 + x,

-2 < x < -1

- x,

-1 < x < 0

0,

x >0

Now g (x) = f (x - 1 ) + f (x + 1)

1 m+1 m +1

0,

G

c m _i

m<-> _

= I (m) - I (m) + m C 0 - 1 mC1 + 1 •^C2 +....(- I f

9.

m

m

0,

x<0

x,

0 < x <1

2 - x,

1
0,

x >2

(1, oo)

0,

x < -2

2 + x,

-2 < x < -1

- x,

-1 < x < 0

x,

0< x <1

2 - x,

1
0,

x>2

It is easy to check that g (x) is continuous V x G R and non differential at x = - 2 , - 1 , 0, 1 , 2 and differentiable elsewhere.

6.

Xi

Yi

Zi

If the given vectors are co-planer then x 2

y2

z2 = 0

x3

y3

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XiX + yiy+ ZiZ =0 x2x + y 2 y+ z 2 z =0 X3X + y 3 y+ z 3 z =0 have a non-trival solution. Let the given set have a non-trivial solution x, y, z without loss of generality, we can assume ihat x > y > z . For the given equation X1X +y1 y + Z1Z = 0 , we have x ^ = - y ^ -z^z => |xix| = |yiy + z-fzj < |y,y | + |ziz| => |x,x| < |yix| = |z-ix| => |xi| < |yi| +

.

Which is a contradiction to the given inequality i.e, |xn|> ly11 +|zi|.

Similarly the other

inequalities rules out the possibility of a on-trivial solution. Hence the given equation have only a trivial solution. Hence the given vectors are non coplanar.

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FIITJ€€ ALL INDIA TEST SERIES IIT - JEE, 2002

FULL TEST - IV (Mains) MATHEMATICS SOLUTIONS 1

10

1(a).

I=

x+-

2

Jx-e -10

Now

x+

dx=

-x+—

2

J(-x).

dx

-10

r

-I x + ^ | + 1

1 if x * odd multiple of —

1 x+ -

2

[2]

1

10

=>l=

1

10

x+—

2

J-x-e

dx = - I => I = 0.

[2]

-10

k

(b).

I ( k ) = f[x]{x 2 }dx = j [ - x ] { x 2 } d x -k

-k

k

k

2

= J(-1-[x]){x }dx - - J{x 2 }dx - I (k) -k

(x is not an integer)

[3]

-k k

=> I (k) = J ( [ x 2 ] - x 2 ) d x 1 ' . -12 ja V4 = Jodx + Jldx + | 2 d x + J*3dx + 0 1 ^2 V3

k

+

3

J(k2-1)dx- — Vk7^

I (k) + — = 0 + 1(V2 - 1 ) + 2(V3 - V2) + 3(V4 - V3) + o l(k)+ ^ - = - 1 - V 2 - V 3 - V 4 o i(k)= (k-i)k(k

+

i)-^-k^;1VF. 3

2.

+ (k2 - 1)(k - V k ^ l )

- V l ^ - 1 + k ( k 2 -1) [3]

r=1

Total number of years listed = 50 50 Number of leap years among them = — - (the years 1700, 1800 and 1900) = 25 - 3 = 22. So the number of ordinary years = 50 - 22 = 28 [3] Now favourable cases for a prime or odd number from {1, 2, 3, 4, 5, 6} are {1, 2, 3, 5}. FIITJC€

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AITS2002-FT-4-MA(S)-2

4 2 So probability of getting a prime or odd number = — = — 6 3 Let A be the event that the superman is on mars. Bi be the event that a year is leap. B 2 be the event that a year is ordinary. 22 11 Now, P(B0 = — = — 50 25 P(B

-

28

[2]

14

11 So, P(B 1 /A) =

P(B 1 )P(A/B 1 ) P^JPCA/BO+PCBaWA/Ba)

1

X —

25 3 1 1 ^ 1 4 ^ 2 25 3 25 3

Let P : (h, k) then for points A and B 2

2

(h + rcos©) (k + rsinO) , , , . , —— + — — = 1, where tan 0 = 1 2 2 a b r^cos 2 0 b 2 + a 2 sin2 0) + 2[b2h cos 0 + a 2 k sin 0]r + b2h2 + a2k2 - a2b2 = 0

=>

=>

2[b2h cos0 + a 2 ksin0] b ^ 5—5— = ±— b cos G + a sin 6 2 2

[3]

n

11

[3]

39

I

y

I

Ao/

V

A'0

/

0

\

x

Bo/

2

V2(b b => — - h + a k) = + — 2 2 b +a "2 , ,2 2, b(b2+a2) => hb2 + a2k = ± — jJ2-2-V2 Hence locus of P is xb2 + a2y = ±

b(b2+a2) 4V2

which will represent A0 A'0 and BoB^

Also if in the given figure AqB0 =

b

[3]

then P can lie any where on the line A 0 B 0 except line

segment A0B0. Similarly P can also lie any where on A' 0 B' 0 except line segment A' 0 B' 0 . Hence, for A0B0 and A'0 Bq b 2ab V(1 + m 2 )(a 2 m 2 + b 2 - c 2 ) [2] 2 a2m2 + b2 a 2 +b 2

= V 2(a 2 + b 2 - c 2 )

4a

(a 2 + b 2 ) 2 2x 16a = > c

2

= a

2

Co = ±

2

+ b

(as m = 1)

= a2 + b 2 - c 2 2_(ai+b^ 32a

= (a2 + b 2 )

(a + b )(31a - b )

32a 2 - a 2 - b 2 32a 2 •(1)

32a2 Hence locus of P is line segment A 0 A' 0 and B 0 B' 0 and lines A 0 B 0 and A' 0 B' 0 except line segments A 0 B 0 and A' 0 B' 0 . Equation of A0B0 and A' 0 B' 0 is y = x + Co where c0 is given by equation (1). [2]

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AITS2002-FT-5-MA(S)-9 \

Let P : ct,,

V

and Q : ct 2 ,

2j

Also ct1 = — => tit 2 = 1

(1)

Equation of normal at't' is ty — c = t3(x - ct) Let (2) passes through (h, k)

[2]

(2)

=> ct4 - ht3 + kt - c = 0 (3) will have 4 roots, let the roots be t1, t2, t3 and t4

(3)

then I t ! = c m2 = o

(4)

2tlt2t3 = - c

(6)

UUzU = - 1 Form (7) and (1), t3U = - 1 From (5),

(7)

(5)

[3]

(ti + t2)(t3 + U) + tit 2 + t3t4 = 0 =i> (ti + t2)(t3 +14) = 0 [as Uh = 1 and t3t4 = - 1 ] = > U = - t 2 or t3 = -t4 Clearly t, * - 1 2 otherwise tf = - 1 => t 3 = —14

t | = 1 => t3 = ± 1 or t4 = + 1.

from (4), t, + t 2 = c

.(a)

and from (6), t3t4(t1 +12) + M 2 (t 3 + U) = - => ti +12 = c [as t3 + t4 = Oand Ut2 = 1] from (a) and (b) h

k

•(b)

U L,

c c or locus of R is x = y. 5.

[3]

Consider g (x) = x2 + 2x. [2] 2 Clearly t ' will be a negative number. If 't* increases then '—t2' will decrease or graph of g (x) will come down by the quantity '-t 2 '. Also F (t) is algebraic area bounded by x-axis and the curve and will be negative. So if we are increasing't', F (t) will decrease. Hence maximum value of F (t) will be corresponding to t = 0 and this value is equal to F(t)| max = J(x 2 + 2x) dx =

x" 3

g(x)

-2 \

2

+X

- 2

3

3

F (t)|max = — for t = 0 and clearly [F (t)]min does not exist. 3

J 0

X

[2] [4]

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AITS2002-FT-5-MA(S)-9 \

Clearly z, 2z, 3z and 4z will lie on the same line, also P (z) will be minimum when AP + BQ + CR is minimum. Now for any fixed |z| = r, P (z) will be minimum when line PR coincides with line AB. So for minimum value of P (z), PQ and AB should coincides so for minimum value of P (z), we can minimise Pi (x) = |2x - 11 + |3x - 2| + |4x - 3|. [5]

R(Az)/ Q(3z/

A P(2j) J

/ C(3 3)

/

2)

'

/ / A ( 1 , 1)

(from the graph) Clearly P, (x)]min = O 2 2 2 P (z)]min = — and corresponding z = - + — i. 3 3 3 (As AB : x = y) [3]

Let position vector of C, A,

B and

D be

(6) a, b and 2Xb respectively (where X e

1

0,

[2]

Clearly E = Xb, B., = Equation

of

lines

C, = BB^

D(2Xb)

. AE

and

^D

B b

are

r = b +11 b - - ,r = a + t 2 (xb - a ) v 2, a+b

+ t. 2Xb

and r =

respectively. = a + t2(xb-a)

For the point 'F' we have b +1, V

[2]

2Y

b(l + t , - M 2 ) + a f t 2 - ^ - - 1 - 0 _ t2_

1 2-X'

E_ h

lb

(1-X)

"(2-X)

a+

(2-X)"(2-A.)

a + b, -(l-t3)+2Xt3b For the point 'G' we have b +1, b - - = . 2j 2 \ f + a - ti (1-tJ -0 1+ t,-2^3-l(l-t3) 2 2 4X-2 G = — - — (b - (2A. - l)a) ti = 3-4X Thus FD = 2Xb

2-X

((1 - X)a + Xb) = —((31 2—X

[2]

- 2X2 )b - (1 - A.)i)


AITS2002-FT-5-MA(S)-9 \

EG =

= ^[(4A

2

- 3 A + l)b-(2A-l)i]

Now area of quadrilateral DEFG, (4A.2 - 3A + l)b - (2A - l ) i

A „ = i EGxFD

axb

axb Let yJ =

(3A - 2X2 )b - (1 - A,);

(3 - 4 A.)

2-X

(2X2 - 3A)(2A - 1 ) + (4A2 - 3A +1)(1 - X)' (3-4^X2 A +A -1

-X) A +A -1

J

o_

2

4A -11X + 6

4A2-HA+6

A2 + A - 1 , 4A22 -11A + 6 ^

dy _ - 5 ( 3 7 i - 1 X 2 1 - 1 ) +

Since X e (0, 1/2), 1 1 at X = — y attains minimum; y1/3 = — Also as A, -» 0 or A, ye

1 5'

6

1 —, y 2 |y|e

1/3

1/2

1 — 6

1

v6 ' 5

1 A0 1 Hence — < — < 6 A 5 8.

[2]

[2]

.E(-1,2)

(-1.0)

As p can lies any where on the line x = -1, so first we will find the range of p so that line through P(-1, p) will cut the curve y = {x} at 4 distinct points. We can see from the diagram for a line through (-1, p) cut curve at 4 points, then p must lie within EF. Equation of BD is y = - — - ( x - 3 ) =>x + 2 y - 3 = 0

[2]

•• E(-1,2) Equation of AC is x = 4y F(-1, -1/4) => to cut y = {x} at 4 distinct points we must have f A \ Pe •(1) Now any line through P(-1, p), with slope tan0 is

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x+1 _ y-p - . . - n, r 2 , r3, r 4 cos0 sin 9 Now r-i lies on line AD i.e. y = x r 2 lies on line GH i.e. y + 1 = x r3 lies on line IJ i.e. y + 2 = x r4 lies on line BC i.e. y + 3 = x as r4 > r3 > r 2 > r, i.e. (-1 + r1 cos 9, p + r 2 sin9) lies on x = y i.e. -1 + n cos 0 - p - r-i sin9 = 0 => r-, (cos 9 - sin 9) = 1 + p Similarly r 2 (cos 9 - sin 9) = 2 + p, r 3 (cos9 - sin9) = 3 + p, r 4 (cos9 - sin9) = 4 + p Now we have 2r, - r 3 2(1 + p) - (3 + p) P-1 < 0 <0 3r 2 - r 4 3(2 + p) - (4 + p) p+1 o p e ( - 1 , 1) ....(2) From (1) and (2), 1 Pe|--,1

11

9.

sin2 k x | = jsin 2x| + |sin 4x| +

[2]

[3]

[3]

+ |sin 2 n x|

k=1

Now |sin 2 x| is non-negative V x, k. Apply AM > GM Isin 2x| + Isin 4x| + Isin 8x| +

+ Isin 2n x

n

sin 2x| (sin 4x|

|sin2 n x|J /n

[4]

11

X'

sin2k

k=1

x I > |sin2x||2sin2xcos2x|

|2(2sin2xcos2x)cos4x|

}/n

> [[sin 2x|n x 2 1 + 2 + + n~1 x (|cos 2x| n " 1 |cos 4x|n~2 |cos 2n~1 x|1)]1/n > [|sin 2x|n x 2n(n"1>/2 x (|cos 2x|"- 1 |cos 4 x f |cos 2 n _ 1 x|1)]1/n Using the fact that if p > r, then |cos 2k x|p < |cos 2k x|r Equality holds for x = 0 Using above, we can write (1) as sin2 k x | > n[|sin 2x|n x 2n(n~1)/2 x {|cos 2x|n |cos 4x|n

•(1)

|cos 2n~1 xjn>]1/n

[2]

k=1

> n[2n(n_1>/2 {|sin 2x| |cos 2x| |cos 4x| > n [2 n(n " 1)/2 {Isin 2x| |cos 2x| |cos 4x| Using 2 sin A cos A = sin 2A we get n-1

> n

|cos 2n~1 x|}n]1/n |cos 2 n " 1 x|}]

|sin2 n xf

I

i

" sin2 n x k We get, 2 ^ | s i n 2 x | > n ' n - 1 ' k=i 2 ~y Write above in terms of fk (x) nf n (x) Efk(x) > n-1 k=1

2

2

[2]

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AITS2002-FT-5-MA(S)-9 \

(b).

We can write above as (sin8 x - 2 sin4 x + 1) + (cos8 y - 2 cos4 y + 1) + 2sin4 x + 2 cos4 y - 4 sin2 x cos 2 y = 0 => (sin4 x - 1 )2 + (cos4 y - 1 f + 2(sin2 x - cos2 y)2 = 0 [2] => sin4 x = 1 and cos 4 y = 1 and sin2 x = cos2 y => sin2 x = cos2 y = 1 71

x = n7i + — 2 y = rrm n, m G I. 11.

f (x) = e|x2~6x+81 Let g (x) = |x2 - 6x + 8| = |(x - 2)(x - 4)| As ex is an increasing function so if argument of x is increasing ex will be increasing. So f (x) will be increasing if g (x) is increasing and f (x) will decrease when g (x) decreases. [3] From graph it is clear that g (x) is increasing in (2, 3) and (4, oo) and decreasing in (-co, 0) and (3, 4). Hence, (a) (2, 3) u (4, oo) (b) H o , 2) u (3, 4).

12.

We have Vsin2AcosB - V s i n 2 B c o s A + sinBV2sinA - s i n A V 2 s i n B = 0

[2]

=> Vsin A cos A cos B - Vsin B cos B cos A + sinBVsinA - s i n A V s i n B = 0 => (VsitrA - V ^ B ) V c o s A cosB - V i i ^ V s h B (V sin A - VsinB) = 0 => (VsinA - VsinB^VcosAcosB - Vsin A sin B) = 0

[3]

=> (Vsin A - V s i n B ) ( l - V t a n A t a n B j ^ 0 As A * B => tan A tan B = 1 => A + B = - => C = - . 2 2 • • •

[3]

1 J


FIITJ€€ ALL INDIA TEST SERIES I I T - J E E , 2002

FULL TEST - V (Mains) MATHEMATICS SOLUTIONS -2 < x < -1

X,

f(x) =

1.

(a).

x 2 + 2x,

-1 < x <0

2

2x-x ,

0 < x <1

2-x,

1
From the graph, it is clear that f (x) is continuous every where in the given interval. [1] And clearly, f (x) is not differentiate at x = 1,-1.

For x = 0 f(x)U=[2x

+

2]x=0_=2

and f'(x) lx=0+ = [2 - 2x] x=0+ = 2 Hence f (x) is differentiate at x = 0. Clearly f (x)|max = 1 at x = 1. [2]

[2] f(x), (b).

f(f(x)) =

- 2 < f(x) < - 1

(f(x)) 2 + 2f(x), - 1 < f(x) < 0 2 f ( x ) - ( f ( x ) ) 2 , 0 < f(x) < 1 2-f(x),

[2]

1 < f(x) < 2

f(x),

-2 < x < -1

f(x)[f(x) + 2],

-1 < x < 0

f(x)[2-f(x)],

0
x,

-2 < x < -1

(x 2 + 2x)(x 2 + 2x + 2), - 1 < x < 0 ( 2 X - X 2 ) ( 2 - 2 X + X 2 ), 0 < x < 1

( 2 - x ) ( 2 - 2 + x), FIITJC€

1
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-2 <

X, 2

_ j x(x + 2)(x + 2x + 2), x(2-x)(x

2

X

< -1

-1 < x < 0

- 2 x + 2),

0 < x <1

x(2-x),

1
for continuity f (f (-1")) = f(f (-1 + )) = - 1 f(f(0~)) = f ( f ( 0 + ) ) = 0 f ( f ( r ) ) = f(f(1+)) = 1 Hence f (f (x)) is continuous every where in the given interval.

