Mock Test Jee Main

ETOOS & MOTION

JEE-Main, 2016 SAMPLE PAPER 1 Duration :3 Hours

Max. Marks : 360

_______________________________________________________________________________ INSTRUCTIONS In each part of the paper, Section-A contains 30 questions. Total number of pages are 32. Please ensure that the Questions paper you have received contains ALL THE QUESTIONS in each section and PAGES. SECTION - A 1.

Q.1 to Q.90 has four choices (A), (B), (C), (D) out of which only one is correct & carry 4 marks each. 1 mark will be deducted for each wrong answer.

NOTE : GENERAL INSTRUCTION FOR FILLING THE OMR ARE GIVEN BELOW. 1.

Use only HB pencil or blue/black pen (avoid gel pen) for darkening the bubble.

2.

Indicate the correct answer for each question by filling appropriate bubble in your OMR answer sheet.

3.

The Answer sheet will be checked through computer hence, the answer of the question must be marked by shading the circles against the question by dark HB pencil or blue/black pen.

4.

While filling the bubbles please be careful about SECTIONS [i.e. Section-A (include single correct)]

Booklet Test Code

D

1 0 0 1

A

0 0 0 0 1 1 1 1

B C D E

2 2 2 2 3 3 3 3 4 4 4 4

Batch

Roll Number

10+1

2 8 3 2 3

10+1 10+2 10+3 Crash Paper

G H

5 5 5 5 6 6 6 6 7 7 7 7

I J

8 8 8 8 9 9 9 9 Paper 2

F

Name

1 Paper 1

0 0 0 0 0 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8 8

For example if only 'A' choice is For example If Correct match correct then, the correct method for (A) is P; for (B) is R, S; for for filling the bubble is (C) is Q; for (D) is Q, S, T then A B C D E the correct method for filling For example if only 'A & C' the bubble is choices are correct then, the P Q R S T correct method for filling the bubble is A B C D E

9 9 9 9 9

the wrong method for filling the F I R S T N A M E M I D D bubble are

L E N A ME Test Date

L A S T N A ME

D D MMY Y

The answer of the questions in wrong or any other manner will be treated as wrong.

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Ensure that all columns are filled. Answers, having blank column will be treated as incorrect. Insert leading zero(s) if required : '6' should be filled as 0006

'86' should be filled as 0086

0 0 0 0 1 1 1 1 2 2 2 2

0 0 0 0 1 1 1 1 2 2 2 2

3 3 3 3 4 4 4 4 5 5 5 5

0 0 0 0 1 1 1 1 2 2 2 2

3 3 3 3 4 4 4 4

3 3 3 3 4 4 4 4

5 5 5 5

6 6 6 6 7 7 7 7

6 6 6 6 7 7 7 7

8 8 8 8 9 9 9 9

5 5 5 5 6 6 6 6 7 7 7 7

8 8 8 8 9 9 9 9

8 8 8 8 9 9 9 9

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SAMPLE PAPER - 1 (JEE-Main)

Page # 2

PART - I [MATHS] SECTION - A

SECTION - A [ oLrqfu"Bizdkjdsiz'u]

[STRAIGHT OBJECTIVE TYPE]

iz-.1 lsiz-.30 rd pkj fodYi (A), (B), (C), (D) fn;s gS ftuesa^^dsoy,d^^lghgSA

Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct

1.

If [x]2 + [x – 2] < 0 and {x} =

1 , then the 2

1.

(A) 4 (C) 2

Which one of the following best represent

the graph of

y =

x  x2 2

x x

1 gks, rks x 2

dslaHkoekuksadhla[;kgS [fVIi.kh: [x]rFkk {x}Øe'k% x ds egÙke iw.kk±d Qyu rFkkfHkUukRedHkkx Qyu dks fu:firdjrsgSA]

number of possible values of x, is [Note : [x] and {x} denote greatest integer less than or equal to x and fractional part of x respectively.] (A) 4 (B) 3 (C) 2 (D) 1 2.

;fn [x]2 + [x – 2] < 0 rFkk {x} =

2.

(B) 3 (D) 1

fuEuesalsdkSulk y =

x  x2 x2  x

dsvkjs[kdksmi;qDr

rjhds lsfu:fir djrk gS \

?

y y=1

y y=1 (A) (A)

O(0, 0)

x

x=1

O(0, 0)

x

x=1

y y (0, 0)O (B)

(0, 0)O x=1

x

(B)

y = –1 (SPACE FOR ROUGH WORK)

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x=1

x y = –1


Page # 3

y

(C)

y = –1

O(0, 0)

x

x=1

y

y=1

(D)

y

y=1 (C)

y = –1

y = –1

O(0, 0)

x

x=1

O(0, 0)

(D)

4.

O(0, 0)

(A)

1 2

(B) 0

(C)

 4

(D)

has a

3.

 2

The value of x satisfying the equation

 

·

 

log 2 x 

·

 

log 4 x 

·

 

log 8 x 

4.

(C) 3



(A)

1 2

(B) 0

(C)

 4

(D)

 2

lehdj.k log x

·

 

log 2 x 

·

 

log 4 x 

·

 

dkslUrq"Vdjus okys x dkekucjkcjgS (B)  (D)

x

;fn lehdj.k sin 2 + cos 2 = p dk ,d gy gks, rksfuEuesalsdkSulkpdkekuughagksldrkgS

 

...=3

is equal to (A)

x=1

y = –1

If the equation sin 2 + cos 2 = p solution then p cannot be

log x

y = –1

y=1

y = –1

3.

x

x=1

y

y=1

y=1

y=1

(A)

1 3



(C) 3 (SPACE FOR ROUGH WORK)

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(B)  (D)

1 3

log 8 x 

...=3

SAMPLE PAPER - 1 (JEE-Main) 5.

Page # 4

If cos2 x – (c – 1) cos x + 2c  6 for every

5.

cos2 x – (c – 1) cos x + 2c  6

x  R, then the true set of values of c is

6.

7.

(A) [2, )

(B) [4, )

(C) (–, –2]

(D) (–, –4]

If ,  be the roots of 4x2 – 17x +  = 0, 

;fn izR;sd xR ds fy, gks, rks c ds

ekuksadklghleqPp;gS (A) [2, ) (C) (–, –2] 6.

(B) [4, ) (D) (–, –4]

;fn ,  lehdj.k 4x2 – 17x +  = 0,  R ds

 R such that 1 <  < 2 and 2 <  < 3,

ewy bl izdkj gaS fd 1< <2 rFkk 2<  <3gSa,rks

then the number of integral values of  is

 dsiw.kk±dekuksadhla[;kgS

(A) 1

(B) 2

(A) 1

(B) 2

(C) 3

(D) 4

(C) 3

(D) 4

If the cubic equation x3 + 2x2 – 4x + 5 = 0 has roots



3



,  3



7.

;fn ?kuh; lehdj.k x3 + 2x2 – 4x + 5 = 0 ds ewy

and  , then the value of 3

5  5  5



13 

,  ,oa 

gks, rks



3





 5 3  5  3  5 13 



dk

is equal to

ekugS

8.

(A) 13

(B) 9

(A) 13

(B) 9

(C) 8

(D) 5

(C) 8

(D) 5

Let f (x) = ax2 + bx + c, where a, b, c  R

8.

ekuk f (x) = ax2 + bx + c, tgk¡ a, b, c  R

and a  0. If a2 + c2 – b2 + 2ac < 0 then

rFkk a  0 gSaA ;fn a2 + c2 – b2 + 2ac < 0 gks,

the equation f (x) = 0 has

rks lehdj.k f (x) = 0 ds

(A) roots real and equal

(A)ewyokLrfod,oacjkcjgSa

(B) non-real roots

(B)ewyokLrfodughagSa

(C) both roots greater than unity

(C)nksuksaewybdkbZlscM+sgS

(D) roots real and unequal

(D)ewyokLrfodgSijUrqcjkcjughagS (SPACE FOR ROUGH WORK)

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Page # 5

9.

The graph of y = f (x) is shown in figure y

9.

x

O

fuEu esa ls dkSulk y = f(–|– x|) ds vkjs[k dk lcls lghfu:i.kgSA

y

y x

O

(A)

y

x

O

(B)

y

x

O

(C)

O

x

O

y

y

(D)

x

O

y

(C)

x

O

y

(B)

x

O

Which of the following best represent the graph of y = f (– |– x|) ?

(A)

y=f(x) dk vkjs[k n'kkZ;k x;k gS y

x

(D)

O

(SPACE FOR ROUGH WORK)

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x


11.

Page # 6

A cubic polynomial P(x) is such that P(1) = 1, P(2) = 2, P(3) = 3 and P(4) = 5, then P(6) is equal to (A) 7 (B) 10 (C) 13 (D) 16

10.

If f : R  R is defined as

11.

(A) 7 (C) 13

x 2  kx  3 for x  0

f (x) =

12.

  2kx  3

,d ?kuh; cgqin P(x) bl izdkj gS fd P(1) = 1, P(2) = 2, P(3) = 3 rFkk P(4) = 5 gS, rks P(6) dk eku gS

;fnf:RRifjHkkf"kr gS  f (x) = 

for x  0

(B) 10 (D) 16

x2  kx  3 x  0 d sfy , 2kx  3

x  0 d sfy ,

If f (x) is injective then 'k' can be equal to

;fn f (x) ,dSdh gS rc'k' cjkcj gksxkA

(A) 0

(B) 1

(A) 0

(B) 1

(C) 2

(D) 3

(C) 2

(D) 3

Let xi (i = 1, 2 ..... n) be the roots of the

12.

equation xn + 3xn–1 + 5xn–2 + .... + 2n+1 = 0

ekuk xi (i = 1, 2 ..... n) lehdj.k xn + 3xn–1 + 5xn–2 + .... + 2n+1 = 0 ds

ewy gS rFkkf (x)= ax2 + bx + c = 0 (a, b, c  R rFkk a  0) ds ewy  + i

and f (x) = ax2 + bx + c = 0 (a, b, c  R & a  0) possess  + i as a root (  R and   0),

n

gS (  R rFkk  0) rc  f ( x i ) gksxkA

n

then

 f ( x i ) is

i 1

i 1

(A)ges'kk _.kkRed ;fnnfo"ke gS (A) always negative if n is odd

(B)ges'kk/kukRed;fnnfo"kegS

(B) always positive if n is odd

13.

(C) may be positive or negative if n is odd

(C)/kukRedrFkk_.kkRed gksxk;fnnfo"kegS

(D) may be negative or positive if n is even

(D) /kukRed rFkk _.kkRed gksxk ;fnnle gS 13.

Suppose f(x) = x2 – 2x. The value of

f  f (2i 1)  is equal to (where i =

1 )

ekuk f(x) = x2 – 2x, f  f (2i 1)  dk eku cjkcj gksxk(tgk¡i=  1 )

(A) 15

(B) 5i + 5

(A) 15

(B) 5i + 5

(C) 35

(D) 15i – 10

(C) 35

(D) 15i – 10


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Page # 7

14.