[3]

Let z = x + iy and z 2 = z - z^ = (x - 1) + i(y - 1) Now f (z) = (Im (z 2 )). (Im (z - z^ 2 ) 2i

As lm(co) =

2i -z.

_ (z-z)(z+z)

2i

2i

co - ft) 2i

A

(z 2 + z 2 )

= y(2x)(y-1)(2(x-1)) = 4xy(1 - x)(1 - y)

As Re(co) =

co + co

[2]

= 4x(1 - x ) y ( 1 - y ) Now as z lies in first quadrant satisfying \z\ < 1 =5> 0 < x < 1 and 0 < y < 1 1 1 The maximum value of x(1 - x) = — at x = — 4 2 1 1 and maximum value of y(1 - y) = — at y = — 1 , 1 +i = — for z = 4 '4 4 2 Also f (z) is always > 0 minimum value of f (z) does not exist. f (Z)|max = 4. 1 1

[2]

[3]

Let P be (-a, a). As P lies on the directrix of the parabola y 2 = 4ax, the chord of contact of P i.e. AB (say) will pass through the focus. Also, from the property of tangent of parabola ZPSA = 90°. [2] Let Q be the reflection of P in AB, then h + (-a) . . — - = a => h = 3a and^

k + a

= n0

k +a =0

Hence, locus of P will be x = 3a. Let L: y +

21 4

[3]

f 9) n m x - — =0 I 4j

L intersects with C at P and Q (given) Then, the equation of any circle passing through P and Q is C + A,L = 0 (1) ' Now let (1) passes through A, then A will satisfy (1) (-3) 2 + (0)2 - 4(-3) + X 0 + — - m - 3 - 4 4

[2]

=0

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21 + A

'21

(1 + m) = 0 => A = -

1 + m.

Hence, equation of the circle passing through A, P and Q is C - f - ^ - V =0 U + mJ

(2)

[2]

We have to show that B lies on (2) and for that B should satisfy (2) for all values of m for which L intersects with C. f oV\ o 21 2 2 3+ m 0-— = 0 0 + (-3) -4.01+m 4 v 4 J) (1 + m) = 0 1 + m ^4 /

(m * -1.)

=>0 = 0 Which is true. Hence, B always lies on the circle through A, P and Q.

[2]

In the given figure let ZAOB = 0, then 1 0 0^ Area (AOAB) = —xrsin—-rcos— x 2 U 2 2 1 9 = — r 2 sin 9

2

r2 n => Area (AOAB)|max = — for 0 = [2] 2 2 Hence, area (AOAB) will be maximum if OA 1 OB. Hence, according to the question two positions of P is possible namely P' and P" (In the figure) From the figure, OB' = ( Z1 - z 0 )e k/2 And OB" = (Zi - z0)e~'*12 => B is z0 ± i(Zi- zo) [2] Similarly, AP' = j ^ j . (z, - z0)eM2 = (z, - z0)e'nl2

^

As |AP'| = |OA| P' is Z1 + (z\ - z0)i Similarly, P" is Zi - i(z, - z0)

[2] Hence, z = z, ± i(zi - ZQ) and z 2 = z0 + i(z, - ZQ). [where z represents P and z 2 represents B] Let At be the event that ball drawn from D is blue. Then as min {m, n, p} > 7 and only 7 balls are withdrawn, in every throw, the transferring of a ball from A, B or C to D is equally likely. [3] Let Br be the event that D contains r number of blue balls.

B n-Green

1/3

1/3

D

Then P (A-|/Br) = and P (Br) = 7 C r

Blue ball

3

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AITS2002-FT-5-MA(S)-9 \

=> P (AI) = ^P(B R ).P(A 1 /B R ) r=0

7.3

f 7 S7Cr2-.r Vr=0

[2]

•(1) J

7

Now (1 + x ) 7 = j y c r x r r=0

Differentiating with respect to x •(2)

7(1 + x ) 6 = J V C r x r _ 1 r=0

1 Putting x = — V'2 from (1)

=J]r7Cr2-r+1 ^ J ] r 7 C r 2 7 - r r=0

=7.3 6

r=0

P(AI)=Y^(7.36) => P (AO =

|

Now let C be the event that D contains balls of all colours and equal number of red and green balls. Then '

7! 2

(3!) 1!

P ( C N AO

7!

~1

7 • 3 7 36

3

H

1 8

7!

^

7

1

x—-

2

(2! ) 3!

7

7

1

n

6!

24. ~ 3

7

f 14 ]

7 I2J s

?!

3

f 1 5

2

-i-

[(1!) 5lJ 7

[2]

37

140

~ 37

Hence the required probability P-(C/A0 =

P(C N A ) f

P(A-i)

140 _ 140 36

729

[3]

v3 y 7.

(a) For the graph, we will start with y = - x 2 and using transformation we will get the graph of |y| + (1 - |x|)2 = 5.

(V 0

(0

FIITJCe

A 1 I

(W5, 0

O \(V5, 0)

k

(ii)

(0,5)

(W5, 0)j

= 5-y

(V5, 0)

X2 = 5-|yl

(iii)

'(0, -5)

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AITS2002-FT-5-MA(S)-9 \

x=-1

,(0, 4)

O

(-1-V5, 0)

(V5-1, 0) < — >•

(1+V5, 0)

'(x+lf = 5-|y| (1-x) = 5 - | y |

'(0, -4)

(iv)

o

(1-V5, 0)

i

(1+V5, 0)

x —-1

(0,

-£)

(vi)

[3]

As the graph is symmetrical about x and y-axes, we can find the area in the Ist quadrant and multiply it with 4 to get the required area. 1 + ^5 1+V5 Hence, the required area, A = 4 Jydx = 4 j ( 5 - ( x - 1 ) 2 ) d x = 4 | 5 x ( 1 + V 5 ) - | ( 5 V 5 + 1 ) j = | ( 7 + 5 V 5 ) s q . units.

[2]

Let P be (x, y) and QR be the tangent. Then equation of QR ^(X-x) dx For R : Y = 0

Y

-y=

R : x--f-, dy

0

dx For Q : X = 0 Q: 0 , y - x f dx d

AR =

FIITJCe

y dx dy dx

yv w

y dy dx



dv dy and BQ = y - x — - y = - x — . dx

dx

Given, | Area (OAPB) = 2Area (APAR) + Area (APBQ) 3 1 => — xy = 2 x — x y x 2 2

f

\

dy 1 +— 2 - x dx (x)

•(1)

v dx j 3 x y

^ = -2y 2 - x2 r dy dx v dx

[3]

dy all three terms in (1) are positive as — < 0 and x > 0, y > 0 dx dyV 2y = 0 x2|^]2+3xy' dx dx ' r

dy x™ +y dx

f

dy x — + 2y dx

=0

dy Now if x — + y = 0 dx dy dx , n => — + — = 0 => xy = k y x => xy = 9

•(2)

(as curve passes through (3, 3))

.(3)

[3]

And if x — + 2y = 0 dx ^ ^

2^=0=>yx y x 2 => yx = 27 (b).

+

2

=k

Minimum distance of P on (2) is 3 V2 corresponding to the point (3, 3) as (2) is a rectangular hyperbola. For the minimum distance on (3), we can minimize D = x2 + y2 27- + y22 = — y , . . dD 27 . A For maxima and minima, — = 0 = —52 + 2y dy y y=

3

V2

and for y =

^

d D dy 2

> 0 =5- minima

Now, if y = ->1/3 .2 _ 27 21/3 = 9 x 2 1/3 Hence, the minimum distance of P 3V3 d = VD = J 9 x 2 1 / 3 + ->2/3 -.1/3 units.

[2]

PI3TJ€€ Ltd, ICES House, Sarvapriva Vihar (Near Hauz Khaz Bus Term.) New Delhi-16 Ph: 6515949, 6865182, 6569493, FAX: 6513942

AITS2002-FT-5-WlA(S)-7

x(1 + xy(x2 + y2))dy = y(1 - xy(x2+ y2))dx => (x + x2y(x2 + y2)) dy + (xy2(x2 + y2) - y) dx = 0 => x dy - y dx + (x2 + y2) xy(x dy + y dx) = 0

9.

=>

xd

y ~ ydx + x y ( x dy + y dx) = 0 x2 + y 2 tan"1 - J ! + xy . d(xy) = 0

[3]

Integrating tan -1 y

Jxy d(xy) = 0

=> tan"1 [ — I + ^l^L + c = 0 2 tan" 10.

-1 y

+ x2y2 + k = 0.

[3]

Use cosine rule in AABt, 2a2-b2 f 3TC^ 5

= cos

2a2 , 1

1b

2

L 5y 3TT

= cos— 2a 5 T 2

n b2 o n1 - c o s3— 1 = -rr => 2

3jt

Now let

[2]

26 = 3n - 39

=> sin 26 = sin 36 => 2 cos 6 = 4 cos2 6 - 1 => cos t) =

2 ± V 4 + 16

3%

But — is in II cos a2 b 11.

2

3 ti 5

b2

1±V5

quadrant.

[2]

1-V5 4

1 -

3 + V'5 2

1-Vs

9 +5+6 ^ +4 •

+

3 + V5

2(3 + V5)

3 + V5

= 3.

[2]

f (x) = (3 - V 4 - x 2 )2 + (1 + V 4 - x 2 )3 (given) Let t = ^ 4 - x 2 , Clealry 0 < t < 2 => F (t) = (3 -1) 2 + (1 + t)3 for maxima and minima => F' (t) = 0 => - 2(3 - t) + 3(1 + t)2 = 0 3t2 + 8t - 3 = 0 => (3t - 1 )(t + 3) = 0 1 => t = -3,

[2]

Also, F" (t)|t=_3 = - 1 0 => maxima £1ITJ€€

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is Term.) New Delhi-16 Ph: 6515949, 6865182, 6569493, FAX: 6513942

and F" (t)|t = 1/3 = 10 => minima as t * - 3 hence, maximum value of F (t) will occur at the end points for which F(0) = 10 F (2) = 28 Hence, maximum value of F (t) = 28 for t = 2 => maximum value of f (x) = 28 for x = 0. >2 1 Also, minimum value of F (t) = 3 - 1 + 3 + at t = — 3 3 = 256 27 minimum value of f (x) = for x = ± V4 - 1 2

=

+

256 27

V35

3 = 28 for x = 0 t=1 / 3

Hence, f (x)|max

..... 256 , v35 and f (x>|min = —— for x = ± - — dL.1 O 12.

[3]

L.H.S terms can be written as X

1-x2 w3

1-xe 1-X

10

R> C = X + X3 + X +

oo

= _ X 3_ X 9_ X 15_ = x5 + x 15 + x25 +

00

[5] st

Now, summation of 1 terms of the above series = x

3

x+ x

5

7

x +

1 + x2

[first term of RHS] Sum of 2nd terms of the above series

x3 - x9 + x15 -

[second term of RHS] and so on Hence, LHS - RHS. 13.

1 + xfc

[5]

Let the given lines be OP and OQ and the point of intersection at P and Q be R : (h, k), then equation of PQ : yk = 2(x + h) For OP and OQ y = 4x . 1 yk-2x y2 = 4x [3] 2h => 8x2 + 2hy2 - 4xyk = 0 Given equation of OP and OQ is 5x2 + 3y2 + axy = 0 Now (1) and (2) represent the same lines

FIITJC€

•(1)

• (2)

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8 _ 2h 4k 5 " 3 a 12 h = 5 Hence, the required locus is 5x - 12 = 0. 14.

[3]

In the figure PQ is the given tower and QD be the flagstaff. Let first position of man be A and second position be B, then ZQAD = ZQBD = 9 (given) [2] Clearly A, B, Q and D are concyclic points. Again let S be the centre of this circle and M and N be the mid points of chord QD and AB respectively. Then ZQSM = 6 and ZRQM = 30° (given) From figure ZMST = 30° ZTSQ = G - 30° [3]

Now in triangle USQ, SU _ a + b cos(e-30°)= s u [as PA = a and PB = b (given) => PN = US = (a + b)/2 ] SQ 2SQ Now in AMQS, MQ (a + b) o^ • a (a + b).sin0 sin 6 = MQ = SQ. sin 6 = v ' SQ 2cos(9-30°) 1 + V3cotG Hence length of flagstaff = 2MQ =

2(3 +

1 +V3cot0 '

[5]

• ••

FIITJCe

l


\

FIITJ€€ Rankers Study Material IIT-JEE 2002 P H A S E - 1

T E S T

PHYSICS SOLUTIONS From geometry velocity vector of the particle is horizontal at the time of hitting the plane. Hence, with respect to the horizontal plane passing through the point of projection, P is the highest point in the path of the projectile. Hence,

/(a)wtime of hitting = — v 0 sin9 = 20sin60° = _2 V 3 2 Q 10 v /

sec

... D _ v 2 sin 2 9 v(20) 2 sin 2 9 . „ 3/4 300 (b) PQ = - ^ - r = — = 4 0 0 x - ^ - = — = 15cm 2x10 20 20 2g velocity of car at the time of firing =(2) (4) = 8 m/s i Now let us assume that velocity of projection of the bullet with respect to the car v x i + v y j + vzk Therefore, velocity of the bullet with respect to ground = (v x +8)i + v y j + v z k Now (vx+8)4 = 40 => v x =2m/s ; (vy)4 = 80; v)4 = 80; vy = 20m/s; vz (4) =120 = 30m/s

vz

Hence v 0 = 2i +20j+30k (a) v = 3 + 6t + 9t2;

— - = 6 + 18t dt Acceleration at t = 3 sec; a = 6 + 1 8 s 6 0 cm/s (b) Displacement = j 4 6 (3+6t+9t 2 )dt = [3t + 3t 2 + 3 t 3 1 = 6 + 60 + 456 = 522cm / \ a , -t jvdt (c) Average velocity = J

JD«

3,8

[3t + 3t 2 + 3t 3 ]5

= 3 + 39 + 387 = 429 cm/s

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RSM12-P1-T-PH(S)-2

4. (i) for 10 kg mass T- 0.2x1 Og = 10 a For 5 kg 5g-T = 5 a

(1)

(2)

On adding 1 and 2 a=g/5 given s=4m so, v= 4m/s (ii) Applying conservation of energy l ^ x1 2 v

=

2

+ ^ k 22 x 2 = I m v 2 +\{x/4)2 2 2 2

+lk2(x/4)2 2

_7l5(k1+k2).

5. (i) • When A loses contact with ground Ti = m A g = 1 x 10 = 10 N . . . .(i) T = 2Ti and F = 40 t = 2T = 4 Ti Hence, Ti = 10 t . . .. (ii) from (I) and (ii) t = 1 sec, Hence A will lose contact at t : 1 sec similarly for B Ti =M B g = 2 x 10 = 101 t = 2 sec, Hence B will lose contact at t = 2 sec. Similarly, for C T = 3 x 10 = 20 t t = 1.5 sec, hence C will lose contact at t = 1.5 sec. Ti - m A g = m A a dv 10 t - 1 x 10 = 1 x — . . . (iii) dt (ii) • Velocity of A when B loses contact with ground V

F = 401

v /s /\

Ti N '

Ti

o n . A B

Ti

2

j d v = J(1 Ot — 10)dt 0 -I which gives, v = 5 m/s

mAg

OI Velocity of A when C lose contact v = J (1 Ot - 1 0 ) dt = 5/4 m/s (iii) • For block C T - mcg = mca dv 20 t - 3 x 1 0 = 3 dt"

T = 20t

nicg

3 dv = 20 t dt - 30 dt 20 ' v

3 2 2 P 10t Jdy = f [ ^ - i o t

+

30 ^]dt

10 2 „ n + 30 = — t -10t + — 3 4

H = 0.14 m.

3/2

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RSM12-P1 -T-PH(S)-3

Let time taken by the car and their final velocities are t-i, t2 and v1f v2 respectively. Given ti = t2 - 1 and Vi = v2 + v •Si = âtf

= s 2 - ^ a 2 t j = s (say)

=> a-itf = a 2 t 2 = 2s • also v-i = a-it-i, v2 = a2t2 => v ^ i =2 sa-^ 2 = 2s 2and v2t2 = a2t22 = 2s s • ^• => U = — and t2 = — so t2 - t-i = 2s J - - J . I - . VV 2

r

2s 2s

V =

v,

i

V-|V 2 V V,V2

2s

v1v2

V,

)

=t

t=

I vfv2

t = Va1a2 t

The ball's position is at time t, (7.5)t. i + (10)t j Suppose the fielder runs from his position with constant velocity 5 [(i cosG + j sine) ] m/s. relative to the wicket. At interception of the ball by the fielder the position must coincide so equating the components we get 7.5 t = 46 + 5t cos 9 ...(I) and 10t = 28 + 5t sin 9 . . . (ii) \2 7.5t - 46 10t-28 These give + =1 5t 5t I

116

which simplifies to t = 4 sec, or - ^ - s e c . Hence for earliest interception t = 4s. The system is illustrated in the figure. Let the tension in the string be T and the accelerations of the system be a. The equation of motion for the masses are for mass 2m, T - 2mg sin 30° = 2ma (0 and for mass 5m 5mg sin 45° - T = 5ma 5mg - T = 5 ma (ii) V2

5mg


RSM12-P1 -T-PH(S)-4

5mg

Adding (i) & (ii), or

mg= 7ma

J2

. . . (iii)

g = 7a

42

Hence the acceleration of the system is a =

42

from (i) T = mg + 2ma 2mg T = mg + —

42

- 1

J

mg(5 + 5V2) t

/

_ 5mg(1 + V2) 7 The force on this pulley is the resultant of the tension in the string on the two sides. The angle between the two tensions is (60° + 45°) = 105. Therefore the force on the pulley is 2T cos (105/2)° = 2T cos 52 1 9.

= 10 ^

If at point P, tension is zero. then mg cos 9 =

#

(1 + V2) cos 52 | . O rj»

I

mv /111

from Conservation of energy, v2 = g/ (1 - cos9) mg cos9 =

r

i v

(1 - cos 9)

=> 9 = cos"1 (2/3) .-. height of point P = — + — cos 9 = — , from lowest point. 2 2 6

gi

v = V

gi

V3 Now the particle describes parabolic path. The height attended by the particle, from point P. _ (vsin9) 2 _ 51 2g

" 54

:. highest point from lowest point will be '5/ 51_ 501 ~6~r~54

54


RSM12-P1 -T-PH(S)-5

10.

Let the angle of projection be p and velocity be u. The velocity parallel and perpendicular to the planes are u cos(p - a) and u sin (p - a). The component u sin (p - a) becomes zero at the first plane where as the component ucos (p - a) at the second plane. If the time required in the first case is ti and the second is t2. usin(p-a) Then 0 = u sin (p - a) - g cos at] => ti = gcosa t2 =

and 0 = u cos (p - a) - g sin at 2

. . . (i)

ucos(p-a) gsina

(ii)

As the particle is mid way between the planes 1 1 u sin (p - a)ti — g cos a t 2 = - {(u sin (p - a) t2 - — g cos a t 2 } => u sin (P - a) (ti + t2) = — g cos a (tf +

. . . (iii)

putting (i) and (ii) in (iii) we obtain u sin (p - a)

usin(p - a )

+

gcosa

1 \( u 2 sin 2 (p - a ) = -gcosa 2 g cos a

ucos(p - a ) gsina u 2 cos 2 (p - a) g2 sin2 a

canceling u2/g from both the sides and rearranging we get sin2(p-a)

sin2(p-a)

sin(p-a)cos(p-a)

cos2(p-a)cosa _

sina 2 cos a cos a 2sin 2 a 2 2 => sin (p - a) sin a + 2 sin (p - a) cos (p - a) sin a cos a - cos2 (p - a) cos2 a = 0 dividing each side by cos2 (p - a) sin 2 a tan2 (p - a) + 2 tan (p - a) cot a - cot2 a = 0 . - 2 c o t a ± v 4 c o t 2 a + 4cot 2 a , , tan (P - a) = ——— = - cot a ± V2 cot a since (p - a) is an acute angle tan (p - a) is + ve p - a = tan' 1 (cota ( V 2 - 1 ) ) P = a + tan "1 {cot a (V2 - 1)}. *

*

*



IIT -JEE 2002 PHASE-I TEST PHYSICS SOLUTIONS (a) Kfrans = -ITIV 2 Krot = —I®2 = — mv 2 2 5 K - Ktrans

+

Krot ~

7 2 — ITIV

10

K trans _ 5 K 7 Krrot„, 2 K (b)

mg

7

mg(R-r)=-Lmv f

2

mv 2\ 10 = —mg R^r

N = mg +

( mv112

N

17

ymg

(a) In case of fundamental vibrations of string QJ2) = L, i.e. A, = 2 x 1 = 2m. Now as v = f and f = 750 Hz VA = 2 x 750 = 1500 m/s. (b) Now as in case of a wire under tension so

Â _

—

1

•

T,

vt

wj

ftA

IT* i.e. vB = 1500 I-2-

*'TA

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M

RSM12-P2-T-PH(S)-2

mg'

or, vB = 1500 V

= 1500 g [ i - p / p j

as g' = g 1 - °

mg

or, v B = 1500^1 - — = 1000m/s Further as here X = constant = 2m vR

fB = -S- =

1000 2

nv so from f = — 2£

= 500 Hz

i.e., in this situation, X = 2m, f= 500 and v = 1000 m/s. If f is the frequency of tuning fork then according to given problem at 30°C f = fc and f = f0 VQ 4L

VA 2L

.e.,

fc = f0

i.e.

vG = 2vA = 2 x 360 = 720 m/s

or

=

Now as in the case sound v oc VT 273 + 30 303 So 273 273 273

v 0 = vG

i.e.,

720 x 0.95 = 684m/s

303

The motion of simple pendulum is angular SHM, so its equation of motion will be y = A sincot

with

© = ,J(g / L)

(a) When p < a i.e., when angular amplitude is lesser, the pendulum will oscillate with its natural frequency so that 2N

L

T - — = 271 1 T-i = — co vg

(1)

(b) Now as in case of simple harmonic motion y = A sin tat. Here A = If time taken by pendulum to move from equilibrium position B to A,t' then £a = sin cot' /

1 :__—1 => t' = — sin co

\ /

So time taken by pendulum to move from B to A and back to B t2 = 2t = 2,

—

sin

1

[dj

a

\

las

,

-

M

IP,

So time period of motion T 2 = ti +12 =

f-

. _i a 7t + sin ' —

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RSM12-P2-T-PH(S)-3

Let x be the displacement of the mass from the equilibrium value and let Xi and x2 be the deformation of spring ki and k2 , respectively , from the equilibrium. Then , using constraint relation, X,

+x

2

or Xi + x2 = 2x (1) Also, tension in the string should be same, therefore, kiXi = k2x2 (2) Solving (1) and (2), we get 2k« 2k 0 x (3) Xi = x2 = ki + k 2 ^ +k2 The energy of the system is E = ^-kiX 2 + ^ k 2 x 2 + ^ m v 2 E =

Using equation (3),

2kiko 1 ki+k;

-x

2

1 2 +-mv

Differentiating w.r.t. time, we get dE

4kik 2 dx dv _ — — x — + mv — = 0 dt ki + k 2 dt T= 2n

or

d2x

4kjk 2 •

dt"

+

(ki + k 2 ) n

x = 0

(ki +k 2 )m 4kik 2

As the collision is elastic, velocities & v2 of spheres 1 & 2 after collison (see fig. will be) 2m(-2v)+v(0) 2m and

v2 =

2mv

=

_2v

=v

2m Thus the magnitudes of velocities are interchanged while direction of velocities get reversed after collision. As far as angular velocities are concerned , the collision will have no effect on them. Let Vi' and v 2 ' be the speeds of the bodies when they start pure rolling again. Using conservation of angular momentum for body A about the axis passing through instantaneous axis of rotation, we have

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I

RSM12-P1 -T-PH(S)-4

f ' \

+ M(2v)R = lc h

•mViR

vRy —MR2 — + 2Mv00R = — MR2 — + mv11R 5 R 5 R 40 Vi1 = + — v 7

=> +8MRv = — MRv, 1 5 2v

Similarly for body B,

"R"

M(v)R = lc

R

+ Mv 2 R

2 o 2v 2 o Vo —MR . — - M V R = —MR — + Mv 9 R R R vR 7_ • v2 = — = — Rv29 => 5 5

7.

M = mass of the sun r = distance between the two stars m2 2 —.r = - r U = MI + M 2 3 r2 =

MI=M/3

mi r —r =— MI + M 2 3

M2=2M/3

Centripetal force on m2 _ G m ^ z _ G(2 / 9)M2

Now o>2 = or T2 =

2 GM 2 GM

= m2r2co

2

/

•M

r

\

co

T - time period of revolutions. 2%

GM

T 47T2r3

GM Since time period of earth around sun

-j-2 _ 4u 2 R 3 GM comparing r = R .


RSM12-P1 -T-PH(S)-5

8.

(a) The cycle shown in PV diagram shows the various processes. (b) Number of mc'&s - n, say = 4 For path AB W AB = n R T A l n ^ VA AUAB

= nRTA ln(2.5)

V

=0

AQAB = W A B + AU a b

= nRTA ln(2.5)

For path BC W B C = 0 , AUBC = n C v ( T C - T B ) =>

AQBC = n C v ( T c - T B )

For path CD f WCD = NrT c In

\

v2-5y

= -nRT c ln(2.5)

For path DA W DA = 0 , AU = AQ = nCv (T A - T D ) Efficiency of cycle = r\, say For diatomic gas Cv = (5/2)R , T C A OB - T ) + T l n 2 . 5 c c |AQBC + A Q C D vRy = 1= 1 fn \ AQ a b + AQ,DA T a In2.5 + (TA-TD) vRy Putting the values we get r\ = 12.8%

9.

= V l , P2 = V 2 , M? = RTi V22 = RT2

Pl

T =

PV = R T

R

. ^ ocV?(TI 2 -1)1 R.AT = j P.dV = J a V d V = — ^ = —— W

y-1 Q = U+ W CT = R

c=

f—1 \y

R y-1

RAT

-1J

R + —. 2


RSM12-P1 -T-PH(S)-6

10.

•.• heat is readially outwards dQ . x dT — = -K(27ir.1.) v ; — dt dr For outer and inner layers 1 dQ " 1 2% dt Jl_ dQ 2TT dt

K2

-In r V 2.

T„-T,

-In — = T. - T„ K«

ill

On dividing (1) by (2), we get T0-T2 2 Ti-T0 3 K, 3T0 - 90 = 200 - 2T0 5T0 = 290 T0 = 58°C *

FIITJCC,

•k

*

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FIITJCC RANKERS STUDY MATERIAL IIT - JEE, 2002 PRACTICE TEST PHASE -1, II PHYSICS SOLUTION The distance x, where gravitational pull of each stars becomes equal and opposite can be obtained ^ MM

G

X

_

R- - G

2

(16M)M

IR (10a-x)

=> x = 2a so the body will reach the smaller star due to stars gravitational pull if the body just crosses point P. By using COE, energy at 0 = energy at P G.16M,m GMm 1 G(16M)m GMm 2 2a Vmin ~

8a -

2 ™ mi

8a

2a

5GM

In equilibrium net torque on the pulley is zero. => 2TiR-T2R = 0 => 2Ti = T 2 (1) Also Ti = mg and T2 = 4mg sin 2mg = 4mg sin Xo = 0 Hence the spring is relaxed. As total energy is always constant. 1 2 1 12 1 kx +mEgxsin
j—x

(8m+

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RSM12-PT-PH-III-IV(M)-PH(S)- 2

I = ^m(2R) 2 = 2mR 2 Hence co2 =

k

10m

Hence T = 2% 3.

Let density of baii be p0 and density of liquid be p Velocity of ball when it reaches to B

f

hi = 35m

•Vi = -J2g\\ = V 2 x 1 0 x 3 5

Jo

Now buoyant force on ball = vpg Net acceleration of ball downward in tank. a = g V ( p ° •P) Vp0

1

OA

h2 = 5 m

H e

P_

I

Po

When bail reaches to C, its speed, v 2 = v 2 + 2ah2 Now due to elastic collision ball just reverses its direction Velocity of ball at B again v 2 = v 2 - 2ah? = v 2 + 2ah 2 - 2ah 2 = v 2 Let height reached by ball from B be h v32 - 2gh = 0 2g 2g 2g Net height reached by ball h + h2 = 35 + 5 = 40 m.

4.

For gas in the right part Po (Adf = P' (A£/4)y ...(1) The right part contains 28 gm of nitrogen i.e. 1mole out of 1/3 of molecules are dissociated into atoms. Due to dissociation of each diatomic molecule, 2 monoatomic molecules are formed, therefore number of moles of monoatomic molecules or dissociated nitrogen atoms are 1 2 n n-j = 2 x — -= — 3 3

2

2

Number of moles of diatomic molecules n2 = - x 1 = — 3 3 Molar specific heat of the mixture at constant volume Cv =

+n 2 Cv 2 n1 + n 2

_ _ 2/3x3R/2 + 2/3x5/2 _ OD Uv — 2K 4/3 Cr CP = Cv + R = 3R Hence y = - f - = 1.5 Cv Hence from (I) P' = 8P0 Work done during compression by gas in right part p V„ - P V W = -2—9 = -4P 0 A£ (-sing indicates that work done on the gas) y-1 Considering the F.B.D. PA=^+P'A 4 FIITJCC, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Ph: 6865182, 6854102. Fax: 6513942

RSM12-PT-PH-III-IV (M)-PH(S)- 2

Pressure at left part: P = 8P0 + 3 Initial temperature T 0 :

ki_ 4A

PpAl

and final temperature T _ =

R v

-A t R

5M(or, 3k^ 8Pn + 0 4R , 4A

Increase in interna! energy of gas in left part AU = nCv (T - T0) 5A 3 k ^ PoM 8P0 + R 2 4R ^ 4A AU = — PoA^ + — xKf '"2 2 32 Work done by gas in left part to compress the string and the gas in right part PA.

r 3 ^2 W= -k + 4P0A^ = 2.9 x 103 J 2 V4 , Hence heat supplied by heating coil Q - W + AU

3kt74+P'A

= — k£2 + 4P0Ai' + — k^2 + — PoA^ = 1.3 X 104 J. 32 32 2 2 mg sin 9 + F r - 3JJ.I mg cos 9 = 2ma p

mg sin 9 - F r = mas

Fr x R = - mR 2 a . . . (iii) 5 For pure rolling at sphere and plank surface aP = a s - Ra . . . (iv) F 3 so, g sin 9 — [ ^ g cos 9 a 2m 2 Fr = a sin 9 - — m •

A

(I)

. . . (ii)

mg sine

5F 2m

N2 2mg cose

3 c Fr = - mî g cos 9

8

We know Fr < N2 = (i2mg cos 9 3 - m ^ g cos 9 < \i2 mg cos 9

8

8

B. = H2 3

Let the equilibrium angle be 6=> 2Kx cose + Ky= mg Where x & y are the compression of C & elongation of A & B given a s x = f (cosec9 - 1 ) & t cot9 respectively. => 7 K I (cosec9 -1) cos9 + K£ cot9 = mg 3 mg putting 9 = sin"1 — we obtain k= ^ ( 3 - 2 sin 9) cot 9

FIITJCC,

->i


RSM12-PT-PH-III-IV (M)-PH(S)- 2 5mg 21 A£ — - , x = —, y = — 12I 3 3 work done by the external agent = Wext = AU = -mg y + — Kx 2 + - Kx2 + 1 Ky2 2 2 2 Putting values of k, x, y We get; Work done = 4 W e „ =

-

^ Fixed axis

Ti - 20 = 2a 40 - T2 = 4a = ®a = 4a 2 40 - 20 = 10 a =>

T,n

j2. j

AT2

A c b 1k

a = 0.2

9

r~]B

2kg

. Suppose time taken by wave to reach the pulley on A and B side be ti and t2 respectively. Velocity of pulse on A side = / - - = V VVelocity of pulse on B side = / — = V^

— — = V2040 0.01 39.2 ooT

= a/3920

1 - - X 0.2 X t? = Viti = a/2040 ti 2 0.1 t,2 + a/2040 h - 1 = 0 ti =

- a/2040 ± a/2040 + 4 x 0.1 x 1 0.2

= 0.02 sec.

1+ - x 0.2 ^ = a/3920 t2 2 0.1 t? - a/3920 t2 + 1 = 0 t2 = ti

a/3920 ± A / 3 9 2 0 - 4 x 0 . 1 x 1 0.02 0.016

tan 6 =

0.2 20 _ 5 16 4 _

gt v0

20 10x2

= U.UTb

= 1

e = 45° => ball hits the wedge perpendicularly, conserving linear momentum in horizontal direction 5v^ mv mv0 = Mv 5v0 = 20 v 42 72 v0 = 4v -

42

. . . (i)



1 _ ( v / V 2 ) + v' V2 ~ 0 - 2 0 / V 2 20 = ~ +V V2

. . . (ii)

solving (i) & (ii) v =

2 0 ( 2 + V2) 9

At maximum compression 1,2 1 — kx = — mv2 2 2 (m I 20 vV2 20 rr.V2 x = vJ— = v . = = —(2 + V2)—m Vk V1000 10 9 10 =

2V2(2 + V 2 ) m 9

9.

1 9 y = ( — + cos2 7it)sin 5007tt (1) 4 1 1 1 = - sin 5007tt + - sin 4987rt + - s i n 5027it 4 4 4 => The sources are three having the frequencies 250,251 & 249 hz. If we eliminate the lowest frequency, the other two sources vibrate to make the necessary equation given as:y= ^ (sin 500TT t + sin 5027it) = — 2sin 501711. cos7it 4 = — sin 501 Ttt. cos7tt 2

. . .(2)

0 1 9 dl From eq.(1) I = K (amptitude) =K=( — + cos 2tc t)2 when — =0 4 dt => (1+ 2 cos7rt) sin 2nt = 0 cos 2jit = - — => t = n ± - where n=0,1,2. 2 3 sin 27tt=0

t = - where n=0,1,2 2 1 2 1 => we obtain t= - , — for maximum intensity and t=0, — ,1 3 3 2 For max intensity => beat frequency f b = 2 as intensity becomes 2 times maximum and 2 times minimum in each second. From eq. 2 we obtain beat frequency = twice the frequency of alternation of the amptitude fb2 = 2 x | = 1 h z = - =2


7

•

RSM12-PT-PH-l-ll (M)-PH(S)-6

10.