Let a, b, c be real numbers with a > 0 such that the quadratic equation ax2 + bcx + b3 + c3 – 4abc = 0 has non real roots let P(x) = ax2 + bx + c and Q(x) = ax2 + cx + b which one of the following is true

14.

ekuka, b, c okLrfod l[;ka, rFkka > 0 bl izdkj f}?kkr lehdj.k ax2 + bcx + b3 + c3 – 4abc = 0 vokLrfod ewy j[krk gS ekukP(x) = ax2 + bx + c rFkkQ(x) = ax2 + cx + b rc fuEu esa ls ,d lR; gksxkA (A) P(x) > 0  x  R rFkk Q(x) < 0  x  R (B) P(x) < 0  x  R rFkk Q(x) > 0  x  R (C) u rks P(x)> 0  xRvkSj u gh Q(x) >0  x R (D)Bhd ,dP(x)rFkkQ(x) /kukRed gksxk  xR

15.

|n|x|| = |k – 1| – 3, 4 fHkUu

(A) P(x) > 0  x  R and Q(x) < 0  x  R (B) P(x) < 0  x  R and Q(x) > 0  x  R (C) neither P(x) > 0  x  R nor Q(x) > 0  x  R (D) exactly one of P(x) or Q(x) is positive

xR 15.

16.

|n|x|| = |k – 1| – 3 have 4 distinct roots then k satisfies (where |x| < e2, x  0) (A) (–4, –2)  (4, 6) (B) (–4, –3)  (5, 7) (C) (–3, –1)  (2, 6) (D) None of these

If

f



3x



=

1 x

is true for exhaustive

lUrq"V djrk gS(tgk¡ |x| < e2, x  0) (A) (–4, –2)  (4, 6) (B) (–4, –3)  (5, 7) (C) (–3, –1)  (2, 6) (D)blesalsdksbZugha

16.

domain of 'f', then domain of f(x) is



(A)  3, 3



(C) 0, 3  17.



 (C)  0,

(A)  3, 3



Let P(x) be a polynomial with degree 2009 and leading co-efficient unity such that P(0) = 2008, P(1) = 2007, P(2) = 2006, .... P(2008) = 0 and the value of

;fn f  3  x  =

1 x

, 'f',ds izkUr ds fy, lR; gS] rc

f(x)dkizkUrgksxkA

(B) 0, 3  (D) 0, 3

17.



3 

(B) 0, 3  (D) 0, 3



ekukP(x) ,d cgqin gS ftldh ?kkr2009 rFkk egÙke xq.kkad,dbdkbZblizdkj P(0) = 2008, P(1) = 2007, P(2) = 2006, .... P(2008) = 0 rFkkP(2009) = ( n) – a dk eku] tgk¡

P(2009) = ( n ) – a where n and a are natural

n rFkk a izkd`r l[;ka gS rc (n+a)dk eku gksxkA (A) 2008 (B) 2009 (C) 2010 (D) 2011

number than value of (n + a) is (A) 2008 (B) 2009 (C) 2010 (D) 2011


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The range of the function

18.

4  x 2  x 2  1 is

f(x) =

19.

Page # 8

(A)

 3, 7 

(C)



2, 3

 3, 5  (D)  3 , 6 

x : x 



19.

2 or x  3

x : x   2 or x  3

Suppose, f (x, n) =

k

 log x  x  , then the

If  and  be the roots of the equation x2 + 3x + 1 = 0 then the value of

20.



(D)

 3, 6 

2, 3



ekuk x rFkky okLrfod la[;k lehdj.k4y2 + 4xy + x +6=0 dk lUrq"V djrh gSxds eku ds fy, lEiw.kZ leqPp;gksxkA

; k x  3 ; k x  3

n k ekuk f (x, n) =  log x   , rc x dk eku lehdj.k x k 1

f (x, 10) = f (x, 11), dks lUrq"V djrs gS &

21.

(A) 9

(B) 10

(C) 11

(D)buesalsdksbZughaA

;fn rFkk lehdj.k x2 + 3x + 1 =0 ds ewy gS rc 2

2

           dk eku Kkr djks & 1      1 

2

           is equal to 1      1  (A) 15 (C) 21

(C)

(D) x : x   2

value of x satisfying the equation f (x, 10) = f (x, 11), is (A) 9 (B) 10 (C) 11 (D) none of these

2

 3, 5 

(B) x : x  2

k 1

21.

(B)

(C) {x : – 2  x  2}

n

20.

 3, 7 

(A) {x : – 2  x  3}

(C) {x : – 2  x  2} (D)

(A)

(B)

Let x and y be real numbers satisfying the equation 4y2 + 4xy + x + 6 = 0. The complete set of values for x, is (A) {x : – 2  x  3} (B)

Qyu f(x) = 4  x 2  x 2  1 dk ifjlj gksxkA

(B) 18 (D) none of these

(A) 15

(B) 18

(C) 21

(D)bueslsdksbZughaA


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Page # 9

22.

Let f (x) = h(x) =

2 ; g (x) = cos x and x 1

22.

x  3 t hen the ran ge of t he

ekuk f (x) = h(x) =

2 ; g (x) = cos x rFkk x 1

x  3 rc la;qDr Qyu fogoh dk ifjlj

gksxk&

composite function fogoh, is (A) R+ (B) R – {0} (C) [1, ) (D) R+ – {1}

(A) R+ (C) [1, )

(B) R – {0} (D) R+ – {1}

23.

Consider the functions f: X Y and g : YZ then which of the following is/are incorrect? (A) If f and g both are injective then gof : X  Z is injective (B) If f and g both are surjective then gof : X  Z is surjective (C) If gof : X  Z is bijective then f is injective and g is surjective. (D) none of these

23.

ekuk Qyu f : X  Y rFkk g : YZ rcfuEuesalsdkSulkvlR;gksxk? (A) ;fn f rFkk g nksuks ,dSdh gS rcgof : X  Z ,dSdhgSA (B) ;fn f rFkkg nksuks vkPNknd rcgof : X  Z vkPNkndgSA (C) ;fn gof : X  Z ,dSdh vkPNknd rc f ,dSdh rFkkgvkPNkndgSA (D)buesalsdksbZughaA

24.

If  and  are the roots of the equation ax2 + bx + c = 0 then the sum of the roots of the equation a2x2 + (b2 – 2ac)x + b2 – 4ac = 0 in terms of  and  is given by (A) – (2 –  2) (B) ( + )2 – 2 (C) 2 + 2 – 4 (D) – (2 +  2)

24.

;fn  rFkk lehdj.k ax2 + bx + c =0 ds ewy gS rc

The quadratic equation x2 – 1088x + 295680 = 0 has two positive integral roots whose greatest common divisor is 16. The least common multiple of the two roots is (A) 18240 (B) 18480 (C) 18960 (D) 19240

25.

25.

a2x2 + (b2 – 2ac)x + b2 – 4ac = 0 ds

;ksxfn;sx;srFkkdsinksaesagksxk& (A) – (2 –  2)

(B) ( + )2 – 2

(C) 2 + 2 – 4

(D) – (2 +  2)

f}?kkr lehdj.k x2 – 1088x + 295680 = 0 nks èkukRediw.kkZafd;ewyj[krkgSAftldkegÙkemHk;fu"B Hkktd16gSArcnksuksaewyksadky-l-e-gksxk& (A) 18240

(B) 18480

(C) 18960

(D) 19240


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Page # 10

The solution set for [x] {x} = 1 where {x} and [x] are fractional part & integral part of x, is (A) R+ – (0, 1) (B) R+ – {1}

26.

tgk¡{x}rFkk[x],xdkfHkUukRedrFkkiw.kkZadh;HkkxgS& (A) R+ – (0, 1) (B) R+ – {1}

1   / m  I  {0} (C)  m  m   (D)

27.

28.

1   m  / m  N  {1} m   27.

The set of values of 'a' for which the inequality, (x  3a) (x  a  3) < 0 is satisfied for all x  [1, 3] is: (A) (1/3, 3) (B) (0, 1/3) (C) ( 2, 0) (D) ( 2, 3) If f (x) = px + q and

30.

(C)

1   m  / m  I  {0} m  

(D)

1   m  / m  N  {1} m  

leqPp; 'a'ds ekuksa ds fy, vlfedk (x  3a) (x  a  3) < 0 lHkh x  [1, 3] ds fy, lUrq"VdjrhgS& (A) (1/3, 3) (C) ( 2, 0)

f  f  f ( x )  = 8x + 21, 28.

where p and q are real numbers, then p + q equals (A) 3 (B) 5 (C) 7 (D) 11 29.

[x] {x} = 1 ds fy, leqPp; dk gy

;fn f (x) = px + q rFkk f  f  f ( x )  = 8x + 21, tgk¡ p rFkkq okLrofd la[;k,sa gS] rc p + qcjkcj gksxk& (A) 3 (C) 7

If x be the real number such that x3 + 4x = 8, then the value of the expression x7 + 64x2 is (A) 124 (B) 125 (C) 128 (D) 132

29.

Given f (x) is a polynomial function of x, satisfying f(x). f(y) = f(x) + f(y) + f(xy) – 2 and that f (2) = 5. Then f (3) is equal to (A) 10 (B) 24 (C) 15 (D) None of these

30.

(B) 5 (D) 11

;fn x okLrfod la[;k bl izdkj gS x3 + 4x = 8, rc O;atd x7 + 64x2 dk eku gksxk & (A) 124

(B) 125

(C) 128

(D) 132

fn;kf(x), Qyu x dk ,d cgqin gS, f(x). f(y) = f(x) + f(y) + f(xy) – 2 dks

lUrq"V djrk gS rFkk f(2) = 5 rc f(3)cjkcj gksxk & (A) 10 (C) 15


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(B) (0, 1/3) (D) ( 2, 3)

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(B) 24 (D)buesalsdksbZughaA


Page # 11

PART - II [PHYSICS] SECTION - A




Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct

1. 1.

,dlh/khjs[kkesa,dd.kfojkelsvkjEHkgksrkgS]

Starting from rest a particle moves in a

ftldkRoj.k

straight line with acceleration

1/2

a  25  t

2

1/2



2

m/s

a  25  t2 

for 0  t  5s

a a

3 m / s2 for t>5s 8

3 m / s2 , t>5s 8

t = 7s ij d.k dk osx gksxk :

The velocity of particle at t = 7s is:

(A) 11 m/s

(B) 22 m/s

(A) 11 m/s

(B) 22 m/s

(C) 33 m/s

(D) 44 m/s

(C) 33 m/s

(D) 44 m/s 2.

2.

m/s2 , 0  t  5s

d.k Aurry ds vuqfn'k osx uls fcUnqPls rFkk mlh

The particle A is projected from point P with velocity u along the plane and simultaneously

le;nwljkd.kBm/okZ/kjls''dks.kijvosxlsiz{ksfir

another particle B with velocity v at an angle

fd;k x;k gSA d.k ry ij fcUnq Qij Vdjkrs gSA rc &

'' with vertical. The particles collide at point Q on the plane. Then : B P B

A

P Q

A

Q

(A) v sin = u cos (B) v cos ( – ) = u (C) v = u (D) v tan ( – ) = u

(A) v sin = u cos (B) v cos ( – ) = u (C) v = u (D) v tan ( – ) = u


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Page # 12

6.

Pushing force making an angle  to the horizontal is applied on a block of weight W placed on a horizontal table. If the angle of friction is , the magnitude of force required to move the body is equal to : (A) W cos /cos (–) (B) W sin /cos (+) (C) W tan /sin (–) (D) W sin /g tan (–)

6.

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of static friction between the block and the slab is 0.60 while the kinetic coefficient is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g=9.8 m/s2| the resulting acceleration of the slab wil be : (A) 0.98m/s2 (B) 1.47 m/s2 2 (C) 1.52 m/s (D) 6.1 m/s2

7.

8.

In a cricket match the fielder draws his hands backward after receiving the ball in order to take a catch because : (A) His hands will be saved from getting hurt (B) He deceives the player (C) It is a fashion (D) He catches the ball firmly

8.

fØdsVeSpesafQYMjdSpysrsle;xsanizkIrdjusdsckn viusgkFkksadksihNsdhvksj[khaprkgSa]D;ksa\ (A)mldsgkFkksadkspksVyxuslscpkusdsfy, (B)ogf[kykM+hdk /;ku gVkusds fy,A (C) ;g ,d QS'ku gSA (D)xsandksetcwrhlsidM+usdsfy,A

9.

A block of mass 20 kg is suspended through two light spring balances as shown in Figure then : (A) Both the scales show 10 kg reading (B) Both the scales show 20 kg reading (C) The upper scale will read 20 kg while the lower zero (D) Their readings are in between 0 and 20 kg with their sum equal to 20 kg 20 kg

9.

æO;eku20kgdk,dxqVdkfp=kkuqlkj nksgYdhfLiazxrqykvkssalsyVdkgS rc (A)nksuksaiSekus10kgikB~;kadcrkrs g SA (B)nksuksiSekus20kgikB~;kad crkrsgSA (C) mijh Ldsy 20kg ikB~;kad tcfd fupyk'kwU; (D) budk ikB~;kad 0 rFkk20 kg ds chp] ftudk ;ksx 20kgds cjkcjA

7.

{kSfrtlsdks.kcukrsgq,,d/kdsyusokykcy]{kSfrtest ijfLFkrWHkkjdsxqVdsijyxk;ktkrkgSA;fn?k"kZ.k dks.kgS] oLrq dks xfr djkus ds fy, vko';d cy dk ifjek.kcjkcjgS& (A) (B) (C) (D)

W W W W

cos /cos (–) sin /cos (+) tan /sin (–) sin /g tan (–)

,d 40kgdhifV;k,d ?k"kZ.kghuQ'kZij fojke ijgSA ,d 10kgdk xqVdk ifV;k ds Åij fojke ij gSA xqVdk oifV;kdschpfLFkfrt?k"kZ.kxq.kkad0.60tcfdxfrt xq.kkad 0.40gSA10kgdsxqVds ij100Ndk,d{kSfrt cy yxk;k tkrk gSA ;fng=9.8 m/s2|ifV;k dk Roj.k gksxk& (A) 0.98m/s2 (C) 1.52 m/s2


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(B) 1.47 m/s2 (D) 6.1 m/s2

20 kg


Page # 13

14.

Two blocks of equal mass 2 kg are placed on a rough horizontal surface as shown and a force F is applied on the upper block. The system is initially at rest. Find acceleration of the lower block in m/s2. (g = 10 m/s2)

(A) 3 m/s2 (C) zero

14.

(A) 3 m/s2 (C) zero

(B) 5 m/s2 (D) none of these

15.

A physics text book of mass m rests flat on a horizontal table of mass M placed on the ground. Let N a-b be th e contact force exerted by body ‘a’ on body ‘b’. According to Newton’s 3rd Law, which of the following is an action-reaction pair of forces? (A) mg and Ntablebook (B) (m+M)g and Ntablebook (C) Ngroundtable and Mg + N booktable (D) Ngroundtable and Ntableground

16.

A side view of a simplified form of vertical latch B is as shown. The lower member A can be pushed forward in its horizontal channel. The sides of the channels are smooth, but at the interfaces of A and B, which are at 45° with the horizontal, there exists a static coefficient of friction  = 0.4. What is the minimum force F (in N) that must be ap plied horizontally to A to start motion of the latch B upwards if it has a mass m = 0.6 kg? B (A) 10 N (B) 0 (C) 14 N F A (D) 22 N

lekuæO;eku2kgdsnksxqVdsfp=kkuqlkj ,d[kqjnjh {kSfrt lrg ij fLFkr gSA rFkk ,d cy FÅijh xqVds ij vkjksfirgSAfudk;izkjEHkesafojkeijgSAfupysxqVdsdk Roj.k m/s2 esa Kkr dhft, & (g = 10 m/s2)

(B) 5 m/s2 (D)buesalsdksbZughaA

15.

/kjkry ij fLFkr æO;ekuMdh,d {kSfrtestijæO;eku m dh ,d HkkSfrd foKku dh fdrkc fLFkr gSA ekuk Na-b oLrq ‘b’ij oLrq‘a’ds }kjkdk;Zjr lEidZ cygSA U;wVu ds3rdfu;edsvuqlkj]fuEuesalsdkSulkcyksadkfØ;k izfØ;k;qXegS? (A) mg rFkk Ntablebook (B) (m+M)g rFkk Ntablebook (C) Ngroundtable rFkk Mg + N booktable (D) Ngroundtable rFkk Ntableground

16.

;gm/okZ/kjdqa
B

F

A

(A) 10 N (C) 14 N


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(B) 0 (D) 22 N


Page # 14

The displacement of y a article as a function of time is shown in Figure. The figure 20 indicates that : (A) The particle starts 10 t in sec. with a certain velocity but the mot ion is o 10 20 30 40 50 retarded and finally the particle stops (B) The velocity of the particle is constant throughout (C) The acceleration of the particle is constant. (D) The particle starts with a constant velocity, the motion is accelerated and finally the particle moves with another constant velocity.

21.

22.

23.

,d d.k dk foLFkkiu le; ds y

bafxrdjrkgSfd

20

(A) d.k fdlh osx ls

10

t in sec.

vkjEHkgksrkgSfdUrqxfreafnr

o 10 20 30 40 50

gSrFkkvUresad.k:drkgSA (B) d.k dk osx fu;r gSA (C) d.k dk Roj.k fu;r gSA (D)d.kfu;rosxlsvkjEHkgksrkgS]xfrRofjrgSrFkk

vUr esa d.k vU; fu;r osx ls xfr djrk gSA

The graph between the displacement x and time t for a particle D moving in a straight line S is sho wn in t he diagram. During the C intervals OA, AB, BC A B and CD the acceleration of the t o particle is : OA AB BC CD (A) + 0 + + (B) – 0 + 0 (C) + 0 – + (D) – 0 – 0

22.

Two bullets are fired simultaneously from the same point parallel to an inclined plane. The bullets have different masses and different initial velocities. Which will strike the plane first ? (A) The fastest one (B) The heaviest one (C) The lightest one (D) They strike the plane at the same time

23.

ds fy, foLFkkiu xo le; t ds chpxzkQfp=kkuqlkjgSA lek;kUrjky OA, AB, BC oCD dsnkSjkud.k dkRoj.kgksxk&

|

D

,dlh/khjs[kkesaxfr'khy,dd.k S C A

B t

o

(A)

OA +

AB 0

BC +

CD +

(B) (C)

– +

0 0

+ –

0 +

(D)

–

0

–

0

nks xksyh ,d leku fcUnq ls urry ds lekUrj ,d lkFk NksM+htkrhgSAxksyh;ksadkæO;ekufHkUu&fHkUugSrFkk izkjfEHkdosxfHkUu&fHkUugSA/kjkryijigysdkSuVdjk;sxk\ (A) rst okyk (B)Hkkjhokyk (C) gYdk okyk (D) ;s leku le; ij ry ls Vdjkrs gSA


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s in m

Qyuesafp=kkuqlkjgSAfp=k

s in m

21.



Page # 15

27.

28.

29.

30.

A particle moves on a rough horizontal ground with some initial velocity say v0. If 3/4th of its kinetic energy is lost in friction in time t0. Then coefficient of friction between the particle and the ground isv0 v0 (A) 2gt (B) 4gt 0 0 3v 0 v0 (C) 4gt (D) gt 0 0

27.

A parabolic bowl with its bottom at origin has the shape y = x2/20. Here x and y are in metres, the maximum height at which a small mass m can be placed y on the bowl without slipping (coefficient of static friction is 0.5) is(A) 2.5 m (B) 1.25 m x (C) 1.0 m (D) 4.0 m

28.

For the arrangement shown in figure the coefficient of friction between the two blocks is . If both the block are identical, then the acceleration of each block isF (A) – 2g 2m m F F m (B) 2m F (C) – g (D) zero 2m A particle moving in a straight line has velocity and displacement equation as

29.

,dd.k,d[kqjnjh{kSfrt/kjkryijdqNizkjfEHkdosxv0 ls xfr djrk gSA ;fn bldh xfrt ÅtkZ dk 3/4th le; t0esa?k"kZ.kesa[kpZgksrkgSArcd.ko/kjkrydschp?k"kZ.k xq.kkadgksxk& v0 (A) 2gt 0 3v 0 (C) 4gt 0

v0 (B) 4gt 0 v0 (D) gt 0

,d ijoy; dVksjk ftldk iasnk ewyfcUnq ij gSA ftldh vkd`fry=x2/20;gkaxoyehVjesagSAvf/kdreÅ¡pkbZ ftl ij ,d NksVk æO;eku m y fcukfQlysdVksjsijjgldsA (LFkSfrd?k"kZ.kxq.kkad0.5gS) gksxh(A) 2.5 m (B) 1.25 m (C) 1.0 m (D) 4.0 m

x

fp=kesaiznf'kZrO;oLFkkdsfy,nksxqVdksadschp?k"kZ.k xq.kkadgSA;fnnksuksa,dlekugS]rcizR;sdxqVdsdk Roj.kgksxkF – 2g 2m F (B) 2m F (C) – g 2m

30.

m

F

(D)'kwU;

,dd.klh/khjs[kkesaxfr'khydkosxofoLFkkiulehdj.k v  4 1 s ,

v  4 1 s , where v is in m/s and s is in m. The initial velocity of the particle is : (A) 4 m/s (B) 16 m/s (C) 2 m/s (D) zero

ds vuqlkj gS] tgk¡ vm/s rFkk s m esa gSA d.k dk izkjfEHkdosxgksxk&: (A) 4 m/s (C) 2 m/s


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m

(A)

|


(B) 16 m/s (D)'kwU;


Page # 16

PART - III [CHEMISTRY] Use Data : 1H1, 2He4, 3Li7, 4Be9, 5B 10, 6C12, 7N14, 8O16, 9F19, 11Na23, 12Mg 24, S32, 15P31, 17Cl35.5, 19K 39, 20Ca 40, 26Fe 56, 29Cu63.5, 30Zn65, 24Cr 52, 23V51 16

SECTION - A

SECTION - A


[ oLrqfu"Bizdkjdsiz'u]

Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct 1. The rate constant of a reaction depends on (A) Temperature (B) Initial concentrations of the reactants (C) Extent of reaction (D) All of these 2.