DB = 2 x — = 5 0 m 4 If he starts at angle 9 as shown in figure 50 _ 100

200 m

4sin0 - 2 4 cos 9 cos G = 2 sin G - 1 =» 2 cos2 0/2 - 1 = 4 sin 9 / 2 cos 6/2 -1 2 cos 8/2 [cos 9/2 - 2 sin 0/2] = 0 cos 9/2 = 0 => 0/2 = re/2 =>

tan 0 / 2 =

Total time =

100

1

9 = % (Not acceptable)

0 = 2tan 1 (1/2)

r

+

=>

100 4cos0

=

200 3

sec.

200 Time taken if he would have denied = —— sec. 4 Ratio = - .

*

*

*

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HITJCC Rankers Study Material IIT - JEE 2002 PHASE - III

PHYSICS SOLUTIONS Let at any time speed of the connector is v, Hence current through it is I =

Bv/ R

Net force on the connector is = F-

B2 / 2 v

[As Fm = Bi/J

R

dv

F

B2/2v

dt

m

Rm dv

A t V - NOMINAL ,

vt =

(1)

dt"

FR

B2/2 From (1) and (2)

(2)

i . L . BI 2. (i)

With switch S closed, potential difference across C3 = 0 and hence charge on C3 = 0 If I be the current through the resistors , I-

5 =™=2A R-i + R 2 + R 3 50

=> p.d. across R, = (20 Q)(2A) = 40 V => p.d. across branch containing Ci & C2 = 40 V Charge on Ci = charge on c2 = Q, say

c, C3 I 1|AF

— V s AR/ W 2

-AAAAAA-H R,

R3

IOOV

Cl 2 ° \ 4 0 V ) = f — x 4 o \ i Coul C-I + C 2J 8 6 = 60 x 10' Coul.

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RSM12-P-3-T(M)-PH(S)-3

Hence p.d. across

BO C : = — V = 10V 6

and p.d. across

C 2 = — V = 30V

(ii) With switch 'S' is open , as there is no current in any branch of the given circuit P.d. across C 3 = 100 V and charge on C 3 = (100) (1) p. Coul = 10"4 Coul. Similarly charge on C2 = charge on C-, = [3/2 jaF] [100 V] = 150 |uC P.d. across Ci = (150 jic)/(6jiF) = 25 v and p.d. across C 2 = (100 - 25) V = 75 V . The element of width dx at a distance x from the axis, while rotating about the shown axis , constitutes a current loop. For this current loop, associated dipole moment _ (?tdx)co7tx 2

Hence total dipole moment associated with the entire rod would be n = Jdî = —

j x dx

a3X

or J L I

-co

q = q0e•t/RC RiC

U R2C

Where Ri = 4MQ, t, = 30 sec., t 2 = 7.5 sec.

••• R 2 = 1MQ Let the three plates be denoted by A, B & C and the six surfaces as Ai, A 2 , Bi , B 2 , Ci , C2 as shown in figure. Taking into account the capacitive effects, one can easily get the charges on A2, Bi, B 2 & Ci. If Ci and C 2 be the capacitance of capcitors formed by plates A & B and B & C respectively, then C l = ôA & C z = eoA a b

Ai

Aj

Bi

Bi

Ci

Ci

Where A = area of each plate as Ci & C 2 are in parallel, C

1

Ci + c 2

&

n 2 -= n Q Q

c

2 Ci + C 2

Solving we get,

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Qi =

Qb

Q2 =

&

a+b

Hence , charge on charge on

B? — Q?

charge on A 2 = -Qi = -

Qa a+b Qb

a +b Qa

—

a+b Qb

a+b Qa charge on Ci = -Q 2 = a+b As net charge on plates A & B must be zero, charges of At & C 2 together should be negative of charges on A 2 & Ci together i.e. Q. Using the fact the electric field must be zero at any point inside an isolated conductor, we get, charge on At = Q/2 charge on C 2 = Q/2 u '-'mmax ax 2

q_

2C 1 Q_

2C

2

Q q

2C

as q = Q sincot, where co = . . sincot =

(a)

(b)

1

.71 = sin — V2 4

' - T 2

1

VLC

V2

= Q sincot

71 ttVlC t= — = 4co

— Bco/ 2

2

Bco I 2

IR-L— = 0 dt

Bco /

r

2R

1-e"

(c) In steady state I =

F m = BIl

L

Bco/

2R

^mg = mg-^cos0(k x

F m

=BI/|k

If applied torque is T then ^mg + ^Fm + * = 0

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RSM12-P-3-T(M)-PH(S)-4 C

=>

t

=

Here

mg/cos9

1 A

B / co

+

\

4R 9 = cot

When the frame has turned through an angle 9, = BA cosG Where A = J2xdy - - j L JVydy = j ^ ^ '

2

Since y1 = - a t 2 2 ... 4 = ® J I a 3 / 2 t 3 c o s e 3 VK do Faraday's law, Eind = dt or Eind =

B a 3 / 2 12

t 3 sin 9

d9 dt

-3t2cos9

When the frame turns through n/4, t _ 9 _ 71 co Eind -

4co Ba 1/2 r j 2 ; — J—t [cot sin 9 - 3 cos 9] V K

i

or Eind =

jt^Ba48co2

IJfn \k[_4

V2

P A3/2 tind —- ^ ( ^ ~ 1 2o) B7= 192co VK Eind _ n 2 ( n - 1 2 ^ a a / 2 R

192co2FWK

When Ki is closed first time and outer sphere is earthed and the potential on it becomes zero. Let the charge on it be q / . V / = Potential due to charge on inner sphere p that due to charge on outer sphere. 0 =

1 47tSn 2R

2R

=> qi' = -q When K 2 is closed first time, the potential V?' on inner sphere becomes zero as it is earthed . Let the new charge on inner sphere be q 2 ' 0 -

1 d2 , 1 ( - g ) 47te0 R 47ts0 (2R)

=> q 2 ' = q/2 Now when K•, will be closed second time charge on outer sphere will be -q 2 '

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i.e. -q/2 Similarly when K, will be closed nth time, charge on outer sphere will be

as

each time. Charge will be reduced to half the previous value. After closing K 2 nth time charge on inner shell will be negative of half the charge on outer shell. i.e. |

and potential on it will be zero. 2n For potential of outer shell 1

v

°

2R

v - q f - M -1 °

1

I s Pv

47ts0

2R

i-q/2"),

4TTS0

47:s 0 2 n + 1 R

q

47ts 0 2 n+1 R

Potential difference = V 0 - V, =

—:

47ts 0 2 N + 1 R

0=•

47is 0 2 N + 1 R

10. (a) At t = 0, there is a steady current downward through the inductor: iD = VQ/R. For t > 0, Kirchhoffs voltage law for the L-C circuit is dt C Here q is the charge on the lower plate of the capacitor. Since i = dq/dt M

dî — 2- = — = -ccci 0 ir dt'4-2 LC

=>

Where co0 = - j ^ .

The general solution of the above equation is a sinusiod:

i = imaxcos(co0t +
and cp = 0, so current is i = —-cos(co0t) R (b) it is told that the charge on the lower capacitor plate is zero initially; at time t > 0 is given by t

.

t

y

q = q 0 + J - ^ d t = 0 + jidt = 0dt 0

R

q 0 = 0. The charge

t

jcos(o)0t)dt 0

\ f s i n M I , =-\sin(o30t)

©0R

11.

co0R

vx = v 0 / = v0t vy = 0 + at = ' e E V I ^ V m

A

v

0

y

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I


y

in the magnetic field qv y B = R

_ my

mvf

(

4-4-4-4-

4-

4

F

v

R =

fm

eE

I

•

I e _ B M v.

qB R =

i

E/ 1

BvJ

V„

"

E

= 5 mm

Pitch of the helix = s = v x .T = vx. (2Tt/'co) (V0)2TCR _ (v027i)m 3.6 mm. eB

12.

Motional emf induced across the infinitesimal element dx is given by de = vB.dx = (x-OrofIdx I 2TCX

fl 1

emf induced across PQ 31/2

e =

J(x - /)». M o dx 2TIX

1/2

_ (®Mgjo [x - / lnxY/7/2 ~ { 2n 2n J =

cofa0i0/(ln3-l)

2n with P at higher potential w.r.t. Q. *

*

*

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FIITJCC Rankers Study Material IIT- JEE 2002 PHASE -IV TEST

PHYSICS SOLUTION (a)P(4000A°)=^=

2 x 4 x 1 0 - 7

10 - 4

= -8 x—10a m.

-7 0. _ 2 x 6 x 1 0 p (6000 A ) = = 12 x 10 m. 10"

dx 10 - 4 x 4 x 10~3 = — m D 2

(b) Path difference = A = S 2 A -

9n 2nx 10" 4 x 4 x 10 - 3 phase difference, fa (4000 A 0 ) = — ^ =—A — = 71 X 4x10" o, 2ti . 2nx phase difference, 2 (6000 A°) = = X 6x10 If the individual intensities are U and l 2 at P, then Ii + l 2 = 9I0 11 cos 2 (fa/2) + l 2 cos2 (2/2) = 2I0 or, Ii + l 2 = 9I0 12 (1/4) = 2I0 l 2 = 8I0, Ii = l 0 Ii : l 2 = 1 : 8 2.

(a) For ^ = 4000 A 0 , E, = ~ ^

=

4000

10~ 4 x4x10~ 3

= 71 =

27t

= 3.1 eV

X2 = 5000 A0,

E 2 = ^ ^ eV = 2.48 eV 5000 12400 and for X3 = 14000 A0 E 3 = eV = 0.89 eV 14000 The energy of excitation of H atom (n = 2 to n = 3) 1 1 eV » 1.9 eV AE = 13.6 The The T. The

work function of the material = (3.1 -1.9) eV = 1.2 eV maximum energy P.E.'s have energies of 1.9 eV and 2.48 - 1.2 = 1.28 eV , D .. . 12400 Ao 12.400 a 00 «„ a 0 de-Broglie wavelengths are — 5 — ; = A 0 , , A or 9A°, 11 A0 3 10 XVT^9 Vl28 (b) Only the first two wavelengths causes any photoemission. Number of P.E. /sec,

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1

N P E = 1.44 x 10 2 x — x 0 . 2 x 2 x 10"4

The photocurrent = NRE X 1.6 x 10

3.1x1.6x10~

19

+-

1

2 . 5 x 1 . 6 x 1 0 -19

19

1_

1 ^

= 1.44 x 102 x 1 x 0 . 2 x 2 x 10"4 1,3.1 + 2.5 3 « 0.14 x 10"2 A.

(c) If the work function was 50 % lower, the third wavelength would also cause photoemission. Iphoto = 0.2 X 10"2

' 1

1

1 ^

A = 0.36 x 10" 2 A

P

1^3.1 2.5 0.89 The stopping potential will be, vstop = (3.1 - 0 . 6 ) V = 2.5 V.

Let |u & F be the R.I. of the liquid and focal length of covex lens (in air) respectively. From lens makers foumula, we have 2^ , where R = modulus of radius of curvature. f=R U2 R ,, AA R For the concave plane lens in air

M)

When both the lenses are in contact f eq

f

v

fl

f

For the pin to coicide with its image, rays that come out through the last refracting surface of lens - system must fall normally on mirror. Hence, for the lens system image must form on infinity. 1 1 1 Using — = — — we get f eq = -u oo U f,eq feq = (-45)cm = 45 cm. => — — = 45cm (1) 2-n In absence of the liquid lens f = 30cm (2) From (1) and (2), we have 30 (a, = 4/3 45

(2-n)

(a) If nth maxima of 750nm radiation coincides with mth maxima of 900 nm radiation , we have n(750) = m(900) The smallest integral values of n & m satisfying the above equation are 6 and 5. Hence, the required seperation

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= 6

750 x 10~9 x 2 2 x 10~3

= 4.5 mm

(b) For first minima asinG = X , where a = slit width, sinG = Xla For small value of G, G = {Xla) - (Ay/2) D ' where Ay = separation of the first minima on two sides of central maxima =>

a

= ^ Ay

Putting the values we get, a = 0.3 mm. 5.

(a) When the curved surface is in contact (real depth)/(apparent depth) = 4/3 When the plane surface is in contact, I,i2 m = [12 - |g-| Using we get v u R

R = -25 cm

1 f 1 using for the plano-convex lens - = (p. - 1 1 f {Rj

Again

1N R2)

we get f = 75 cm 1 1 1 (b)(i) — - J - = _L v uu. L e e 'e Here v e = -25 cm, f e = 6.25 cm Solving we get u e = -5 cm Hence v0 = 15 - |ue| = 10 cm

1

1

1

We have — = v L o

u0

Putting the values we get (ii)

Ve =

CO

=>

u0 = -2.5 cm ue =

-fe

=> v 0 = 15 - |ue| = 8.75 cm 1 1 1 —= — we get, u 0 = -70/27 cm v 'o 0 uu„ 0 The magnifications for the two cases are as follows Using

_ v0 mi = u0

fey

2.51

6.25

= 20

m 2 = — — = 13.5 u0 fe 6.

(a) d (sepration between the slits) = 3a, where a = slit width For, interference maxima d sin G = n d For diffraction first minima

. . . (i)

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a sin 9 ~ ct => taking ratio of (i) to (ii) n = d/a

. . . (ii)

2d Total no. of maxima, = 2n = — a = 6.

(b) n p+e+ v Let E-i and E 2 be the kinetic energies of the electron and anti-neutrino, respectively. The energy released is Q = (mn - m p - m e )931 = (1.0087 - 1.0072 - 0.00055)931 = 0.884 MeV Using energy conservation Q = E, + E 2 = 0.884 . . . (i) A/2m1E1

and momentum conservation

=

c

Eo or 2m 1 E 1 = - f Eliminating Et from (i) and (ii), we get E i + 1.024E 2 -0.905 = 0 E 2 = 0.568 MeV (for anti-neutrino) and E, = 0.316 MeV (for electron) 7.

Moseley's law l = R ( z - f f l - i for K- lines where n =2,3,4,. n2; (a) For K-absorption edge (z - 1 ) = J —

V X.R

or z =

(o.171x 1 0 _ 1 0 j l . 0 9 7 x 107 j

+ 1 = 74

The element is Tungsten. (b) K a -line

— = R(74-1 f 1 - 4

= 0.228A ° K

p

-line ^ - = R(74-1) 2 1 - 4

^p = 0.192 A 0 — = R(74 - 1 f 1

Ky - line

7T

Xy = 0.182A 0 (c) Cutt off wavelength he _ (6.63 x10~ 34 )(3x 10 8 ) 'Vnin

_

E

—

100 x 1.6 x 10

1Q

124A



8.

Activity = XH = Ol / 2 ) Where N 0 =

(i) A =

—

6 x 1 Q

215

0 6 9 3

X

x 6.023 x 10 23 = 16.8 x 10 19

16.8 X 1019 = 1.165 X 1024 Becqurel

100 xlO" 6 (ii) A(t) = XN0e"xt =

Becquro!

N dN ... f dN V. n-(n-XN0)e"M K1 4t = (b) — = n A.N => = dt => N * X —— dt NJn-A.N 0J 0

K, • Where N is maximum, IA/U

dN n — = 0 => n = — dt X

* *

*


FIITJ€€ RANKERS STUDY MATERIAL IIT - JEE, 2002 PRACTICE TEST PHASE- III, IV PHYSICS SOLUTION When key is open charge on plate 1&3= ^ i ^ a n d on c,+c2 2&4= -

eC,C2

c,+c2

After key is closed charge on plate 3= e c2 and on 4= sc2 Charge on 1 &2=zero Hence charge flown through path 2=

eCjC2 Ci+C 2

Charge flown through path 1= s c2 2.

Equation of the motion for the rod : T-F=m1a . . . (1) For the body m 2 : m2g - T = m2a . . . (2) (1) + (2) =>m2g-F = (rrh + m2)a B2£2v putting F = R we obtain,

-i

E

^ F

T m,

v

X f T

/v T

m2g

m?g

B f v . .dv — = (m,1 + m 7 ) — R dt

• (m1 + m 2 )dv _ d t B2£2V

m2g-

R



dv

integrating both sides we obtain (m, + m2)

m2g-

dt

BVv

R (m,+m2)R

.

After few manipulation we obtain v=

1-e

B <

When t -> co , v = vterminal

B

_ m 2 gR B 2f2

The distance of mth bright fringe from the central fringe = y m = mA — = mp d where p = fringe width. Y9 = 9p ....(i)

(

Distance of the mth dark fringe from the central fringe = y' m = m — v y'2=|p

p

2J

....(ii)

From (1) and (2), we get y 9 _ y 2 =9P- | p =

yP.

^

or

= 8.835 mm 2 p = 1.178 x 10" 3 m

Now, A =

4.