3.

iz-.1 lsiz-.30 rd pkj fodYi (A), (B), (C), (D) fn;s gS ftuesa^^dsoy,d^^lghgSA 1.

The C – Cl bond energy of the following compound in decreasing order. (a) CH3–CH2Cl (b) CH2 = CH – Cl (c) CH  C – Cl (d) Ph – Cl (A) a < b < d < c (B) a > b > d > c (C) c > d > b > a (D) a > b > c > d

2.

In the following reaction, BrO3–(aq) + 5Br–(aq) + 6H+ (aq)  3Br2() + 3H2O(aq) Select correct relation

3.

(A) 

,d vfHkfØ;k dk nj fu;rkad fdl ij fuHkZj djrk gS \ (A) rkieku (B)fØ;kdkjdksadhizkjfEHkdlkanzrk (C)vfHkfØ;kdkizlkj (D)mijksDrlHkh fuEufyf[kr ;kSfxd esaC–ClcU/k ÅtkZ dk ?kVrk Øe gksxkA (a) CH3–CH2Cl (b) CH2 = CH – Cl (c) CH  C – Cl (d) Ph – Cl (A) a < b < d < c (B) a > b > d > c (C) c > d > b > a (D) a > b > c > d

fuEuvfHkfØ;kesa BrO3–(aq) + 5Br–(aq) + 6H+ (aq)  3Br2() + 3H2O(aq)

lghlEcU/kdkp;udjsa

 3

d[BrO ] d[Br2 ]  dt dt

(B)  1 3

(A) 

d[BrO3 ] d[Br2 ]  dt dt


(B)  1 3

 d[BrO3 ] 1 d[Br2 ]  dt 3 dt (D) None of these

(C)

(C)


 d[BrO3 ] 1 d[Br2 ]  dt 3 dt

(D)buesalsdksbZugh (SPACE FOR ROUGH WORK)

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Al27,

13

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Page # 17

4.

The reactivity order of following compounds with AgNO3. CH2Br Br

4.

(b) Ph —CH — CH3

(a)

The reactivity order of following compounds with AgNO3. CH2Br Br

CH2Br

5.

CH2Br

(c)

(d) (CH3)3C–Br

(c)

(d) (CH3)3C–Br

Cl (A) a > b > c > d (C) b > c > d > a

(B) b > d > a > c (D) b > a > d > c

Cl (A) a > b > c > d (C) b > c > d > a

(B) b > d > a > c (D) b > a > d > c

Select the correct statement (I) –OH > – NH2 > –Cl , – I effect

5.

(A) I, II & III (B) II & III (C) II only (D) I & III For the reaction, 2N2O5  4NO2 + O2, the rate equation can be expressed in two ways

6.

d[N2O5] d[NO2 ] = k [N2O5] and + = k[N2O5] dt dt k and k are related as : (A) 2k = k (B) k = 2k (C) k = 4k (D) k = k If 3A  2B, then the rate  (A)



(C)



(B)



2 d(A) 3 dt

(D)



d[N2O5] =k[N2O5]rFkk dt

d[NO2 ] = k[N2O5] dt krFkkk fuEuizdkjlEcfU/krgS: (A) 2k = k (B) k = 2k (C) k = 4k (D) k = k +

d(B) is equal to: dt

2d(A) dt

vfHkfØ;k, 2N2O5  4NO2 + O2 ds fy, nj lehdj.k dks nks inks –

–

7.

lgh dFku dk p;u fdft,s (I) –OH > – NH2 > –Cl , – I effect (II) –NH2 > –OH > –Cl , + M effect (III) – Cl > –OH > –OR > –NH2 – I effect (A) I, II & III (B) II & III (C) II only (D) I & III

(II) –NH2 > –OH > –Cl , + M effect (III) – Cl > –OH > –OR > –NH2 – I effect

6.

(b) Ph —CH — CH3

(a)

7.

;fn 3A2BgS]rc nj 

d(B) fuEu ds cjkcjgksrhgS: dt

1 d(A) 3 dt

(A) 

(B) 

3 d(A) 2 dt

2d(A) dt

1 d(A) 3 dt

(C) 

2 d(A) 3 dt

(D) 

3 d(A) 2 dt


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Page # 18

Among the following the one which has largest carbon – Carbon bond is: (A) CH2 = CH2 (B) CH2 = CH – NH2

8.

(A) CH2 = CH2 (B) CH2 = CH – NH2 (C) O2N – CH2 = CH – NH2 (D) HO – CH = CH – NH2

(C) O2N – CH2 = CH – NH2 (D) HO – CH = CH – NH2 9.

In which of the following compounds + M effect is present

OH (I)

N=O (II)

fuEufyf[kresalsdkSuls;kSfxdesavfrl;qXeumifLFkr g SA

9.

fuEufyf[kresalsdksuls;kSfxdesa+MizHkkomifLFkr gksxkA OH

CH3 (I)

(III)

(IV)

11.

(III)

CHO

O–C–R

O–C–R (IV)

(V)

(V)

(A) I, II & V (C) I, III & IV

(A) I, II & V (B) I & IV (C) I, III & IV (D) I & III 10.

(II)

CH3

O

O

CHO

N=O

The rate of a gaseous reaction is given by the expression K [A] [B]. If the volume of the reaction vessel is suddenly reduced to 1/4th of the initial volume, the reaction rate relating to original rate will be: (A) 1/10 (B) 1/8 (C) 8 (D) 16

10.

The rate of a reaction is expressed in different ways as follows :

11.

xSlh; vfHkfØ;k dh nj O;atd K [A] [B] }kjk n'kkZ;h tkrhgSA;fnvfHkfØ;kik=kdkvk;ruvpkudizkjfEHkd vk;ru dk 1/4thdegksrkgS]rksokLrfodnjlslEcU/kj[kusokyh vfHkfØ;knjgksxh: (A) 1/10 (C) 8

1 d[C] 1 d[D] 1 d[ A ] d[B] – – =  2 dt 3 dt 4 dt dt The reaction is : (A) 4A + B  2C + 3D (B) B + 3D  4A + 2C (C) A + B  C + D (D) B + D  A + C 

(B) 1/8 (D) 16

,dvfHkfØ;kdhnjdksfuEurjhdksa}kjkçnf'kZrfd;k tkrkgS 

1 d[C] 1 d[D] 1 d[ A ] d[B]    2 dt 3 dt 4 dt dt

mDrvfHkfØ;kgS% (A) 4A + B  2C + 3D (B) B + 3D  4A + 2C (C) A + B  C + D (D) B + D  A + C


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(B) I & IV (D) I & III

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Page # 19

12.

13.

Units of rate constant for first and zero order reactions in terms of molarity (M) are respectively. (A) sec–1, M sec–1

(B) sec–1, M

(C) M sec–1, sec–1

(D) M, sec–1 13.

(B) 0.01 kg sec–1

–1

(C) 0.6 kg hr

'kwU;rFkkçFkedksfVdhvfHkfØ;kdsfy,njfu;rkad eksyjrk(M)ds:iesaçnf'kZrgSA (A) sec–1, M sec–1 (C) M sec–1, sec–1

The rate of production of NH3 in N2 + 3H2  2NH3 is 3.4 kg min–1. The rate of consumption of H2 is : (A) 5.1 kg min–1

14.

12.

(B) sec–1, M (D) M, sec–1

vfHkfØ;k N2 +3H2  2NH3 esa NH3 ds mRiknu dh nj 3.4kg min–1. H2 ds miHkksx dh nj gksxh % (A) 5.1 kg min–1

(B) 0.01 kg sec–1

(C) 0.6 kg hr–1

(D)buesalsdksbZugha

(D) None of these 14.

In which of the following molecules -electron density in ring is minimum?

fuEuesalsfdlv.kqdsoy;esa-bysDVªkWu?kuRoU;wure gS\

OCH3

NO2

(A)

(B)

(A)

OCH3

NO2 (B)

NO2

NO2

NO2

NO2 (C)

H2N

15.

(C)

(D)

(D)

H2N

NO2

15.

How many ml of 0.150 M Na2CrO4 will be required to oxidize 40 ml of 0.5 M Na2S2O3.

NO2

0.5 MNa2S2O3 ds 40ml dks vkWDlhd`r djus ds fy, 0.150M Na2CrO4 ds

fdrus vk;ru

CrO42– + S2O32– Cr(OH)4– + SO42–.

vko';drkgksxhA

(A) 225 ml

(B) 355 ml

CrO42– + S2O32– Cr(OH)4– + SO42–.

(C) 455 ml

(D) 555 ml

(A) 225 ml

(B) 355 ml

(C) 455 ml

(D) 555 ml


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|


(ml) dh


Page # 20

Select the correct statements (I) All mesomeric effect are resonance (II) All resonance effect are mesomeric effect

16.

(III) CH2 = CH – Cl

CH2 – CH = Cl This shows – M effect of Cl

(III) CH2 = CH – Cl

CH2 – CH = Cl Cl dk–MizHkkon'kkZrsZgSA

CH2 – CH = Cl

(IV) CH2 = CH – Cl

18.

19.

20.

Number of moles of 1 mole of NO3– ions NH2OH is: (A) 2 (C) 5

Cldk+M izHkkon'kkZ;ktkrkgSA (A) I, II vkSj III (B) I, II vkSj IV (C) dsoy III (D) I vkSjIV

electrons take up when is reduced to 1 mole of

17.

1eksy NO3– dks 1eksyNH2OH esa vipf;rdjusa ds

fy,vko';dbySDVªksuksadhla[;kgksxh&

(B) 4 (D) 6

(A) 2 (C) 5

1 g equivalent of a substance is the weight of that amount of a substance which is equivalent to : (A) 0.25 mol of O2 (B) 0.50 mol of O2 (C) 1 mol of O2 (D) 8 mol of O2 When one gm mole of KMnO4 reacts with HCl, the volume of chlorine liberated at NTP will be: (A) 11.2 litre (B) 22.4 litre (C) 44.8 litre (D) 56.0 litre

18.

fdlhinkFkZdk1xzkerqY;kadHkkjfuEuesalsfdlds rqY;gksrkgSA (A) O2 ds 0.25 eksy (B) O2dk 0.50 eksy (C) O2 dk 1 eksy (D) O2 ds 8 eksy

19.

tc 1xzkeeksy KMnO4,HClds lkFkvfHkfØ;k djrk gSa rksfudyus okyhDyksjhu xSl dkvk;ruNTPij D;kgksxkA

20.