D

= 5 . 8 9 x 1 0 ~ 7 m = 5890 A0

(a)

(a. = NiA = y

=>

JJ.

e 27tr

=

—

7t.r

2

711-2

=

evr — 2

we know that L = mvr = (1) & ( 2 ) = > ^ =

R-

F

nh

..(2)

2n

neh 47rm

(b) Magnetic induction • B = J

..(1)

2r

-

8v

2r 2nr

E V

(3)

ond nu Newton's 2 law and Coulomb's law e2 v2 -

mv 2

e 47ie0mr

(2) & (4) => v =

e2 2he n

(4)

n2h2e & r = 8 n

7ime

2—7 putting all these values in (4) we obtain B= — 8s0h n FIITJCC, ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 16. Ph: 6865182, 6854102. Fax: 6513942


For the convex lens L, |i = 3/2, r1 = +30 cm, r2 = -40 cm. l_i/3_1Yj_ fg I 2 A30

+ +

j j 40;

, 240 fa = cm 7 For water lens, 4 „ > 3 1 i 4 •1 f,. = -120 cm = 40 13 t Focal length of combination, due to the mirror, is given by f A 1 J_ 1 f = -24 cm = -2 f 7 -ve sign is due to the mirror. A l ^ = 2mA AIb 20jaA

6.

=10Q

(b) The input resistance RBE = 20mV 20|iA

AVBE

AIR

= 1 kQ. 2mA

ALR

= 0.1 mho. AV be 20mv (d) The change in output voltage is RL ALC = (5 kQ) (2mA) = 10 V The applied signal voltage = 20 MV (c) Transconductance =

Thus, the voltage gain is

10V

20mv

= 500.

i at hypotenuse = 45° Hence ray is totally reflected, if 45° > C sin 45° > sin C => \i >V2 M-min = 1-414 Let r be the angle of refraction in water. Then V2 sin 45° = - sin r 3

r = 48.6°

E = -20V6 i - 20 V6 j NIC"1, |EA| = |EB|

\ c

|E| = 2 |Ea| COS 6 = V(20A/6) 2 +(20^6) 2 1 Q Vr 2 - 1 = V4800 47te0 r 2 r 2x1 Q = 80 and V = 4nsn „

..(i) . . .. (ii)

ER

45°



from (i) and (ii) 2Vr2-1 r

V4800

2

80 ,

. 2 16 (r - 1 ) = 3r which gives r = 2 m, - = m. V3

=>

2 r = — m is not possible as 9 = 60° v3 Hence r = 2m, and CO = V3 m. from (ii) substitute r = 2 m

Q = —^ x 10"9 C 9 C is the mid point of A and B, let coordinate of C be (x The unit vector along CO =

CO

y').

- x'i - y'j

CO I

x/3

, . E -20V6i-20V6j 1 . 1 , and a l s o — = , =——i- — j IE| V4800 V2 V2 , , CO E and also — — = ——— ! C O | |E | x'i - y'j 1 1 V3

V2

which gives x' =

V2J , y' = J |

Hence the x co-ordinate of A =

V2

- AC sin 45° =

y co-ordinate of A = J - + AC sin 45° = V2 (V3-1) (

+ 72

(V3+1)

V2

similarly B(

9.

V2

V2

72 hr

(i) Energy of each photon = E = — = 3.975 x 10"19 J A. No. of photons falling on surface per second & being absorbed, n= — — = 2.52 x 1019. 2.48eV (ii)

h

hv

The linear momentum of each photon = p = — = — X c /.. Total momentum of all photons (falling in one sec.) nhv 10J = 3.33 x 10"B N-s c Rate of change of momentum = Force = — = 3.33 x 10"8 N. dt



10.

(a) According to Moseley's equation for k„ radiation \i

i-^-Wi-.il

A. IJ^ 2 Let X is the wavelength of tungsten k« radiation and ^ and X2 the wavelengths of two unknown k,( radiations then 2

>M _

(z-1) X (Z1-1)2 But for tungsten z = 74 i 21 3 (Zi - 1 ) = 7 3 J — V 71 '21

3

(z2 - 1 ); = 7 3 ,j—— \ 198 (b)

Zi = 41

=> 10

^

z 2 = 25

(i)

R=

(ii)

P = (10 - 0.7) 4 x 10"3 = 37.2 mW.

2 x 10

=4.65 KQ

*

*

*

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IIT- JEE 2002 PHASE - I - IV

PHYSICS SOLUTIONS Heat received by ice is Qt = mL + mCAT = 10700 cal. Heat given by the container is Q 2 300

j m c ( A + BT)dT = -m c AT +

BT2

500

300

= +21600m r 500

By principle of calorimetry, Ch = Q 2 => m c = 0.495 kg.

2.

(a) Let total distance moved by the block is S = (I + 2)m Where I is the distance moved by the block before touching the spring. Now, work done by gravity on the block is W g = mg S sine = 10 x 10 x S sin30 J => Wg = 50 S J (1) Work done by spring on the block is Ws = - I k x 2 Here and

K=100N/m x = 2m

=>

Ws = ~ x 1 0 0 x 4 J 2 ^ W s = -200 J Total work done W = W g + W s => W = (50 s - 200) J

(2)

Since change in K.E. of the block is zero, as W = AK.E. => 50 s - 200 = 0 => S = 4m (b) as S = I+2 => I= S- 2 = 2 m Work done by gravity over this path length is 1 W g = mg x 2 sine = 1 0 x 1 0 x 2 x - = 1 0 0 J

WQ = A K.E.


RSM-12-P-l-IV-T(M)-PH-(S)-2

100 = — mv 2 - 0 2

v2 = 1 ^ 10

= 20

v = 2V5m/s 3.

Magnetic field at the origin, B KNM + ^KLM ~ (a) Required force, F = q

4R

(-Vobx^(I-i) 4R

~~ ' )

4R

L

J

(b) Force on semicircular loop KLM due to external field B j is same as the force experienced by straight current carrying conductor along KM = - I ( 2 R k ) x B ] = 2IBR? Similarly force on the semicircular loop KNM

= 2IBR i

.-. Net force on the loop = F = 4IBRi. 4.

Consider a small element of the rod. developed in this element is = dm co2 r = (Adrp)co2 r Hence total tension in the rod T = JdT

Tension r

>dr pm 1 = nsm

i

= jAdrpco2r = 0 = 17t(0.1x10~ 2 ) 2 x 104 x (400) 2 x 0.5 2 = 6.28 x 102 N. Suppose Ae is the extension in the rod at that point P. . . M stress Then strain — = Y dT 1 x — xr .-. M. ~ area Y l.

Total extension

pco2r2dr = J-

1 p 3,3 1 104 x 16x 104 x.125 -co L = — x — 11 3 Y 3 2x10 4 = 3.33 x 10" m. 5 (a). Let a section be taken at a distance y from free end. fm Tension in the string at the section = yg Let dt be the time taken by the wave to travel a small portion dy of the string.

i

dy

W m / l ,, dy -J(mU) yg Total time to travel the whole length I dt =

=>

d

Jdt= J - ^ L o oVyg

= 2 vg

E. vg

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(b)

I

vb

Wall

Required beat frequency = |fi - f 2 | Where, f-, = apparent frequency for the motorist corresponding to the signals directly coming to him from source, and f 2 = apparent frequency for the motorist corresponding to the signals coming to him after reflection. Now,

f, = f

VJ + V" rn

&

v + vh

f 2 = f'

V + Vm V

where f ' is the frequency at which signals from sources are incident on wall. f' = f => f 2 = f

V V-Vb V V-VK

J VJ +r V m v-vh

Vv + ^ V "rr V

2V h (V + V )f Hence , the beat frequency = l^ - f 2 | = — 2 2

(v -v b )

6.

Let t be the time after which slipping between the sphere and plank disappears. For the sphere, N = mg, (iN = ma s

'

Sphere

=> a s = MS

x = la => jiimgr = ^2/m r 2 a => a =

rn

^sV

Plank

2r

For the plank, JLLN =

ap

Map

M

After time t, velocity of plank, vD = v - — — t M velocity of sphere, v s = jj.gt angular velocity of sphere, co =

1

For no slipping, the point of contact of sphere should have same speed as vs + cor = vD that of plank, Substituting and solving, jimg r =v • t M V. ^ ) t=

PIITJf

€

2v

2v

g(2^ + 5 l i + 2|a n >/)

ng[7 + (2m/M)]

a>

+

vs

ar

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2 2 . , rmr Ze(a) =— — r 47i80r

... ...(i

o mvr = — nh & 271

.... . . . (n)

-7

7.

From (i) & (ii), (b)

r= — ^ ~ Znme

Here I JWB W z = 3 W & WV m S B= B 208 m«"«e» e,

n2h2g

° , = 624nm„e 7xm0e

,.

(c) E n = Total Energy = 1872

mee4

n2

8s§h2

1872 w = - —— x 1 3 . 6 eV 2 n Ei = -25.4 keV ; & E 3 - Ei = 22.6 keV .

• •r .1 JJ-.

^ ^ 6247imee

OJ

n *25

ze 2

z 2Time 4

87ts0r

87iSon2h2

E 3 = -2.8 keV

he Required wavelength = X = — = 55 pm. (a) For the time interval from 0 to 60 seconds rocket accelerates and after then it moves under gravity. Distance moved by it in 60 seconds is given by 1 OHm Si = — x — j - x ( 6 0 s ) 2 = 36000m. 2 s 20m v(60s) = — 5 - x 60s = 1200m I s s If H be the maximum height reached. Then

0 = 11200 —

=>

H = 36000 +

2g(H- 36000), (v2 = u 2 + 2as) 1200x1200

2x10 (b) Time taken to ascend is

m ^

H = 108000m

t-i — 60s + ————s — 180 s, [t = t., + - ] 10 a Let time taken to descend is t 2 then 2 108000 = ~ gt 2 2 ,2 =

"V

2x108000 10 = 146.96s

Total time T = ti + t 2 = 180 + 146.96 = 326.96 s. 9.

Consider the ball in ground's reference frame. It is observed to be doing uniform circular motion for which each ball perpendicular must experience a resultant force of m © 2 1 magnitude directed towards centre of its circular path. => R + rng = mco2^n Where R is the force by rod on one of the balls. =>

R = mco 2 ^n - m g

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R | = V(mco2^)2 + (mg) 2 = m To decide direction of R , consider the following vector diagram. Thus, the angle that R makes with vertical /

equals tan' 1

10.

(a)

CD

2

i— Ttr2 T evr 2 n.r =

-mg

LI = NiA = M = 2nr

•(1)

—

v we know that L = mvr =

nh

• (2)

2n

neh

(1) & ( 2 ) = > H =

4nm

(b) Magnetic induction : B =

2r

= -Hi— 2r 2nr

ev B= M o 2 47ir Newton's 2 nd law and Coulomb's law

(3)

mv 47is0mr v2=-

47ts0mr

(4)

(2) & (4) => v =

2hs„n 7ime putting all these values in (4) we obtain 2 7 B = ju 0 nm e 8e0h 5n- 5 11.

Applying Kirchoff's Law in loop abcdefa E - L(di 1 /dt)-i 1 R 2 = 0 = [dt => i, = {1 - exp(-R 2 t / L)} 0 '1 2 o R2 Potential drop across inductor is L — = E e x p ( - R 2 t / L ) = 12e" 5t volts dt In steady state ii = E/R 2 When the switch S is opened, 1 = (E/R2)exp(-(R1 + R2)t/L) R

Putting the values, we get, i = 6 e"10t (direction of current in R-i is from e to b)

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RSM-12-P-l-IV-T(M)-PH-(S)-6 12.

(a) Since source lies in focal plane of lens so all the emergent rays will be parallel. Focal length of lens = 20 cm Inclination of emergent rays d/2 cl from principal axis : tan a = = — 20 40 initial path difference = ^ ^ = — - — sina 2sina

initial phase difference = — — - — X 2 sin a position of central maxima is shifted above O „ d Dd by 2sina = X 2Xsina (b) Intensity at O = 4I 0 cos 2 (cJ>/2) = 4I 0 cos 2

*

PIITJf €

*

ud v 2A,sina

*

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6865182, Fax: 6513942

FIITJCC

ALL INDIA TEST SERIES PART TEST -1 IIT - JEE, 2002 PHYSICS SOLUTIONS

1. t f

vx = u - v sin 9 v„ = v cos 9

W

45,

v tan 45 o_= y _= 1

[2]

=> VY = VX

u - v sin 9 = v cos 9 u v= sin9 + cos9 u A./2sin(9 + 45°) clearly minimum value of v =

v2

for 9 = 45°.

[3]

Applying C.O.E. m i2 \ V ^ g

h/2

/V

\ [5]

3

6

a = kr2 (say) Moment of inertia of the disc about an axis passing through it's centre and perpendicular to it's plane kR6 —

r IT say = j a27rrdr.r2 =

[2j

Mass of the disc R

m (say) = fo27trdr = \2nkr 1 + mR2 = I

dr =

TtkR4

[2]

(Parallel axis theorem)


AITS2002-PT-ll-PH(S)-2

TtkR6

+

TikR6

. =1

"i=—= ttR

6

61

5ttR

=

6

=>

, k=

57iR

[2]

2j

~ 5

l2 (Moment of inertia about diameter) =

4.

61

2

= - (Perpendicular axis theorem) 5

[2]

The minimum force on the rod required to open the container = mg + pgh A = 10 + 10 = 20 N Therefore minimum elongation required in the spring =

mg+

k

P9hA

= xo say = 0.5 m

[3]

Initial compression in the spring yi = — = 0.25 m k If y2 is the further compression required then applying conservation of mechanical energy j k(yi + y2)2 = | k x 2 + mg (yi + y2 + x0)

[3]

putting values and solving we get y 2 = 1 = o.75 m. 4 5.

[2]

If we consider that a sphere of radius R is placed with centre at Ci of density p! the force on the mass at P is (4/3)7tR3p1m (R + y)

towgrds

2

If we consider that a sphere of radius R/2 is placed with centre at C 2 of density pi the force on the mass at P is _ _(4/3)w(R/2) 3 p 1 m , . . . . F2 = G — I towards the sphere ( R / 2 + R + y)

[21 l J

If we consider that a sphere of radius R/2 is placed with centre at C 2 of density p2 the force on the mass in at P 3

_ G(4/3)7t(R/2) 3 p 2 m ^ 2— (R/2 + R + y)

[2]

By the principle of superposition F = F 1 - F 2 + F 3 = — TCR3 G m

3

6.

FIITJCC,

T . _ v _ v _ 1 IT v = J — ; Vw vw = m ' ~ X ~ £ ~ film v _ 3v V °P 1 ~ nc X~21

Pi (R + y Y

+

JP2_Pl)/8 ((3R/2) + y)

[2]

t = 2m ; x = .



v

=

S

=

V P

v=

/ypRT

=

V pM

V M

COT

1.44x8.31x300

27.7 x10" 3 = 1 . 2 x 3 x 100 = 360 m/s [2] 3 x 360 .. v0p = _ = 270 Hz [2] 2x2 v w = 275 Hz Vw - V o p = 5 275 = — x 2 V

[2]

302.5 m

302500 =

=>

^ = 3AZ2 ; A. = 2^/3

302.5

m m = 10"3 kg/m.

[2]

y component of velocity of the ball before it hits the ground equals v1y= - 7 2 x 1 0 x 0 . 8 =-4 m/s. y component of velocity of the ball after it hits ground, v2y= 7 2 x 1 0 x 0 . 4 5 = 3 m/s. Impulse in y direction = (v2y- v1y) m [2] Impulse in x direction = - |^m(v2y - v1y) [2] = change in momentum in x direction = m Avx => Avx= - f x ( v 2 y - Viy) = - 0 . 2 x 7 = - 1 . 4 m/s. => v2x - vix = - 1 . 4 => v2x = 0.6 m/s [2] _ V;Sin29 _ 2(v 2 sin9(v 2 cos9) _ 2v 2y x v 2x g 2x3x0.6

10

g

g

= 0.36 m = 36 cm.

Process in section C is adiabatic 125 PoVJ = PoVJ (y=~) 27

[2] (PV = C) V

125 PV = nRT

Tc =

25

27

Vn

1xR 9

C

=^V 25

0

5P0VQ 3R

[2]

41V

Final volume in B = V 0 + (V0 - —- V0) = 25 25 125 Final pressure in B = 27 v 41V, N "125

25 1xR 205P0V0 Final Temp, in A = Final temp in B = 27R PV = nRT

TB =

27

2Q5P0V0 27R

[2] [1]

Total heat supplied by the heater = Total increase in internal energy (As net work done on the system in zero)

FIITJCC,



W = nA Cv ATA + nBCvATB + N c C v ATC > 2 x 2R x 368

205P 0 V 0

P 0 2V 0

27R

2R

{ Cv =

+ 1 x2R

PTFV0.