CH2 = CH – Cl

CH2 = CH – Cl

(a)

(b)

(B) 4 (D) 6

(A) 11.2 litre (C) 44.8 litre

The correct stability order is:

(B) 22.4 litre (D) 56.0 litre

LFkkbZRodklghØegksxk% CH2 = CH – Cl

CH2 = CH – Cl

(a)

(b)

CH2 – CH = Cl

CH2 – CH = Cl

(c)

(c)

(A) a > b > c (C) c > a > b

CH2 – CH = Cl

(IV) CH2 = CH – Cl

This shows + M effect of Cl (A) I, II and III (B) I, II and IV (C) III only (D) I and IV 17.

fuEu esa ls lgh dFku dk p;u dhft,A (I)lHkhfelksesfVdizHkkovuqokngksrsgSA (II)lHkhvuqoknizHkkofelksesfjdizHkkogksrsgSA

(A) a > b > c (C) c > a > b

(B) a > c > b (D) b > c > a


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(B) a > c > b (D) b > c > a


Page # 21

21.

21. AB4¯ + C +2  C +3 + A+2

AB4¯ + C +2  C +3 + A+2

If the O.N. of B is –2. Choose the true statement for the above change – (A) O.N. of A decreases by –5

;fn mi;qZDr vfHkfØ;k esa B dh vkWDlhdj.k voLFkk – 2 gks] rks lR; dFku dkSu lk gS–

(B) O.N. of C decreases by +1

(A) A dh

(C) O.N. of A decreases by + 5 and that of C increases by +1

(B) C dh

Identify the cations which is more stable than following carbocation?

vkWDlhdj.k la[;k – 5 ?kVrh gS

vkWDlhdj.k la[;k +1 ?kVrh gSA (C) A dh vkWDlhdj.k la[;k +5 ?kVrh gS vkSj C dh vkWDlhdj.k la[;k +1 cM+rh gSA

(D) O.N. of A decreases by +5 and that of C decreases by +1 22.

mikip;u vfHkfØ;k gS-

(D) A dh 22.

vkWDlhdj.k la[;k +5 ?kVrh gS vkSj C dh vkWDlhdj.k la[;k +1 ?kVrh gSA fuEufyf[krdkcZ/kuk;uesalsfdldkcZ/kuk;udkLFkkf;Ro fn;sx,dkcZ/kuk;udhrqyukesavf/kdgksxkA

CH2 CH2

CH2

(A)

CH2 CH2

(B) (A) CH2

(C)

23.

CH2

(D)

(C)

No. of no–bond resonating structures of the

23.

following compound.

(A) 5

(B) 6

(C) 8

(D) 9

CH2

(B)

fuEufyf[kr;kSfxddhvcU/khrvuquknhlajpukvksadh la[;kfdruhgSA

(A) 5

(B) 6

(C) 8

(D) 9


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(D)



Page # 22

Find DOU for the following compound.

24.

fuEufyf[kr;kSfxdesaDOUcrkbZ,sA OH

OH CH 3

CH 3

C=N

(A) 5

(B) 6

(C) 8

(D) 9

25. Which of the following reactions does not involve either oxidation or reduction – (A) VO2+ V2O3 (C) Zn+2Zn 26.

27.

25.

(B) Na  Na + (D) CrO4–2Cr2O7–2

Which of the following is the least stable? (A)

(B)

(C)

(D)

Which of the following is least stable resonating structure.

C=N

(A) 5

(B) 6

(C) 8

(D) 9

fuEufyf[kr esa ls dkSulh vfHkfØ;k esa u rks vkWDlhdj.k gks jgk gS vkSj u gh vip;u– (A) VO2+  V2O3 (B) Na  Na + (C) Zn+2  Zn (D) CrO4–2  Cr2O7–2

26.

27.

fuEuesalsdkSudeLFkk;hgSa? (A)

(B)

(C)

(D)

fuEufyf[kresalsdkSulhvuquknhlajpuklclsdeLFkk;hgSA RS-1, CH3–O–CH=CH– C H–CH3

RS-1, CH3–O–CH=CH– C H–CH3

RS-2, CH3–O– CH–CH=CH–CH3

RS-2, CH3–O– C H–CH=CH–CH3

RS-3, CH3– O=CH–CH=CH–CH3

RS-3, CH3– O=CH–CH=CH–CH3 (A) RS-1 (B) RS-2 (C) RS-3 (D) RS-1=RS-2

(A) RS-1 (C) RS-3


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(B) RS-2 (D) RS-1=RS-2


Page # 23

28.

The number of delocalised -electrons in the given compound C10H10–4 (A) 14 (B) 16 (C) 8

29. 3A(s)

28.

g SA (A) 14 (C) 8

(D) 12 5B(g) + 7C (g)

If the rate of appearance of B is 5 × 10 –2 mole/lit-sec & the volume of the container is 2 lit. then The rate of disappearance of A is: (A) 3 × 10–2 mole/lit.Sec

29. 3A(s)

(D) buesa ls dksbZ ugh

Which of the following pairs represent resonating structures ? (A) CH2 = CHOH; CH3CHO

30.

fuEuesalsdkSulk;qXevuqukniznf'kZrdjrkgS\ (A) (B)

CH2 = C = O; CH  C – OH O

(C)

5B(g) + 7C (g)

(A) 3 × 10–2 mole/lit.Sec (B) 2 × 10–2 mole/lit/Sec (C) 5 × 10–2 mole/lit.Sec

(D) None

(B)

(B) 16 (D) 12

;fn B ds cuus dh nj 5 × 10–2 mole/lit.-sec vkSj ik=k dk vk;ru 2 yhVjgSA rc A ds xk;c gksus dh nj g SA

(B) 2 × 10–2 mole/lit/Sec (C) 5 × 10–2 mole/lit.Sec

30.

C10H10–4 ;kSfxdesafoLFkkuhÑr-bysDVªksuksadhl[;ka

OH

CH3 C

CH3 ; CH3 C

(C)

CH2 = CHOH; CH3CHO CH2 = C = O; CH  C – OH O OH

CH3 C

CH2

OH +

OH

(D)

CH3 C

OH +

CH3 ; CH3 C

(D)

CH3 C

CH3 ; CH3 C +

OH


|

+

CH3 ; CH3 C

CH3

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CH2


CH3

ETOOS & MOTION

JEE-Main, 2016 SAMPLE PAPER 2 Duration :3 Hours

Max. Marks : 360

_______________________________________________________________________________ INSTRUCTIONS In each part of the paper, Section-A contains 30 questions. Total number of pages are 32. Please ensure that the Questions paper you have received contains ALL THE QUESTIONS in each section and PAGES. SECTION - A 1.


NOTE : GENERAL INSTRUCTION FOR FILLING THE OMR ARE GIVEN BELOW.

1.

Use only HB pencil or blue/black pen (avoid gel pen) for darkening the bubble.

2.


3.

The Answer sheet will be checked through computer hence, the answer of the question must be marked by shading the circles against the question by dark HB pencil or blue/black pen.

4.


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Page # 2

PART - I [MATHS] SECTION - A


[STRAIGHT OBJECTIVE TYPE] of which ONLY ONE is correct


1.

1.

Q.1 to Q.30 has four choices (A), (B), (C), (D) out

2.

If log0.3 (x  1) < log0.09 (x  1), then x lies in the interval (A) (2, ) (B) (2, 1) (C) (1, 2) (D) None of these

The value of (A) 2

3.

( 0. 2)

1 1 1  log 5    ....   4 8 16 

(B) 4

(C) 8

is

3.

2

4.

If a, b, c are integers and b = 4(ac + 5d ), 2 d  N, then roots of the equation ax + bx + c = 0 are (A) Irrational (B) Rational & defferent (C) Complex conjugate (D) Rational & equal

5.

Let

x  1 x  1

4.

5.

(where [] denotes the greatest integer function) The [f(23)] is equal to (A) 2 (B) 3 (C) 4 (D) 3

(B) 4

(C) 8

(B) 8

(C) 16

(D) 64

1 | x | , x  1 ekuk f ( x )   x  1  [ x] , tgk¡[] egÙke iw.kkZad Qyu dks iznf'kZr djrk gSA rcf{f(23)} dk izkUr gSA (A) 2 (C) 4

|

(D) 16

;fn a, b,c iw.kk±d gks vkSj b2 = 4(ac+5d2),dÎN gks rks lehdj.kax2 + bx + c = 0 ds ewy gSa& (A) vifjes; (B) ifjes; vkSj vleku (C)lfEeJla;qXeh (D)ifjes; vkSjleku


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dk eku gS

;fn xy2 = 4 rFkk log3 (log2x)+log1/3 (log1/2y)=1, rcx cjkcj gS& (A) 4

(D) 64 2

( 0. 2)

(B) (2, 1) (D)buesalsdksbZugha

1 1 1  log 5    ....   4 8 16 

(A) 2

(D) 16

(C) 16

1 | x | , f ( x)    [ x] ,

(A) (2, ) (C) (1, 2)

2.

If xy2 = 4 and log3 (log2x)+log1/3 (log1/2y)=1, then x equals (A) 4 (B) 8

;fn log0.3 (x  1) < log0.09 (x  1), rc x fdl vUrjkyesafLFkrgS-


(B) 3 (D) 3


Page # 3

6.

The sum of integers from 1 to 100 that are divisible by 2 or 5 is (A) 3000 (B) 3050 (C) 4050 (D) None of these

6.

[6  2 3  2 2  2 6 ]  1

[6  2 3  2 2  2 6 ]  1 7. (A) 1 (C) 0 8.

is simplify to

(5  2 6 )

7.

The value of 8.

(A) 7

(B) 8 2400

9.

If L =

 log

7

r 7

2011

N=

  4 1  3 2 

r  1  r , M =  

r 2

11.

 1  1 4 6 + log 3/2  

dkekugS

(D) 4

(A) 7



3 2 

(B) 8 2400

1023

 log (r  1) and r

9.

;fn L =

 log

7

r 7

r 2

2011

1

 log

N=

where p = (1 · 2 · 3 · 4 ... 2011) r p

then (A) L + M = 13 (C) L – M + N = 7 10.

(C) 5

(B) –1 (D)bueslsdksbZugh

 4  ......    

1 4 3 2

gy djsA

(5  2 6 ) (A) 1 (C) 0

(B) –1 (D) None of these

 1 6 + log3/2  3 2 

1ls100rddsiw.kk±dksadk;ksxtksdh2;k5lsfoHkkftr gS,gksxk(A) 3000 (B) 3050 (C) 4050 (D)buesalsdksbZugha

1

 log r 2

r

p

3 2

4

(C) 5

1 3 2

 4  ......    

(D) 4

1023 r  1  r  , M =  logr (r  1) rFkk   r 2

tgk¡ p = (1 · 2 · 3 · 4 .... 2011)

rc (B) M2 + N2 = 110 (D) LMN = 60

(A) L + M = 13 (C) L – M + N = 7

If x – a is a factor of x3 – a2x + x + 2, then ‘a’ is equal to (A) 0 (B) 2 (C) – 2 (D) 1

10.

If logab = 2; logbc = 2 and log3c = log3a + 3 then (a + b + c) equals (A) 90 (B) 93 (D) 102 (D) 243

11.

;fnx – a, x3 – a2x + x + 2 dk xq.ku[k.M gS] rks ‘a’ dkekugksxk& (A) 0 (C) – 2

(A) 90 (D) 102

|

(B) 2 (D) 1

;fn logab = 2; logbc = 2 ,oa log3c = log3a + 3 rc (a + b + c) cjkcj &


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(B) M2 + N2 = 110 (D) LMN = 60


(B) 93 (D) 243


13.

14.

Which of the following is graph of the function.

(A)

(B)

(C)

(D)

12.

Two real numbers  &  are such that  = 3 & || = 4, then  &  are the roots of the quadratic equation 2 (A) 4x – 12x – 7 = 0 2 (B) 4x – 12x + 7 = 0 2 (C) 4x – 12x + 25 = 0 (D) None of these

13.

log10

fuEu esals dkSulk ,dQyu dk xzkQgS

(A)

(B)

(C)

(D)

nksokLrfodla[;k,¡,oablçdkjgSfd=3,oa ||=4gks]rksoglehdj.kftldsewy,oagks]gS& 2

(A) 4x – 12x – 7 = 0 2 (B) 4x – 12x + 7 = 0 2 (C) 4x – 12x + 25 = 0

(D)buesalsdksbZugha

The domain of the function f(x) =

15.

Page # 4

14.

Qyu f(x) = log10

3x is x

 3 (A)  0,   2

(B) (0, 3)

3  (C)  ,  2 

 3 (D)  0,   2

(x  1)3 (x  2)5 (x  3)7 0 Solve inequality (x  1)9 (x  2)11 (x  3)13 (A) (–3,–2)  (–1,1)  [3,) (B) (–, –3)  (–2,–1)  [1,2]  [3,) (C) (–, –2)  (2, 3) (D) None

15.

3x x

 3 (A)  0,   2

(B) (0, 3)

3  (C)  ,  2 

 3 (D)  0,   2

(x  1)3 (x  2)5 (x  3)7

vlfedk (x  1)9 (x  2)11 (x  3)13  0 dkgygksxkA (A) (–3,–2)  (–1,1)  [3,) (B) (–, –3)  (–2,–1)  [1,2]  [3,) (C) (–, –2)  (2, 3) (D)dksbZugha


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dk izkUr gS

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Page # 5

16.

17.

18.

x 1 2 x2 (A) (2, 3) (C) ]2, 3]

(B) (1, 2) (D) [1, 2]

Solve |x – 2| < –7 (A) (–, –7)  (7, ) (C) (–, )

(B) (–2, 2) (D) None

x 1  2 dks gy dhft,A x2 (A) (2, 3) (B) (1, 2) (C) ]2, 3] (D) [1, 2] |x – 2| < –7 dks gy

dhft,

(A) (–, –7)  (7, )

(B) (–2, 2)

(C) (–, )

(D)dksbZugha







(C) N

 s

(B) N

s 

4

The value of log 1/3

If



2



6

17  1 2

(C)

1  17 2



 s



s 

729.3 91274 / 3

(C) N is

19.

x2  4

log1/3

4

dk eku cjkcj gS & (B) N

= 1 then the

20.

729.3 91274 / 3

(B) 1/4

(C) –1/4


log ;fn

logx2

x2  4

2

 .log

x

x2  4.log3 x

6

x2  4 x2  4

= 1 rc 'x'

dkog eku tks lehdj.kdkslarq"V djs & (A)

17  1 2

(D) None of these

(C)

1  17 2


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dk eku cjkcj &

(A) 4

17  1 2

(B)

 s


1/x

value of 'x' satisfying the equation is (A)

a

(A) N

.logx x2  4

x2  4.log3 x

 log s N

(a )

 s

(B) 1/4 (D) None of these

log1 / x x2  4 logx2

18.

(D) None of these

equal to (A) 4 (C) –1/4

20.

17.

The value of (a ) logas N is equal to

(A) N

19.

16.

Solve

|


(B)

17  1 2

(D)buesalsdksbZughaaA


Page # 6

Find the solution of the equation

21.

logx3.logx/3 3 + logx/813 = 0

djks&

(A) x = 8, 2/8

(B) x = 5, 1/9

(A) x = 8, 2/8

(C) x = 9, 1/9

(D) None of these

(C) x = 9, 1/9 22.

22.


25.

lehdj.k 9

(D)buesalsdksbZughaA 1 + logx

–3

– 210 = 0; ds

gy Kkr

(tgk¡ logdk vk/kkj 3gS)

(where base of log is 3)

24.

1 + logx

(B) x = 5, 1/9

djks&

91 + logx – 31 + logx – 210 = 0;

23.

lehdj.k logx3.logx/3 3 + logx/813 = 0 ds gy Kkr

(A) 1

(B) 5

(C) 4

(D) None of these


23.

(A) 1

(B) 5

(C) 4


lehdj.k log(log x) + log(log x3 – 2) = 0; ds gy

log(log x) + log(log x3 – 2) = 0;

Kkr djks&

(where base of log is 10 every-where.)

(tgk¡ lc txg log dk vk/kkj 10gS)

(A) 10

(B) 12

(A) 10

(B) 12

(C) 11

(D) None of these

(C) 11



24.

lehdj.k 5logx + 5xlog5 = 3(a > 0); dk gy Kkr djks

5logx + 5xlog5 = 3(a > 0); (base of log = a)

(log dk vk/kkj =a gS)

(A) x = alog5 2

(B) x = alog4 1

(A) x = alog5 2

(B) x = alog4 1

(C) x = a log5 2

(D) None of these

(C) x = a log5 2


If  are the roots of quadratic equation

25.

2

x + px + q = 0 and  are the roots of

vkSj lehdj.k x2 + px – r = 0 ds ewy,  gks rks

2

x + px – r = 0, then ( – ) . ( – ) is equal

( – ) · ( – ) dk eku gSa &

to (A) q + r

(B) q – r

(C) – (q + r)

(D) –(p + q + r)

;fn f}?kkr lehdj.k x2 + px + q = 0 ds ewy ,  gks

(A) q + r

(B) q – r

(C) – (q + r)

(D) –(p + q + r)


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Page # 7

26.

27.

28.

29.

30.

The ratio

2

log

a

2

3

 3log27 (a 1)  2a simplifies to 7 4 log49 a  a  1

21 / 4

26.

vuqikr

2

log

a

2

3

 3log27 (a 1)  2a dks ljy djus ij 7 4 log49 a  a  1

21 / 4

(A) a2 – a – 1

(B) a2 + a – 1

(A) a2 – a – 1

(B) a2 + a – 1

(C) a2 – a + 1

(D) a2 + a + 1

(C) a2 – a + 1

(D) a2 + a + 1

27.

The number of the solution of the equation,

lehdj.k log(–2x) =2 log(x + 1) ds gyksa dh la[;k

log(–2x) = 2 log(x + 1) is

Kkr djks&

(A) zero

(B) 1

(A)'kwU;

(B) 1

(C) 2

(D) None of these

(C) 2


Number of Solution of sinx = x2 + x + 1 are

28.

sinx = x2 + x+ 1 ds gyksa dh la[;k Kkr djks &

(A) 1

(B) 0

(A) 1

(B) 0

(C) 2

(D) 

(C) 2

(D) 

29.

If [x + [x + [x +...........100 times]]] = 100

;fn [x+[x+[x+........100 times]]] = 100 rc x

then x belongs to

laca/kj[krkgS

(where [ * ] denotes greatest integer function)

(tgk¡[*]egRreiw.kkZadQyudksiznf'kZrdjrkgS)

(A) (1, 2]

(B) [1, 2)

(A) (1, 2]

(B) [1, 2)

(C) (0, 1]

(D) [0, 1)

(C) (0, 1]

(D) [0, 1)

30.

If function (x) is defined in [– 2, 2] then

;fn Qyu(x),[– 2,2]esa ifjHkkf"kr gS] rc (|x|+1)

domain of definition of (|x| + 1) is

dksifjHkkf"krdjusokykizkUrgS&

(A) [– 2, 2]

(B) [– 3, 3]

(A) [– 2, 2]

(B) [– 3, 3]

(C) [– 1, 1]

(D) [– 3, 1]

(C) [– 1, 1]

(D) [– 3, 1]


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Page # 8





Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct 1.

2.

A body starts from rest and is uniformly accelerated for 30 s. The distance travelled in the first 10s is x1, next 10 s is x2 and the last 10 s is x3. Then x1 : x2 : x3 is the same as (A) 1 : 2 : 4 (B) 1 : 2 : 5 (C) 1 : 3 : 5 (D) 1 : 3 : 9

1.

(A) 1 : 2 : 4 (C) 1 : 3 : 5 2.

A particle starts moving rectilinearly at time t = 0 such that its velocity ‘v’ changes with time ‘t’ according to the equation v = t2 – t where t is in seconds and v is in m/s. The time interval for which the particle retards is (A) t < 1/2 (B) 1/2 < t < 1 (C) t > 1 (D) t < 1/2 and t > 1

(b) T2 = 25 N

(c) T1 = 25 3 N (A) a, b (C) c, d

(d) T2 = 25 3 N (B) a, d (D) b, c

5kg nzO;eku dk ,d fi.M {kSfrt ds lkFk 60º o30ºdk

(a) T1 = 25 N

(b) T2 = 25 N

(c) T1 = 25 3 N (A) a, b (C) c, d

(d) T2 = 25 3 N (B) a, d (D) b, c


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(B) 1/2 < t < 1 (D) t < 1/2 rFkk t > 1

dks.kcukrsgq,sMksfj;ksalsyVdk;kx;kgS]rks

A body of mass 5 kg is suspended by the strings making angles 60º and 30º with the horizontal -

(a) T1 = 25 N

(B) 1 : 2 : 5 (D) 1 : 3 : 9

,d d.k le; t = 0”ij ljy js[kk ds vuqfn'k bl rjg xfrizkjEHkdjrkgSfdblosx'v'”le;t dslkFk]lehdj.k v=t2–t”dsvuqlkjifjofrZrgksrkgSatgk¡t”lSd.MesarFkk vm/s esa gSA og le; vUrjky tc d.k eafnr gksxkA (A) t < 1/2 (C) t > 1

3. 3.

,doLrq fojkels izkjEHk gksdj30secdsfy, leku:i lsRofjrgksrhgSA ;fnigys10Sec.espyhxbZnwjhx1 gSA rFkk vxys 10seces x2 gS rFkk vfare 10sec.es x3 gS rcx1 : x2 : x3 gksxkA

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Page # 9

4.

5.

Balls are thrown vertically upward in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is 5m, the number of balls thrown per minute will be (A) 40 (B) 50 (C) 60 (D) 120

4.