= 2R }

y-1

205P 0 V 0

P„V„l

27R

R

+

1 > < 2 R x

5 PQV0 3

R

P0VQ R

[5]

Block 'A'. (Ground frame)

Cylinder (Block A frame)

Block 'B' (Block A frame)

Block B goes downward by y Conservation of energy 1 1 , 1 mR 2 mgy = - m v '2 +, - m v 7 - v.A2 r +, y 2 2 , 2 2 2 conservation of linear momentum 0 = - mv + m (v2 - v) - mv constraint equation v2 - Rco and Vi = v2 + Rra v 1 J 03 = v i1 v2 = - i and 2R

putting v2 = ~

2

C)

2+lm(v2+v2)

(1)

[3]

(2)

[2]

(3)

[2]

in equation (2) we get

v i v = —L

(4)

Putting the values of v, v2 and co in terms of Vi from equation (3) and (4) in (1) we get 93 • 2gy = — v 1i 72 fdy J

=

oV7

31h

V48g^

[3]

12g

o 31

Alternate Block A ' (Ground frame)

-41

Block 'B' (Block A frame) N

t

> mA

Cylinder (Block A frame) >T

H

mA

[3] For block A T + f - N = MA For block B mg - T = mai N = mA

Hi

(2)

[1]

(3)

For cylinder T + f + mA = ma2 (T - f) R =

(1)

mR'

-a

(4)

.(5)

EH

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Constraint a2 = Ra a2 + Ra = ai From (1) and (3) T + f = 2mA From (6) and (7) and (5) mR a ma. T-f 2 2R 4 From (4), (6) and (7) _ , . ma, T + f + mA = — 1 2 From (2) and (8) mg + f = 2mA + ma! or f = 2mA + m a 1 - m g and T = mg - ma! From (9) (11) and (12)

(6) (7)

[1]

(8) (9)

(10)

...(11) • • • (12)

ma. mg - mai - 2mA - mai + mg 2g - 2A =

4 From (10), (11) and (12)

...(13)

^ a « rna 2mA + mA = —1 or

A = —

.(14)

6

from (13) and (14) a -— a-]

249

[2]

31 block B displaces height h with acceleration ai.

Therefore h = - - ^ t 2 31 t=

10.

2

[1]

If x is the displacement of container w.r.t the block acceleration of the container = - — — m kx acceleration of the block = — m acceleration of container w.r.t. block F-kx kx F-2kx vdv .„, a= = =—[1] m m m dx Where v is the velocity of the container w.r.t. the block.

^

kx

" F - 2kx j v d v = Jdx m v

F

x

kx;_

2 m m For maximum compression the two bodies will move together or the velocity of container w.r.t. block will be zero. Putting v = 0 we get FIITJCC,



P =

(V 2 + 2Vq - 2W 0 )RT 0

[1]

VV 2

For max/min — = 0 dV

V = 7 2 V0

And pressure for V = 72 V 0 = 2 ( 7 2 - 1 ) ^ Vn 2v„

w=

(Max.)

[2]

2v„

jPdV = ^ V, Vo

J (V + ~~~_

2 RTn ^3V —+ 2Vg In2-2VQ

2V 0 )dV

= RT0 [2ln2 -1/2]

Total work done = RT0 (2ln 2 - ^ ) - 2RT 0 In2 RTn

12.

[5]

At t = 0, the source emits a sound and moves with 1 acceleration a and in time T moves a distance — aT2

Xo

s

2

T' = T +

x0+(1/2)aT

2

\

= T+

V C ,

1

2c

T' a= f=

2cT + aT

2Cf

f

a 2c

2c 2

2

T [2c(1/T) + a]]

2cfn 2cf0+a

°(f0-f)

2Gf

o . . . (1) [5] a + 2cf n For acceleration dv (Ahp) — = V2gh. sV2gh p dt dv 2gs => — =— =a . . . v(2) [3] dt A putting the value of a from equation (2) into (1) we get et 2cfn f= [2] 2cf0 + ( 2 g s / A ) * * *

FIITJCC,


FIITJCC ALL INDIA TEST SERIES IIT-JEE, 2002

PART TEST - II

PHYSICS SOLUTION a) S2P - STP = path difference = a sinG ^ = — x path difference = — a sinG X X = — .aG = — — X X D

b) The three phasors may be represented by:

When added end-on-end one gets, Ares = OR = A (1+2cos<|>)

c) lP oc Ar2es = A (1+2cos(|))2 The intensity minima occur when lP = 0 1 + 2cos<|> = 0 1

or coscj) = — (b = 2nrc + Y or

2n = cos — 3

= 2n n ± -

2n ay 1 = 2n n ± X D XD,

y= — I n ± a v

3



2.

Let radius of the disc be R, if disc is displaced x, the corresponding angular displacement that is the angle rotated by the disc, 6 = x/R The restoring torque x about point of contact of disc with ground T p = (F sin
MR2 22 \3 / 2

2

47T80(h +X )

+ MR2 a

Q2x

a =

[4]

67ts0MR(h + x,2\3/2 ) as x < < h Q2x Q20R a = 67ts0MRh3 67te0MRh a =

A F cos (|)

( - ve sign because a and 8 are opposite)

Q26 67ts0Mh3

Hence co = or T = 2n ,

Q2 67cs0Mh3 l67te0Mh3 q2

h ^—— = 2n-^/6ne 0 Mh

[4]

ALTERNATIVE METHOD Total energy at any instant 1 ^MR 2 ^ ~Q2 + -Mv2 + 2 2 4ns oy /h 2 +x 2 Differentiating w.r.t and simplifying Q x dx _ dv = 2v — dt 37T£0(h2 +X 2 )2£ dt Here, x « h dv — = - co x dt E=

where co =

3.

Q2 67rs0Mh

[4]

[3]

[3]

-Xlt Here we know NT =N0e

[1] -e"" 2 ']

and N2 =

[2]

X2 — .'. p activity after time t = N ^

_

= N0 (e a activity =N2^2 =

fiitjcc,

m„

A,1 X 2

210 {X2

-X,)

[e

m

o 210

-Mt

[3]

[4]


AITS2002-PT-ll-PH(S)-2 (a) Magnetic field at origin, B = Field due to arc ABC + field due to straight wire AC = W 2r _

8r

[1]

— [ S j n 450 + S jn 45° ] 47rrcos45°

2n _

Y

J |

27ir

+

d i r e c t i o n Jj

B = — 1 _ 1 2r 4 71 _ Mo1 (71 - 4) [along +z direction] 87tr

[2]

Since v and B are mutually perpendicular, Hence F = qvB = - ^ ^ ( 7 1 - 4 ) v 0 87tr _ ^0i(7r-4)qv0m/s2 Hence acceleration | a | = 87trm Hence initial acceleration a =

— 8v2rm7i

j

)

m/s

2

[2]

[2]

1 1 1 (b) Area of segment ABC = —7ir2 - - r 2 = - r 2 [1] 4 2 2 Since external field is uniform, hence net force on the loop will be zero. Torque t = j i x B -2 TT Hence torque | x | = — — 1 Bo[2] Apply Kirchoff's loop rule. We get

q

+ iR + e = 0

X

[2]

=

T_ R

iR = — - e

. . . (i)

Differentiating this equation with respect to time, considering that in our case (q decreases) ^ R

di

n.

=-i, di

dt ~ ~ c '

^

i

{ . \

•dt RC

In

1

nt RC

nt RC

i = i 0 e" ' where i0 is determined by equation (i)

[4]

where q0 = eC is charge of the capacitor before its capacitance has changed e «o = ( n - 1 ) -

R

-nt/RC

i = ( n - 1) —eR

fiitjcc,

[2]



6.

Motional emf induced across the infinitesimal element dx is given by de = vB.dx = (x - /)»

dx ..Q

p Tco

[2]

dx

l2rtx emf induced across PQ

_X

/LV.

= J(X"/>D ^ d x 2 nx [x - 1 In*]2

2n

[l-l\n2]

2n

_ C0ôio/(ln2 -1)

[4]

2 71 with Q at higher potential w.r.t. P. 7.

a) n=4 n=3

[2]

E 4 = -3.4 eV E 3 = -6.04 eV

n=2

E 2 = -13.6 eV

n=1

Ei = -54.4eV

[2]

b) Minimum energy for state n = 2 is 13.6 - 6.04 = 7.56 eV

[2]

c) Energy of the electron accelerated through a potential difference of 50 V is 50eV At most, it can excite electron fron n=1 to n=3 The number of possible wavelength are 3 1 K

—

54.4x1.6x10 _ he

19

i-

j" 1 i i _ j nf

ij ->2 \ n j

For transition 3-2, ni =2, n 2 = 3. A.32 — 1645A0 For 3-1; n-, = 1 ;n2 =3 >.3i = 257A0 For 2 - 1 ; ^ = 1;n 2 = 2 X21 = 304A°

[3]

d) Noenergy corresponding to X = 350A° is The E =

hc x

( 6 . 6 3 x 1 0 - 1(3 , 1 0 - ) (o.ssxio^Xi.exio71®)

The minimum excitation energy is 54.4 - 13.6 = 40.8 eV.

fiitjcc,

[2]


AITS2002-PT-ll-PH(S)-2 8. (a) Using Einstein's relation E m a x = hv - W 0

here

Emax = hv - W n =

13.6 eV ..(1) v4y In the second case if the excited level is nth, then n C 2 = 6 fa

\

V^

/

- W 00 = — v(l3.6eV) 16 '

[2] n =4

. . . (2)

From the above two equations, _ 5(2.55Xl -6 x10' 1 9 ) v = — = — (13.6 eV) or 6.63 x 10~34 5 16 15 = "3.08 x 10 Hz [ 4 ] / rt \ = hv 13.6eV = 5x2.55 - 10.2 = 2.55 eV

(b) W 0

l L

1

Ma ,

,

fwater = 4 x

v1/8y 4

16

— = —

[4]

= 4

m

[2]

The bird will pass through its own image if the rays fall normally on the mirror. This will happen if apparent distance of bird from lens =

fwater

[4]

h + — = 16/3 m 3/4 h = 4 m.

[3]

10. (a) At t = 0, there is a steady current downward through the inductor: i0 = V 0 /R. For t > 0, Kirchhoffs voltage law for the L-C circuit is dt C Here q is the charge on the lower plate of the capacitor. d i i 2. Since i = dq/dt [3] = _ =-a>o1 dt^ LC 1 Where coD s . The general solution of the above equation is a sinusiod:

VLC

i = imaxcos(co0t +
[3]

(b) It is told that the charge on the lower capacitor plate is zero initially; time t > 0 is given by t J I w t q = q 0 + f — d t = 0 + fidt = - 2 - fcos(co0t)dt J Ht J R J -

V

d t

° [sin(co 0 t)]l=-^sin(co 0 t) co0R co„R fiitjcc,

q 0 = 0. The charge at

[4]


AITS2002-PT-ll-PH(S)-6 11.

Charge on the capacitor is given by q

.

c

W

^ i = CB 0 /j

i

+

f

[2]

x W — + CB0v dt

[2]

Fne, = F - BU 2 X N dV ^ 1 + — m ^ - F - C B ! / V V dt"

dv dt

CB02

( 1+ -

v

t;

X 4 1+—

F-CB2

V

t

J

2 2

d 1+

x

m + CB ^

v

\2

[3]

—

^

F-3CB^V02 [1]

m + 9CB2j?2

*

*

*


FIITJCC RANKERS STUDY MATERIAL . IIT - JEE, 2002 PRACTICE FULL TEST -1 PHYSICS SOLUTION Velocity of stream at hole v0 = 7 2 g(H - x) horizontally. Let stream strikes the incline after time t at P, at that moment. Vx = v0 Vy = - gt Stream at the point of strike, makes an angle of 60° with horizontal downward. => =>

v

tan (-60°) x

gt = Vo V3

. . . (1)

PQ = ON = OM - MN = x - h = x - — gt2 2 For incline PQ = OQ tan 30° = PN tan 30° = v 0 1 tan 30° =>

x - - gt2 = v 0 1 tan 30°

2 2

g

g x

From (1) and (2)

73

_ 5 v| 2 g 5 50 0 x = _ H = — = 8.33 m. 6 6

x=-2(H-x)=> Velocity of source after time t g

}V3v 0

=Vo

2g The frequency of sound changes due to the Doppler effect ,-M v s + v 0 cos60° 1On0 n' = n 0

11

vs = 5v0. FIITJCC, ICES H o r n ' , (Opp. Vijay Mandal Enclave), Sarvapriya Vibar, New Delhi - 16. PA: 6865182, 6854102. Fax: 6J1J942

RSM12-PT-I-(M)-PH(S)- 2

Heat is transfered from vessel A to vessel B. The temperature of A will decrease while the temperature of B will increase. Let any time the temperature of A and B are 81 and 02 respectively. Then temperature difference between two vessels G = ©i - 02 Hence change in temperature difference d8 = (-de^ - d02 d8 = - (d8! + de2) . . . (i) During an elemental time dt, heat transferred through the rod, d Q _ kA(e, - e 2 ) d t = kA8dt e e The gasses are contained in two adiabatic vessels hence amount of heat d8 will transfer from vessel A through the rod to vessel B in the time dt, also the process on the gases in two vessels are isochoric dQ 2dQ ... dQ 2 dQ Hence d8< = = and d82 = = nCy 3 nR nCv2 5 nR Hence from (i) d8 = from (ii) d8 = -

16

—

w 3

fdt = - 15nR^ i ^ 16kA 16kA

+

—

5 nR v \ fkAG.dt

15nR ^

J

2A

'2

16 dQ 15 nR

£

f

8

J

T,-Ta

16kA (a) Magnetic field at origin B = Field due to arc ABC + field due to straight wire AC 2r Iv 2n / _ jlx0S 8r

i±o|__ [ sin 45 + sin 45 u ] 4n(r/2)

u 0 iV2

[along + z direction ]

4ic(r/2)

8nx

[along +z direction]

Since v and B are mutually perpendicular, Hence F = q v x B —8v27tr Hence initial acceleration a = _ ^ _ ( n _ 4 ^ 2 ) v 0 B o ( i - j ) 8v27irm 1 1 1 1 (b) Area of segment ABC = — n 2 - - r 2 = — r 2 71 2 Since external field is uniform, hence net force on the loop will be zero. Torque x = j l x B Hence torque |x | =

ir

1

K - - Bn

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RSM12-PT-I-(M)-PH{S)- 3

(i) Just after the key is closed the circuit will be as shown in figure 1 then circuit can be simplified as in figure2

The points D and E will be at same potential hence the circuit can be further simplified as shown in figure 3 & 4.

Fig. 4

68 Hence equivalent resistance across A and B, Req =

68

—

7

Hence reading of ammeter I =

x

16 „„

= 6.04 Q

+ 16

12

2A. 6.04 After long time the circuit can be redrawn as 16 Q

D 16Q

32 a -vVAM/-wvwv32 Q

c

JWWV-

A

16 Q

L©

-AW/ 16 Q

h Fig. 2

16 + 32

32 Q

Fig. 4

3

f l f r j f c ICES House, (Opp. Vijay Mandal Enclave), Sc .apriya Vihar, New Delhi - 1 6 . Ph: 6865182, 6854102. Fax: 6513942

RSM12-PT-II (M)-PH(S)-3

Hence the reading of the ammeter I = - A

8

(ii) The current through paths ADC and AEC will be same and will be equal to l2 = 1 9 16 3 . x x— = —A 2 8 48 16 3 Hence potential difference across AD, AE, DC and CE, V' = 16 x — = 3V 16 Hence the charge in the Ci, C 2 , C 5 and C 6 q = 100 x 3 = 300 |xC Also V DB = VEB = 1 2 - 3 = 9 V Hence the charge in the capacitor C 3 and C 4 = 100 x 9 = 900 nC. The potential at B due to the charge q on A = Due to charge -q on the inner surface of B = Due to the charge q' on the surface of B Due to the charge -q' on the surface of C = -

47ts0b 47I80b

47te 0 b

q' 4to0C

And due to the charge q'-q on the outer suface of C The potential is V B =

q'

q

47CE0b

47IS0C

-bq/c

_ q'-q 47CE0C

This should be zero as the shell B is earthed. Thus, q' = (b/c)q 7.

Using, u = - 30 cm, v u f Hence D = 1.6 - 0.6 = 1m

f = 15 cm

v = 30 cm.

and m = — = — = -1 O u Hence after separation, through upper part the rays will coverage 0.25 mm above principal axis. Similarly after refraction through lower halve the rays wiii converge 0.25 mm below principal axis. Hence separation between two coherence sources d = 1 mm. DX, 1x5400x10~ 1 ° Fringe width B = — = 5 = 540 urn d 1x10

"o:S"i

K -

FIITI€€,

30 cm

30 cm

i? L

s2 D = 1m

ICES House, (Opp. Vijay Mandal Enclave), Sar ipriya Vihar, New Delhi - 1 6 . Ph: 6865182, 6854102. Fax: 6513942

RSM12-PT-l-(M)-PH(3)- 5

(b) Maximum intensity at screen I =

+ J J ^ f = 4I0

as U = l2 = lo 3 3 But now intensity at O is — of the maximum intensity, I' = — x 4I0 = 3I0 But intensity on the 4 4 screen I = 2I0 (1 + cos 4>), where <> j is the phase difference between two rays reaching at O. Now intensity at O = 3I0 = 2I0 (1 + cos <)>) cos <)> = ^ ; <> j = 2r\n + ^ where n is an integer. 2%

S

But phase difference = — A x = 2% X p Where S is the p shift of fringe pattern. As fourth maxima lies below O while fifth minima lies above O hence fringe pattens is shifted downward i.e. the slab is kept infront of S2. Also shift DX, lies between 4p and 4.5 p whre p= — (fringe width) 6 From (I) s s = f c j

4+•

)tD

6

25

=

d

6

P

( H - l ) x 4 . 5 x 1 C T 6 x p _ 25 5400 x10" 1 0

~ 6 M=

ji. Sin 6X = constant, 8X = angle made with X-axis at any point, p sin 9X = po sin (%/2) Mo sin 9X = ^Mo _ = M n0{i-x/rr1 1-x/r (1-x/r) dy dx = tanGY = fdy=f dx " ^ - ^ - x / r f J l - (1-x/r)2 Putting z = 1 - (1 - x/r)2 Putting y = d and x - xA,

x 2 + y2 - 2xr = 0 xA = r{1 ±

- (d/r) 2 }

Putting the condition, as r -> oo, xA -» 0, xA = r{1- 7 l - ( d / r ) 2 (i) A = 228 + 4& 92 = z + 2 m v (ii) ^ ^

}

232 z = 90

2

= qv a B

v a = 1.59 x 107 m/s

From COM, m a v a = m y vy Thus, energy released or the sum of kinetic energies of the products

1

'IT) ' 'a

2

m

V v a H

2 2

a—V a — ra,

= 5.342 MeV = 0.0057 amu. Applying COE, Mass of 92 X 2 3 2 = m y + m a + 0.0057 amu. f l l T J € € , ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 1 6 . Ph: 6865182, 6854102. Fax: 6513942

RSM12-PT-I-(M)-PH{S)- 6

= 232.0387 amu Mass defect = 92 (1.008) + 140 (1.009) - 232.0387 = 1.9573 amu = 1823 MeV

10.