(A) 40 (C) 60 5.

The 50 kg homogeneous smooth sphere rests on the 30° incline A and bears against the smooth vertical wall B. Calculate the contact forces at A and B. (A) NB =

1000 3 1000

(B) NA =

3

(C) NA = (D) NA =

100

N, N = B

1000

50kg dk le:i fpduk xksyk] 30° ds urry Ao Å/okZ/ kjnhokjBdschpfp=kkuqlkjj[kkgqvkgSAAoBijlEidZ cy Kkr djksA (A) NB =

A

N

B

N, N = B

(C) NA = (D) NA =

A

N, N = B

500

N

B

30°

3

100

N,N = B

3

500

N

3

1000 3

N, N = B

50

N

3

N The acceleration of a particle which moves along the positive x-axis varies with its position as shown. If the velocity of the particle is 0.8 m/s at x = 0, the velocity of the particle at x = 1.4 is (in m/s) 2

a (in m/s )

a (in m/s ) 0.4

0.4

0.2

0.2

(A) 1.6 (C) 1.4

N

3

3

2

0.4 0.8

500

N

The acceleration of a particle which moves along the positive x-axis varies with its position as shown. If the velocity of the particle is 0.8 m/s at x = 0, the velocity of the particle at x = 1.4 is (in m/s)

O

N, N = A

30°

3 50

3

3

3 500

1000

1000

(B) NA =

6. 6.

(B) 50 (D) 120

N

3 500

N,N = B

3

3

N, N = A

500

dqNxsansm/okZ/kjÅijdhvksjblrjglsQsadhtkrhgSfd tcfiNyhxsanvf/kdreÅ¡pkbZijgksrhgSrksvU;xsan ÅijQsadhtkrhgSA;fnvf/kdreÅ¡pkbZ5m”gksrksizfr fefuVdqyfdruhxsansQsadhx;hgksxhA

O

1.4 x (in m)

(B) 1.2 (D) none

0.4 0.8

(A) 1.6

(B) 1.2

(C) 1.4

(D) none


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1.4 x (in m)



8.

9.

Page # 10

At a given instant, A is moving with velocity of 5 m/s upwards. What is velocity of B at the time (A) 15 m/s (B) 15 m/s (C) 5 m/s  (D) 5 m/s 

7.

(A) 15 m/s (B) 15 m/s (C) 5 m/s  (D) 5 m/s 

A B

A body is dropped from a height h under acceleration due to gravity g. If t1 and t2 are time intervals for its fall for first half and the second half distance, the relation between them is (A) t1 = t2 (B) t1 = 2t2 (C) t1 = 2.414 t2 (D) t1 = 4t2

8.

The displacement-time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the x point (A) C D (B) D F E (C) E C (D) F

9.

A B

,doLrqdkshÅ¡pkbZlsxq:Roh;Roj.kgdsizHkkoesNksM+k x;kgSA;fnt1rFkkt2 igyhrFkknwljhvk/khnwjhesfxjus esyxkle;gSArksmunksuksdse/;lEc}gksxkA (A) t1 = t2

(B) t1 = 2t2

(C) t1 = 2.414 t2

(D) t1 = 4t2

fdlhxfrtd.kdkfoLFkkiu≤xzkQuhpsn'kkZ;kx;k gSArksfdlfcUnqijrkR{kf.kdosx_.kkRedgksxk\ (A) C

x D

(B) D (C) E

E

F

C

(D) F

t

10.

fdlh {k.k ij A5m/s ls osx ls Åij dh vksj xfr dj jgk gSA ml {k.kijB dkosxgksxkA

t

2

The pulley is given an acceleration a0=2 m/s starting from rest. A cable is connected to a block A of mass 50 kg as shown. Neglect the mass of the pulley and friction between block and ground. Then the tension in the cable connected to block A is

10.

,df?kjuhdks fojke lsa0=2m/s2dkRoj.kfn;k tkrk gSA50kgnzO;ekudk,drkjfp=kkuqlkjtqM+kgSaAf?kjuhdk nzO;eku]rFkkCykWdo/kjkrydse/;?k"kZ.kux.;ekfu;sA CykWdAlstqM+srkjesarukogksxkA

a0 a0 A A

(A) 200 N (C) 400 N

(A) 200 N (C) 400 N

(B) 300 N (D) 500 N


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|


(B) 300 N (D) 500 N


Page # 11

11.

12.

13.

A particle of mass 10 kg is acted upon by a force F along the line of motion which varies as shown in the figure. the initial velocity of the particle is 10ms–1. Find the maximum velocity attained by the particle before it comes to instantaneous rest. (A) 40 (B) 60 (C) 20 (D) None of these

11.

10kg nzO;eku ds d.k ij] cy F dks xfr dh fn'kk esa

yxk;ktkrkgSrFkkcyfp=kesafn[kk,savuqlkjifjofrZr gksrkgSAd.kdkizkjfEHkdosx10ms–1gSaAd.kdksrkR{kf.kd fojkevoLFkkesavkuslsigys]blds}kjizkIrfd;kx;kvf/ kdre osxKkrdjksA (A) 40 (B) 60 (C) 20 (D)buesalsdksbZugh 12.

Two blocks A and B of masses 4 kg and 12 kg are placed on a smooth plane surface. A force F of 16N is applied on A as shown. Find out the normal force (in Newton) between grond and Block B. B A (A) 60 N (B) 240 N F=16N (C) 100 N (D) 120 N

4 kg rFkk 12 kg nzO;eku ds nks CykWd A o B fpduh lery lrg ij j[ksa gSaA CykWd A ij 16N dk cy F fp=kkuqlkjvkjksfirfd;ktkrkgSACykWdBrFkkleryds e/;vfHkyEcizfrfØ;kcy(U;wVuesa) KkrdjksA

B

(A) 60 N (B) 240 N (C) 100 N (D) 120 N

A projectile is thrown at an angle of 60º with the horizontal with an initial speed of 20 m/ sec, with H being highest point of its trajectory. Another particle P is now forced to move along the same trajectory as that of the projectile. When the particle P is   2 at O v o  10 m / s, ao  2.5 m / s an d its

13.

A F=16N

,diz{ksI;dks{kSfrtls60ºdsdks.kijvkjfEHkdpky20 m/seclsQsadktkrk gS rcHmldhvf/kdre Å¡pkbZ gSA nwljkd.kPHkhmlhiFkdsvuqfn'kxfrdjrkgSAtcd.k   2 P O v o  10 m / s, ao  2.5 m / s ijgSrFkkbldh

pky

?kV

jgh

gSA

fcUnq

spe ed is de crea sing . At point   2 H v H  20 m / s, aH  50 m / s and its speed

  H v H  20 m / s, aH  50 m / s2 rFkkbldhpkyc<+

is increasing. (g=10 m/s2)  (A) ao = -2.5 î m/s2  (B) ao  2.5ˆj m / s2  (C) aH  (20 6î  10ˆj) m / s2  ˆ (D) aH  (30iˆ  10j)m / s2

jgha gSA(g=10m/s2)  (A) ao = -2.5 î m/s2  (B) ao  2.5ˆj m / s2  (C) aH  (20 6î  10ˆj) m / s2  ˆ (D) aH  (30iˆ  10j)m / s2


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15.

Page # 12

Three block are connected as shown, on a horizontal frictionless table and pulled to the right with a force T3 = 60 N. If m1 = 10 kg, m2 = 20 kg and m3=30 kg, the tension T2 is(A) 10 N (B) 20 N (C) 30 N (D) 60 N As shown in the figure A sphere of 50 kg is kept in between two surfaces A & B. Then the find the normal B force (in n ewto n) exerted by surface B on A sphere. (A) 200N (B)250N 30 (C) 300N 30 (D) 350N If a particle is moving as t = x  1 where x is the position and t is time. Then at t= 3sec velocity of particle is: (A) 6m/sec (B) 5 m/sec (C) 7 m/sec (D) 3 m/sec

14.

(A) 10 N (B) 20 N (C) 30 N (D) 60 N 15.

o

17.

18.

What is relation between VA and VB if pulleys and string are ideal (A) VA = VB (B) VA = VB tan 300 (C) VA = VB cot 300 (D) VA = VB cos 300

A 0

fn[kk, x;s fp=k esa50kgdk xksyk] lrgAo Bds chp lkE;koLFkkesaj[kkgqvkgSrkslrgB}kjkxksys B ij vkjksfir vfHkyEc lEidZ cy ¼in newton½ dk eku Kkr djksA A

(A) 200N (B)250N (C) 300N (D) 350N

o

16.

rhuCykWdtksfdijLij¼fp=kkuqlkj½ca/ksgq,gSrFkk?k"kZ.k jfgr{kSfrtVsfcyijfLFkrgSAmUgsank;havksj T3=60 U;wVucyls[khapktkrkgSA;fnm1=10fdxzk]m2=20fdxzk rFkkm3=30fdxzkgS]rksruko T2gS&

30o o

30

16.

;fn ,d d.k t= x  1 ds vuqlkj xfreku gS tgk¡ x fLFkfr rFkk t le; gSSA rc t =3sec ij d.k dk osx gS (A) 6m/sec (C) 7 m/sec

17.

(B) 5 m/sec (D) 3 m/sec

VA rFkk VB dse/; D;k lEcU/kgS

A 0

;fnf?kjuhoMksjhvkn'kZgSA (A) VA = VB (B) VA = VB tan 300 (C) VA = VB cot 300 (D) VA = VB cos 300

B

B

Displacement (x) of a particle is related to 18. ,d d.k dk foLFkkiu le; (t) ds lkFk x = at + bt2 – time (t) as x = at + bt2 – c t3 where a, b and c are constants of motion. c t3 }kjklEcfU/krgSAtgk¡a,brFkkcxfrds fu;rkad The velocity of the particle when its accelgSAtcdk.kdkRoj.k'kwU;gSrcd.kdkosxgksxkA eration is zero is given by b2 b2 b2 b2 (A) a  (B) a  (A) a  (B) a  c 2c c 2c b2 b2 2 2 b b (C) a  (D) a  (C) a  (D) a  3c 4c 3c 4c (SPACE FOR ROUGH WORK)

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Page # 13

19.

If a particle is moving with relation x = t3 where x is position & t is time then (A) when v = 2 m/sec then a = 6 m/sec2 (B) when v = 4 m/sec then a = 6 m/sec2 (C) when x = 8 m then a = 12 m/sec2 (D) when x = 10 m then a = 12 m/sec2

20.

A particle is moving according to relation x = t2 - 4t. Then during the interval from t = 1 sec to t = 2.5 sec. (A) |Displacement|> Distance (B) |Displacement|= Distance (C) |Displacement|< Distance (D) |Displacement|<= Distance

21.

22.

A body, moving in a straight line, covers half the distance with a speed 15 m/sec, the remaining part of the distance was covered with a speed 20 m/sec for half of the time and with a speed 10 m/sec for the other half of the time. What is the average speed of the body? (A) 15 (B) 30 (C) 45 (D) None of these A particle is thrown from the origin, at an angle (0<<90) such that it just crosses a wall of height 9m. Wall is along the plane x =

19.