(i) During de-excitation, photons of six different wavelengths are emitted, therefore, principal quantum number of highest excited level is 4. he Energy of incident photons EinCident = — Eiincident ' Emin

=

he 304(A )

= 40.8 eV

n=4 V

n =3 \!

\t

\!

n=2

\f

n =1

E4 - E3

Since, some of emitted photons have energy less than 40.8 eV, Therefore E min < 40.8 eV Also, E m a x = E 4 - E 1; but some of emitted photons have energy greater than 40.8 eV. Therefore E m 3 X > 40.8 eV. As the atoms of the gas make transition by absorbing photons of energy 40.8 eV. Since Emax > 40.8 eV, therefore atoms of ground state do not absorb the incident radiation, hence the incident radiation is absorbed by initially excited atom which may belong to either n = 2 or n = 3, Emin ( E 4 - E 3 ) < 40.8 eV, therefore transition 3-» 4 is not possible. Therefore, the only possible value of principal quantum number of initially excited level is n = 2. (ii) The atoms of n = 2 make transition to n = 4 by absorbing photons of energy 40.8 eV. E 4 - E 2 = 40.8 eV . . . (I) _ But

En

=

13.6z 2 n^

w

-eV

._ _ 13.6z 2 _ _ 13.6z 2 fc4 — , fc2 16 4 substituting these values in (i) 13.6 z 2

16 + 4

= 40.8

Z = 4

6 , - 1 ^2- 2 1 7 . 6 e V 1 (iv) As E,max = E 4 - E1 and Emin = E 4 - E3 2 also E 4 = - 1 3 ' 6 y - = -13.6 eV

m

(4) and

e 3 = - 1 M W ! = . 2 4 . 2 eV (3)

Emax = -13.6 + 217.6 = 204 eV Emin = - 1 3 . 6 + 2 4 . 2 = 1 0 . 6 e V .

*

*

*

FIITJ€€. ICES House, (Opp. Vijay Mandal Enclave), Sarvapriya Vihar, New Delhi - 1 6 . Ph: 6365132, 6854102. Fax: 6513942

f l U J C C RANKERS STUDY MATERIAL IIT - JEE, 2002

PRACTICE FULL TEST - II

PHYSICS SOLUTION 1.

Mass corresponding to each hollow sphere _ (4/3MR/4)3m_ M (4/3)R 3 64 (i) As the force at centre of a sphere is zero so •F 4 + F R = O Fi +F 0 Since F, and F2 are equal and opposite so they will balance each other. Similarly F2 and F4Fr

=0

(ii) Gravitational potential at 1 if there were no hollow spaces = VI + V 2 + V 3 + V 4 + V R = -

VR

GM

R/2 4R GM - [V-, + v2 + v3 + v4 ] 4R GM/64 GM/64 GM/64~

GM 04R R/72 GM (17-272). 64R

2.

GM'

GM' _ GM R/2 " 4R

R/72

For pure rolling a = a x 2R • • •(1) Let the force of friction be f backward. Net force along horizontal is F cos 9 - f = 5 M.a (Translational motion)... (2) For rotational motion



Torque about axis of rotation: T = ia F. R + f x 2 R = — MR2 .a 2 17 Applying equation (1) F + 2f = — Ma =>

(3)

4

Adding 2 x equation (2) with equation (3)

u f A-7

=>

\

2F cosG + F = Ma j — + 10 5 7

KA

= —Ma 4 4F(1 + 2 cos 6) 57M For any tension T at a given loss of temp the restoring force is x 4T ' ' F= -2T sine = -2T —— = x (9 is small) ii 2 1 => wosc =

T

'

T

4T Vm* ml

T=iz

GV

!x

(V

T

where m = mass of the bead

The time of propagation of a transverse wave from 1 to 2 is 21 t = — where v=

u

1

m1

21

(2) where mi = mass of the wire j e u (2) T=

T B (

* |m \/2Vm

Tit V2n '

Suppose T-i be the initial temperature and T2, the gas temperature after AQ heat has been supplied to the system. In this case, if the gas expands, work done will be zero so, AQ=AU AU= change in internal energy + Potential energy stored in the spring = ~ R(T2-T1)+ ^ k(x 2 - x 2 ) , .. „ _ f kx Ps Initially, P= - = — => x = — , s s k ^ Where x is the compression in the spring. Also PV= RT RT Psx = RT Ps= — Hence x2 _

RT

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RSM12-PT-II (M)-PH(S)-3

k RT2 AU=-R(T2-T1)+^ [ k

RT,. 3 R ] = | R ( T r T 1 ) + ~ (T 2 -Ti) k

AU= 2R(T 2 -T 1 ) ^ C(T 2 -T 1 )=2R(T 2 -T 1 ) => C = 2R (i) B = A ( l

+

j + k)

A=2a 2 (7r + 1 ) i + — ] + — k The flux through close loop <> j = B.A = 2a (n + 1)

=

,

473(97I

B &

7ia2 B

a2 B

4 S

2 V3

+ 10)

, def> e = — ' dt

=

(97r + 10)a 2 g 4^3

(ii) current in the loop, I =

e r

2

(97t + 10)a a 4^3 Since strength of magnetic field is increasing, therefore flux linked is also increasing. Hence according to Lenz's law induced current should oppose increase of flux linked. Therefore its direction will be as shown in the figure.

Consider that length x of the dielectric is inside the capacitor. The capacitance of the system is c _ e 0 kxl | g 0 l ( l — x)

Charge on the capacitor zj [ ( k - l ) x + /]e Q= dq dt ^

(k~1)8

dx dt

| _ s 0 lsv(k - 1 )

d Heat produced in time t = I2 Rt Let temp, of the water increase by T => mcT = I2 Rt FIITI€€, ICES House, (Opp. Vijay Mandal Enclave), Sar ipriya Vihar, New Delhi - 1 6 . Ph: 6865182, 6854102. Fax: 6513942


2

T =

Rt

mc

The distance between the two distinct images can be given as

_F

D=F 2

= 1

_R

R

n2 - 1

If ni - n 2 « An d=

n, - 1 &

(n., - n 2 ) R

=

(rv,-1)(n 2 -1)

ni « n 2 » n we obtain,

(An)R

(n-1) 2 Let the first maxima occurs at P. For that

d cos 6 = X => cos 6 = X/d =>' 1 - 2 sin2 0/2 = A/d 2 1 - y /2D2 = X / d =>

putting sin 6/2 » tan 9/2 « y/D for large value of D.

y- = V 2 ( l - X / d ) D > = f

1-

(n-1) 2 X AnR

D.

L

A

y2

Using geometry, yi = (190) — = 1.9cm 10 y 2 = (195)

(0.1)

3.9cm

The region in which interference occurs is y 2 - y i = 3.9-1.9 = 2cm /ID The fringe width is co = — 2d Where D = 190 + 5 + 5 = 200 cm ; 2d = 2mm

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c • 3 x 10 X= - = rx = 0.5um v 6x10 CO

_ { > . 5 x i < r «

= 0.5mm

_ ( y 2 - y i ) _ (20) = 40. no.of fringes = CO 0.5

9.

Nuclear reaction 84Po210 — > 82Pb206 + 2 He 4 Mass converted into energy per reaction Am = 209.98264 - 205.97440 - 4.00260 = 0.00564 amu = 5.25 Mev 0 693 0 693 The decay constant X = = = 0.005 / day T1/2 138.6 Let M gms of Po

required per day for the reactor

Number of nuclei in M gm =

210

x NA

(6x10 2 3 )M _ ^ 210 dN

,M n n n (6x10 2 3 )M = X.N = 0.005 x — per day dt 210

. . . (i)

(6x10 2 3 ) So energy produced per day = 0.005 x

^

—- x M x 8.4 x 10~18 Joule

= 12 x 106 M Joule Energy obtained per day = 2 x 103 x 24 x 60 x 60 Joule Hence M = 14,4 gm per day The efficiency of the generator is 10 %. Hence the quantity required will be 144 gm per day.

10.

From (i) = - — = 2.057 x 1021 per day = A dt Let A 0 be the activity before 1386 days (10 half lives) then A = A 0 (1/2)10 or A0 = A x 210 = 2.057 x 1021 x 2 10 Which gives A 0 = 2.10 x 1024 per day. Here we know Nt = N 0 e " M [e'^ 1 ' - e"*-2']

and N2 =

where Ni represents the number of B; atoms after time

X-2 —

t, N 2 represents the number of P 0 atoms, p activity after time t = A , ^ 210 a activity = a.2N2 =

M

° ^ 210 (X2 -

*

*

*

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FIITJCC ALL INDIA TEST SERIES IIT - JEE, 2002 FULL TEST - III (MAINS) PHYSICS

SOLUTIONS 1.

Let m be the mass of the beads. Using energy conservation 1 , 1 mgh = — mv^ + - m v : =>

v 2 + vf =2gh

[2]

(1)

Constraint relation

2 x ^ + 2 x ^ = 0 dt dt or,

v2

(2)

3

[3]

From (1) and (2) we get, v1

4^2 . . 3V2 —— m/s and v2 =

[13

Heat current from A to B _ IAB =

(T

A

-T

B

)KA_

=4.62 J/s

[1]

Similarly, Heat current from A to C, lAc = 3.85 J/s Heat current from B to C, lBc = 3.08J/S Rate of heat supplied by the reservoir at A = IAB + IAC = 8.47 J/s Rate of heat supplied by the reservoir at B = lBC - IAB = -1.54 J/s Rate of heat supplied by the reservoir at C = -l B c - IAC = -6.93 J/s (Negative sign of heat supplied indicates heat being absorbed)

fllT/ee

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AITS-2002 - FT-III- PH(S)-2 For the first ray, net deviation 6 1 = (i - r) + (i - r) = 2(i - r) and for the second ray 52 = i - r + TC - 2r + i - r = 7i 2i - 4r Now, 82 = 35 => n + 21 - 4r = 3[2(i - r)] Putting i = 60°, we get, r = 30° sin 60° = V3 Msin30° Apply COE

Partially silvered

f

2

1 1 1 - m v .2 + - m v „2B +—2 L m — M 2 " 2 I 2) L R J 2gl(1 - cos8)= v 2 + | v 2

[3]

•••(1)

Also from Constraint relation y l +x 2 = t2 2yA

dy

+ 2XB

dt

(£ = length of the rod) dx B

=0

dt

: _X_B •Vb vA = (v y.A => v A = tan9vB from (1) and (2) & V b

=

/2g^(i-cose)

1 + - cot 0 2

i

[2]

. . . (2)

2gl(l-cos9)

VA =

dt

dt

[2]

2

tan 0 + -

2

Velocity of mid point of the rod, v = ^-(v B i - v A j ) Putting the values of i and B we get, v = f - j L - I - J jm/s 5.

E = -(13.6 eV) Z2 (

= -13.6

(Z + 2 ) 2

Z 2 - ( Z + 2)2)

So,

Or,

[2]

Z2 4(Z + 1)x13.6

(Z-2)2

(Z + 2)

Now energy of electron is k = Or,

[1]

h2 2A2m

eV

2

we have X =

...(1)

V2mk

k = 6eV

[3]

4(Z + l ) x 1 3 . 6 = 6 (Z + 2)

(Z + 1) _ 3 (Z + 2)2

+ 4 2

=

1 Q 2 e V

2

16

(Z-2) (3Z+2) = 0

So, the value of Z = 2 (neglecting the negative/ fractional value)

[3]

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AITS-2002 - FT-III- PH(S)-2

If the string slacks at 6 . mg cose

mv 2

2kXQsin0

. . . v(1) ^sinG £ ' Using work energy principle /./sine 2kXQdx 1 (2 z\ -mgfcos6+ = -mlv -uz) it x 2

[2]

- m g ^ cose + 2VX Q In(sine) = - miv2 -u 2 )

[3]

using (1) -2mg£ cose + 4 kXQ In(sin9) = mg^ cosG -2kX,Q - mu2 mu 2 = 3m£ cose - 2kXQ [2ln(sin9) + 1] u-

[2] V V2

m [1]

= 5.7 m/s (putting values and solving) At any given instance of time the slide wire is at distance x from origin, then the

x

A

resistance of the circuit is R = k ( 2 x + a If the velocity of slide wire is V, then the

x

x

X

X

Bâ

Or,

x

X

x

X

x

X

x

X

$

x

X >

pl» X

X

X

X X X

emf generated is B£V so we have B£V - - (2x + £)\ = 0 a

x

X

X

X

x

X

x

x

x

X

X

X

F=Fo V

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

(0,0^ X

V

X

X

X

X

X

X

X

X

X

X

[2]

2x-H

This current exerts magnetic force in the wire given by F' B£a V B 2*2, V So, F' = \ m = ^B = 2x + ^ 2x + Since, F - F ' =

mdV dt 2 2

So

F0V 0 m

f

Jo

B£a

2x + < km F^ m

= mV

1

dV

l v 2x + ^.

dx

B2^2C

B2£2af

1

km {2x + £J

B¥a Fo -2-X2km m

FOTJtl

V

dV

[3]

dx

dx=

j\

dV

r

2x +

In \

I

j)

+v0 =V

[3]

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AITS-2002 - FT-III- PH(S)-2 We first find the area in which the cannon shell can reach. The equation of trajectory for cannon shell is 1

2

—sec2 9

y = x tanG -

(1)

u For maximum y for a given value of x 1 2 . —gx ay __(2tan0) =0 d(tane) = x - 2 u

[2]

tane =

gx Putting in equation (1) 1 2 -gx2

ymax — Xgx

1+

u 1

or

ymax = — • 2g

[5]

g2x2

Danger zone

2

u2

The cannon shell can hit an area given by

y <

y

u2

1 2 29X

2g

u2

250m 5Q(W2m

500V2m x

Given in the problem y = 250m, u = 1Q0m/s, g = 10 m/s. Putting these value we get, „ 2

2000

<250

500V2 < x < 500V2

Plane is in danger for a period of

9.

1000V2 500

= 2V2 sec

[3]

For the charged particle m-, = 3mg, q = 10|aC, v = 1o(i + 2j), B = 0.6-rcTT The particle A will move in a helical path. Time after which it will cut x-axis 2rcm 27i x 3 x 10~6 T = = 1 sec qB 10x10~ 6 X0.6tt

[2]

x-coordinate where the charged particle A will cut x axis is v x T, v x 2T, v x 3T, v x 4T Till it is less than 50cm Given vx = 10cm/s x = 10cm, 20 cm, 30cm or 40 cm Velocity required by second mass for collision at these point, 50-x v= time of motion 50-10

1

- 50cm -

vx

[2]

= 40 cm/s, for x = 10cm

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FIITJ€€ ALL INDIA TEST SERIES IIT-JEE, 2002 FULL TEST - IV (MAINS)

PHYSICS SOLUTIONS Equation of motion for the device ; dv mg - kv = m — y dt

°(g

V m

)

v=

• • • (i)

[2]

• • • (ii)

[2]

0

m

9(i k

C"

kt/nn

)

1

f' = 700 { 330 -660 [330+ v j => v = 20 m/s putting this in equation (ii) e; a i.7t 2 0 = —— x 1 0 ( 1 - e 5 4 ) 1.7 ' 34 — ~ = (1 - e 5 4 ) 54 54 solving t = — - 3 . 2 sec. Vmax

[2]

=M=540=31.7*32m/s k 17

fmm = 700 j ! | | j = 656.25 Hz. 2.

[1]

[3]

(a) W AB = \ Pdv Using PV = RT a V T V = RT R or a d V = — p = d T 2vT

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6513942

AITS-2002-FT-IV-PH(S)-2

4 T

WAB = ~ Wbc

0

[2]

jdT = ^ x 3 T 0 = | R T 0

P CLIAT V/2 1 0> = 4RT 0 In — = 4RT 0 In = 4RT0 In 2 P1 a(T 0 )

[1] [1]

W c a = R ( T o - 4 T O ) = -3RTO AUAB=1X^(4T0-T0) =

^

AUBC = 0

AUca=1X^(T0-4T0) =

Qbc

QAB = | RTO + = 4RT 0 In2

nJCA UcA

-

^ [1]

= 6RTo

[1]

15RT

1o

[1]

(b) Q A B = U A B + W A B C (4T0 - To) = | R T 0 + | R T 0 = C x 3T0 6RT 0 = C x 3T0 C = 2R. 3.

[2]

(a) COLM 2mv 0 = 3MV cm 2m Vcm = 3M COAM

[1]

M 2mv 0 — = l 0 co = 3 +m 2 12 vV3y v3y

(0

= -IW 2 co co =

3mv,

[2]

2Mt Apply COE, lrn(2v0)2=±3Mv2m+±lcmco2 m _ 24

[1]

M ~ 17

Aliter: By Coefficient of restitution 3mv 0 2m 2 v 0 = — v00 + col/2 = Vn 0 + 2IW 3M 3m m 24 M

Vcm =


m

(2

M

17' 48 5l" V °

and co =

72 v 0 34 I

36 v 0 17 t

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6569493,

Fax:

6513942


(b) v

2

21 = v m +|co c m x —

\2

2

VA

2 f48] + f24) L 51J I17J

:

8Vl3, 17

[3]

, 2n 17n£ (c) T = — = co 18v0 Distance travelled by O in time T, __ 8 ni s = v cm T = — . 4.

[3]

(i) From COE mgh = ^ k ( h - L ) 2 solving, ^ __ kL + mg ± ^2mghL + m 2 g 2 k Taking only positive sign _ kL + mg + ^ m g k L + m 2 g 2

h

[3]

(ii) When speed is maximum, acceleration is zero 1 /. » 11 o m< —mv 2 =mg(L + x ) — k x ; x = — 2 ^ ' 2 k L . mo 2 = ii2gL +

V

[3]

(iii) Time to come to rest = time of free fall + time in SHM to stop stretching (2T —

free fall time, tf = V

9

the velocity of block when it enters into SHM = gtf=V2gL=v0

[1]

Period of SHM =

= T

Equation of velocity for SHM, V= Vmax Sin ©t when v = v0 to=-sin-1^CO

v

m

_

[2] v0

Time in SHM till v = 0 = T-to Total time taken to come to rest for 1st time = tf + (T - to)

t=

\2l

+

n - sin

^2g£ + mg 2 / k

[2]

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Length of air column for resonance, 3 2 0

£ = ( 2 n - 1 ) — = (2n-1)x 4f 4x320 , 1 3 5 7 9 ... i = — m , —m, — m , —m , — m . . . (i) W 4 4 4 4 4 Minimum length of air column = 0.25 m Equation of SHM. x = 1 cos cot . . . (ii) First resonance will take place at t = 0

[2]

[1]

Second resonance will take place when length of air column is — m x = 0.5 = 1 cos cot => t, = 1 sec. The third resonance will take place when the length of the air column is 5/4 m x = 0 m, t 2 = 3/2 sec. similarly for fourth and fifth resonances length of the air column will be 3/ 4m and 1/ 4 m and time for these is equal to 2 sec. and 3 sec. respectively. [3] (b)

PMAX = A P 0 + P M A X = APQ + 2 x 1 0 5

It is when the spring is at its maximum elongation.

[2]

Rate of radiation from sun = aTs4(47rR2 ) OT 2 R 2 S

Intensity of sun rays at earth = — — W m ~

2

=I

[2]

R s = radius of sun, r = distance of sun from earth, T s = Temperature of sun. Power incident on the lens P = l7rd2/4 Watt. Rate of incidence of momentum on metal surface - P/c ; c = velocity of light, after incidence momentum becomes zero, rate of change of momentum dP = — = force on metal surface [2] dt c But for a lens

v u formed on the screen = x — .0 u

Radius of image circle

Area of image = a = n j ^ s

'

V

[2]

Intensity on metal surface lm =

dP dt

a

=

^

4f

= 5 . 6 7 x 1 0 - W/m 2

[1]

(b) From Wein's law, AmTs = 3 x 10'3 , 3x10-3 _ ,__7 Am = = 5x10 m 6000

[1] 12400

2.48 eV 5000 According to Einstein formula for photoelectric effect Energy of photon = E P

FIITJCC Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 6515949,

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6513942

AITS-2002-FT-IV-PH(S)-5 2.48 = 1.48 + Kmax Kmax = 1 eV de-Broglie wavelength of photoelectron X= (a)

7h = 1.23 x 10"9 m. ; V2mK

v It just reaches at O' Its velocity at O' is zero .•.U0 = U 0 . ' qq0 + _ n q q o _ !o 7W 4Tts0Va2 + H 2 solving n = 61 [2]

_nqq^ a a 4m 0

H =

qq0 47tE 0 Va 2 +H 2

+

[4]

+ mgy 4ne0â + (H-y) 4mja + {y f •: Potential energy is converted into kinetic energy so it decrease initially but again it increases because at O' total energy is potential only. (b)

2

(a)

2

+

[2]

nqq0

qq0

U

3

m is the mass of electron

2

151 30

K^] H

y

[2]

[1]

^

( 2 nr jl = l A ; Area of sector PQ= — + r2 v 2 , Area of triangle POR = Area of AOQR = r 2 ju. = i r 2 —+ 1 i + r 2 j + r 2 k = lr' v2 , x = jl x B = - i l r 2 V3 (b)

| + 1 i + ] + k l x ( i + j + k)

Flux linked with area PQ =

_B_

Induced emf = e =

r dt

ar

7tr'

V3

(k-1)

[2]

2

V3

— + 3 = <> l v2 , +3

[3]

dB dt

—+ 3 , V3 V3 Current will be in clock-wise sense is P->R-*Q->P.

FIITJCC Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bw

TiBlr2

[3]

-— + r V3 . 2 j

B Flux linked with AOQR = - ^ r 2 V3 D Flux linked with AOPR = - = r 2 V3 g Total flux linked with circuit = — V3

e =

71

— + A1 i + r 2 j + r 2 k

[2] [1]

'er,n.j, New Delhi - 16, Ph : 6515949 , 6865182, 6569493, Fax : 6513942


(a) During the time the satellite is above the horizon, its angular displacement relative to the earth should be 29. where 9 = cos"1(R/2R) hence, 9 =7i/3 required time t = ^ ^ [2] ©-co,o Hence © is the angular speed of the satellite, given by GM ©2 (2R) = (2R)2

or,

GM

= 4.419 X 10^ rad/s 8R VSR3 and ©o is the angular velocity of the earth 2n „ „ = 0.727 x 10- rad/s ©o = 24x3600 putting the values of © and ©0, we get t = 5.67 x 103 sec. = 1.57 Hrs. [1] © =

[2]

(b) (i) When the coin has been removed from the top of the block, the buoyant force of the block should decrease by an amount equal to the weight of the coin (for equilibrium). As a result say x length of block comes out of water, then xApg = Vdg

-

H

Ap Thus, value of I will decreases by x.

(ii) Now h' be the level of water after lifting the coin, then conserving the volume of water, hA, - LA = h'Ai - (t - x)A =>

h

. A,

[ A1P

2

]

V When the coin falis in the water, it will cause a rise in water level equal to — . Thus net Ai change in the water level A V Vd V Ah = [2] A 1 A 1 P A, I Pj 10.

(a)(S 1 P) 0 P = M S 1 P ) ( S 2 P ) O P = Mm ( S 2 P ) + (Hg - JLtm)to

Optical path difference : A0p = (S2P)0p - (S-|P)0p A 0 p = Mm (S 2 P - S I P ) + (Hg - |Im) t0 _ MmYd A0p ' + (P-g - Mm )*() For Central max., A0p = 0 (Mg-Mm)t0D

y = -

Mn (4-t)t0P ( 1 0 - t)d When at'O', y = 0 t = 4 sec. FIITJCC Ltd., ICES House, Sarvapriya

[4]

[2] Vihar (Near Hauz Khas B- s Term.), New Delhi 347, Ph : 6515949 , 6865182,

6569493, Fax : 6513942


(b) Speed of central max. dy 6Dtn v= [2] dt (10 - t ) 2 d When it is at O, t = 4 sec. 6Dt n 1 x 36 x 1 x 10~6 v = 36d 6x2x10~3 = 3 x 10"3 m/sec = 3 mm/sec.

* *

s ->

S;

[2]

*

FIITJCC Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas B- s Term.), New Delhi 347, Ph : 6515949 , 6865182, 6569493, Fax : 6513942

FIITJCC ALL INDIA TEST SERIES IIT-JEE, 2002 FULL TEST - V (MAINS)

PHYSICS SOLUTIONS 1.

Calculation of Centre of mass 'C' 2m x 0 + m x R _ R OC = [1] 2m+ m Conservation of linear momentum mv0 + 0 = 3mv=> v = vJ3 [1] Conservation of angular momentum about C

L2mRD2 + 2m

2R mvn sin30° x — CO = Vo/8R. VRC =

,3,

+ vc=

Loop PQRSP 10 + q1 5 - q ,

2m X. —

1

I

\ I

j

V ~~

2R2 [ ] 10) + mf ] h

I

° \

J

>

J I I JJ 3

[2]

R + - l c o (cos 30°] - sin 30?) ?+^ ^ 7 = 3J 12 12

Vp = Vpc

2.

I

(Rs 2

y

j

J

j

[1]

10 - q 2

|

2+ q

5-q

5 5 2 => 7 ( q + q 1 ) + 19 = 0 Loop PUTSP 1 0 - q 2 | 5 + q2 | 2 + q 5 - q 10 5 5 2 => 3q2 + 7q = 6

= Q

2

q = qi +q2

(1) [2] _

Solving the above equations, we get, 31 ^ 232 q 2 = — nC and q, = ^C. 13 91

5 + q2

(2) (3) [2] [1]

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AITS-2002-FT-V-PH(S)-2 3.

For central maxima d sine = (V3 - 1)d => sine = (V3 - 1)

[3]

VDW . y2 = (D2 + y2) (4 - 2V3) y 2 (1- 4 + 2V3) = (4 - 2V3)D2 4-2V3

[2]

V2V3-3 Distance between the two central maximas.

2

4.

V2V3-3

[1]

°

For lens L-1 1

1-1-1f v

u ' + 50 J

^

_

L1

1

v

-100

(-100, o)

1_

1 // i \)i | !

I

v ~ 50~100

=>

I

30 c m '

t TI

v=100cm

m = — = -1 u

[3]

Because of lens L1 the image will be formed at (100, -1)cm. This will work as an object for lens L2 u = + 70 cm, f = + 50 cm 1

_ 1

1_

+ 50 ~ v

+70

A-_1_ 1 _ 6 + v " 50 70 " 175

v=

175

cm

v 175 m = —= u 6x70 .I = m x r* O =

175

25 « =— x15 cm 6x70 4

So, |y co-ordinatel of final image Coordinate of final image =

FIITJCC

355

[3]

25 27

6 ' 4

1 27 ~ = —cm cm

[2]

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AITS-2002-FT-V-PH(S)-3 0.7V

V A - V B = 1V VA - V G = 0.7V ^

—Of-

V G - V B = 0.3V

=>

0.3 i, = — =0.3A

V C - V D = 1V Let us assume that no current is passing through diode P and Q both. [3]

-Ot-

"wwr W—

A

i

B

1fi

0.7V i.R

1.3

1V Y AAAM/1

-o-

V c - V E = i x (0.3 + 0.4) = — (< 0.7V) 1.3 VF - VD = 0.4i + 0.6i = 1.3 (>0.7) That indicates that the current is not passing through P but passing through Q.

-w-

•WWT^

V F - V D = (0.4 + 0.6)i3 = 0.7 [3] => i3 = 0.7A V c - V D = V c - V F + V f - V d =1V => i2 x 0.3+0.7 = 1V =c> • i 2 = 1 A => Current through the battery = i1 + i 2 = ( 1 + 0 . 3 ) A = 1 . 3 A [2]

E - O -

i2

F

-MWV1

'3

I2-I3

Considering a point at a distance x from the bottom, hence mass of the section of the bottom

I

=

A,0x

Adx '

xf 1+ L, Hence: tension at that point = weight of the hanging part T=

; hence velocity at that point = 1 +

I

*

— = gx 1 + —| V^ V V LJ

[2]

[2]

LJ

Time taken by this pulse to cross a distance 'dx' dx gx

[1]

1+ —

L

Total time required by the pulse to cover the entire length L

, , d * - , = 2 ^ l n ( l + V2) 2 [ J id ' I : Vg[x + (x /L)J 7.

[3]

Let q be the charge on the capacitor after a time t. Then, we can write, dq 1 f q^ +— ...(1) =i(t) R .dt, vw where . . . (2) i(t) = (A.N0e~xt)2e.f = (2e.fAN0 )e~u= i 0 e" xt [3] Solving equation (1), we get, t -xt i«e RC q= -+C., e , where, Ci is an arbitrary constant. RC

FHTJCC Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi -16, Ph: 686 5182,6965626, 685 4102, 6515949 Fax: 6513942

AITS-2002-FT-V-PH(S)-4

Using the initial condition, we get, r

q(t) = -

\

+q0

[5]

fe"t/RC+-

RC RC M - -q, e + q2 e The total charge on the capacitor becomes zero at a time t = T where q(x) = 0 i.e. JRC 0 = -q,e~ + q 2 eXT 1 or, T = -ki 3l [4] 1/RC-l t,RC

8.

The block will initially move along with the sphere in a circular path w.r.t. sphere and than after loosing the contact it will move in parabolic path w.r.t. ground. Part-I Body in contact with the sphere (frame of reference - sphere) mv,2 mg cos 0 - N - ma sin 0 = • •(1) For loosing the contact N = 0 Using work energy principle 1 ~ -mv

• •(2)

= mgR(1 - cos0) + maR sin 0

from (1), (2) and (3) mg cos 0 - ma sin 0 = 2mg (1 - cos 0) + 2 ma sin 0 or 3mg cos 0 - 3ma sin 0 = 2mg v a= g

ma

f0. - rsin « 0)a l =l 2 3 (cos'sin u , 4 jj I 3^2 2 _ >/2 —

sin

[4]

. . . (3)

* - 0

4

3V2~

0

3

and v

[2]

Part - II Body in air (frame of reference - ground) Vertical component of the velocity at this moment = v sin 0 displacement in the vertical after that h = (R + R cos 0) Final vertical velocity v v = V( v sin0) 2 +2gR(1 + cos0) = ^

Rg sin2 0 + 2Rg + 2Rg cos 0

Total KE of the block 1 1 mg 2R = — mvH2 + — mv 2 2 2

[4]

vH = V 4 gR - 1 = ^2Rg -—Rgsin 2 0 - 2Rgcos0

fllT/ee

. n . .TV2 where 0 = — s i n — . 4 3

[2]

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AITS-2002-FT-V-PH(S)-2 9.

Let's say at any instant radius of water is r and in time dt it becomes r - d r . 2nr dr hpL = dQ = Heat flow out in time dt dQ _ _ _ k2rixhde [63 ~dT ~ " dx j ^ - j ^ h k d O H In - = 27thke0 r

27Thke0 _ dQ

H =

d Q =

ln(R/r) ~~dT

2Khke^ d t ln(R/r)

27thk60 dt =27trdrhpL ln(R/r) V2R

ke0t = pL

V2R

I r In — dr = pL

j r(ln R - In r)dr

R V2R

R

r ( i nR R))dd r - |Irr In in r dr = pL

r2 1 = PL — l n R - l n r + -

2

2

lnR-
1V2R

2

2R2

2

= pLR

10.

V2R

«/2R

= PL

=PL

[3]

r 11 lnR-lnV2R + 2

2

2

R^ f 1 J|nR-lnR + 2 | 2 2

4

t =

pLR 2 2k9

. 1 1 -In-^H-— 42 4

[3]

For bullet 1

9

u cos6t = x0 + — at

(1) a: acceleration of chimney

Final horizontal component of the bullet should be equal to the velocity of chimney. ucos0= at (2) 1

h = u sinGt

2

gt

9

[2+2+2=6] (3)

From (1) 10 x - t = x 0 + - at2

2

2

From (2) 10 x -

2

(4)

= at (5)

From (3) ^ = 10x^t-lx10xt 20 2 2 t=

2

[3]

9V3

(6)

10 Put the value of t in (5) 50 9V3 From (1)Xo = 5 t - - a t 2 = 2

FIITJCC

4

~m

[3]


AITS-2002-FT-V-PH(S)-2

11.

Sudden change in magnetic flux produces an electric field and the ring will experience an impulsive torque. Thus, induced emf e = - — = —— flB.ds dt dt J But s = or or

cfE.dl = - — fads J dt J dB E. 2nr = - n r 2 — dt r dB E= 2 dt

[4]

1 J

or 2 dB Force on the ring, F = qE and the torque x = Fr = qEr = - ~ — — or

Now, angular impulse = change in angular momentum jx dt = loo

or

qr 2 fdB , - - — —.dt = mr. © 2

J

or

dt

(o = -

qB 2m

Thus |co| = ^ 2m Now, current I = q/T

[4]

Hence induced magnetic moment p. = iA = — x 7ir2 271 Substituting co = qB/2m, we get (J. =

=

q

2

B r

2 ,

4

m

27t 2m I = 2ti Induced magnetic moment (j. = i A = — x 7tr2 => |j. = 2ti

*

FIITJCC

*

4m

.

[4]

*


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