;fn , d.klEcU/kx=t3 dsvuqlkjxfrdjjgkgS]tgkW x fLFkfr rFkk t le; gSA rc & (A) tc v = 2 m/sec rc a = 6 m/sec2 (B) tc v = 4 m/sec rc a = 6 m/sec2 (C) tc x = 8 m rc a = 12 m/sec2 (D) tc x = 10 m rc a = 12 m/sec2

20.

,d d.k lEcU/k x = t2 - 4t ds vuqlkj xfreku gSA rc le; vUrjky t = 1 sec ls t = 2.5 sec ds fy, & (A)|foLFkkiu|>nwjh (B)|foLFkkiu|=nwjh (C)|foLFkkiu|
21.

ljyjs[kkesaxfreku,d oLrqvk/khnwjhpky15m/sec ls]'ks"kvk/khnwjhdsfy,vk/ksle;20m/secpkylsrFkk 'ks"kvk/ksle;dsfy,10m/secpkylsxfrekugSoLrq dh vkSlr pky D;k gSA (A) 15 (C) 45

22.

12m. Speed of projection is n 30 & particle strike the ground at x = 48 m, value of n is (g = 10m/sec2) (A) 3 (B) 4 (C) 6 (D) None of these 23.

A particle moves with a constant speed u along the curve y = sin x. The magnitude of its acceleration at the point corresponding to x = /2 is

u2 (A) 2 (C) u2

(B) (D)

,d d.k dks ewy fcUnq ls ,d dks.k (0<<90°)ij bl izdkjiz{ksfirfd;ktkrkgSfd;g9mÅ¡pkbZdh,dnhokj dksdsoyikjdjikrkgSAnhokjryx=12mdsvuqfn'k gSAiz{ksi.kosx n 30 gSrFkkd.k/kjkryijx=48m ij Vdjkrk gSA n dk eku gS (g = 10m/sec2) (A) 3 (C) 6

23.

u2 2

(B) 4 (D)buesalsdksbZugha

,d d.k oØ y = sin x ds vuqfn'k fu;r pky v ls xfr djrkgSAbldsRoj.k dkifjek.kx=/2dsle:ihfcUnq ijKkrdjksA (A)

u2 2

(C) u

2

2 u2


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(B) 30 (D)buesalsdksbZugha

|


(B) (D)

u2 2

2 u2


Page # 14

Acceleration-time graph of a particle moving along X axis is as shown. The particle will have the velocity same as its initial velocity at

24.

cjkcjfdlfcUnwijgksxkA

a (m/s2)

O

25.

26.

27.

X-v{k ds vuqfn'k xfreku ,d d.k dk Roj.k & le; (at)xzkQfp=kesaiznf'kZrgSAd.kdkosxvkjfEHkdosxds

a (m/s2)

10

20

t(s) O

(A) 10 sec

(B) 20  10 3 sec

(C) 10  5 3 sec

(D) 10  2 3 sec

A water tap situated at a height h ejects n droplets of water per second in vertically upward direction. If the first drop reaches the ground when nth drop is about to emerge, then the height h is (g = 10 m/s2). (A) h > 5 m for any value of initial velocity (B) h < 5 for initial velocity is 3 m/s (C) h  5 m for initial velocity is 4 m/s (D) None of these Which of the following sets of displacements might be capable of bringing a car to its returning point ? (A) 5, 10, 30 and 50 km (B) 5, 9, 9 and 20 km (C) 40, 40, 90 and 8 km (D) 10, 20, 40 and 90 km A rubber ball escapes from the horizontal roof with a velocity v = 5 m/s. The roof is situated at a height h = 20m. If the length of each car is equal to x0 = 4 m, with v which car will the ball hit ? (A) 4 h (B) 2 (C) 3 (D) 5

25.

10

20

(A) 10 sec

(B) 20  10 3 sec

(C) 10  5 3 sec

(D) 10  2 3 sec

hÅ¡pkbZijfLFkr,dikuhdhVadhizfrlSd.Mn cawnsÅ/

okZ/kjÅijdhvksjfudkyrhgSaA;fnigyhcwan/kjkryij rcigq¡ps tcnthcwanfudyusokyhgksrksÅ¡pkbZhgSa (g = 10 m/s2). (A) vkjfEHkd osx ds fdlh Hkh eku ds fy, h>5m (B) 3 m/s ds vkjfEHkd osx ds fy, h< 5 (C) 4 m/s ds vkjfEHkd osx ds fy, h  5 m (D)buesalsdksbZugha 26.

fuEu esa ls dkSulkfoLFkkiu leqPp; ,d dkj dks mldh vkjfEHkdfLFkfrijiqu%ykusesal{kegSA (A) 5, 10, 30 rFkk 50 km (B) 5, 9, 9 rFkk 20 km (C) 40, 40, 90 rFkk 8 km (D) 10, 20, 40 rFkk 90 km

27.

,d jcj dh xsn v =5 m/s ds osx ls ,d Å/okZ/kj Nr lsfxjrhgSANrdhÅ¡pkbZh=20mgSA;fnizR;sddkj dh yEckbZ x0 = 4 m v gksrksxsanfdldkj lsVdjk;sxhA (A) 4 (B) 2 (C) 3 (D) 5

h


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t(s)

|



Page # 15

28.

29.

30.

Referring to the 2 – s diagram of a particle, find the acceleration of the particle during v2(m2/s2) motion. 2 (A) 4 m/sec 2000 (B) 5 m/sec2 1500 (C) 2 m/sec2 2 s(m) (D) 1m/sec O 50 100

28.

According to diagram a particle is projected with 50 m/sec at angle 53° from the horizontal then find the time when initial velocity and velocity at time t becomes perpendicular to each other. (A) 25/4 50 m/sec (B) 25/2 (C) 3 53° (D) None of these

29.

,dd.k ds2-svkjs[k}kjkd.kdkRoj.kxfrdsnkSjku KkrdjksA v (m /s ) 2

2

(A) 4 m/sec (B) 5 m/sec2 (C) 2 m/sec2 (D) 1m/sec2

30.

20 m/s (D) velocity at hightest point of its motion is 10 m/s

|

2000 1500

O

50 100

s(m)

50 m/sec 53°

,difg;kxkM+h4m/sdsosx ls+xfn'kkesaxfrekugSA xkM+hijcSBk,dO;fDr6m/sdsosxls,diRFkjQsadrk gSAxkM+hdsfunsZ'kra=kesaiRFkjÅ/okZ/kjZv{kls30°ds dks.kijy-zryeasQsadkx;kgSArc/kjkryij[kM+s,d izs{kddslkis{k (A)d.k dkizkjfEHkd osx10m/sgSA (B)d.k dk izkjfEHkd osx2 13 m/s gSA (C) xfr ds mPpre fcUnq ij osx 20 m/s gSA (D) xfr ds mPpre fcUnq ij osx 10m/s gSA


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2

fp=kkuqlkj ,d d.k dks {kSfrtls53°ij50m/secosx lsiz{ksfirfd;kx;kgSArcogle;KkrdjkstcizkjfEHkd osx rFkk fdlh le; tij osx ,d nwljs ds yEcor~ gks tk,xsaA (A) 25/4 (B) 25/2 (C) 3 (D)buesalsdksbZugha

A cart is moving along +x direction with velocity of 4m/s. A person on the cart throws a stone with a velocity of 6m/s with respect to himself. In the frame of cart the stone is thrown in y–z plane making 30° with vertical z-axis. Then with respect to an observer on the ground. (A) the initial velocity of the stone is 10m/s (B) the initial velocity of stone is 2 13 m/s (C) velocity at hightest point of its motion is

2



Page # 16

PART - III [CHEMISTRY] Use Data : 1H1, 2He4, 3Li7, 4Be9, 5B 10, 6C12, 7N14, 8O16, 9F19, 11Na23, 12Mg 24, S32, 15P31, 17Cl35.5, 19K 39, 20Ca 40, 26Fe 56, 29Cu63.5, 30Zn65, 24Cr 52, 23V51 16

SECTION - A

Al27,

13


[STRAIGHT OBJECTIVE TYPE] of which ONLY ONE is correct


1.

1.

Q.1 to Q.30 has four choices (A), (B), (C), (D) out Which of the following is a redox reaction : (A) NaCl + KNO3  NaNO3 + KCl

(A) NaCl + KNO3  NaNO3 + KCl

(B) CaC2O4 + 2 HCl  CaCl2 + H2C2O4

(B) CaC2O4 + 2 HCl  CaCl2 + H2C2O4

(C) Mg (OH)2 + 2 NH4Cl  MgCl2 + 2

(C) Mg (OH)2 + 2 NH4Cl  MgCl2 + 2

NH4OH

NH4OH

(D) Zn + 2 AgCN  2 Ag + Zn (CN)2

(D) Zn + 2 AgCN  2 Ag + Zn (CN)2

H |

2.

fuEuesalsmikip;hvfHkfØ;kdkSulhgS%

H

C 4H9

|

|

The IUPAC name of CH3 CH2  C  C  CH3 |

2.

|

CH3CH2  C  C  CH3 dk IUPAC uke gSA |

CH3 CH3

|

CH3 CH3

is -

(A) 3,4, 4-VªkbZesfFkygsIVsu

(A) 3, 4, 4-trimethylheptane

(B)3, 4, 4-VªkbZesfFkyvkWDVsu

(B) 3, 4, 4-trimethyloctane

(C) 2-C;wfVy-2-esfFky-3-,fFkyC;qVsu

(C) 2-butyl-2-methyl-3-ethylbutane (D) 2-ethyl-3, 3-dimethylheptane 3.

C 4H9

|

(D)2-,fFky-3, 3-MkbZesfFkygsIVsu

The oxidation number of nitrogen atoms in

3.

NH4NO3 are (A) +3, +3

(B) +3, –3

(C) –3, +5

(D) –5, +3

NH4NO3 esaukbVªkstudhvkWDlhdj.kla[;kD;kgksxh(A) +3, +3

(B) +3, –3

(C) –3, +5

(D) –5, +3


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Page # 17

4.

Which of the following chemical reactions depicts the oxidizing behaviour of H2SO4? (A) 2HI + H2SO4  I2 + SO2 + 2H2O (B) Ca(OH)2 + H2SO4 CaSO4 + 2H2O (C) NaCl + H2SO4 NaHSO4 + HCl (D) 2PCl5 + H2SO4 2POCl3 + 2HCl + SO2Cl2

4.

fuEuesalsdkSu&lhjklk;fudvfHkfØ;kH2SO4 dsmipk;d O;ogkjdksiznf'kZrdjrhgS\ (A) 2HI + H2SO4  I2 + SO2 + 2H2O (B) Ca(OH)2 + H2SO4 CaSO4 + 2H2O (C) NaCl + H2SO4 NaHSO4 + HCl (D) 2PCl5 + H2SO4 2POCl3 + 2HCl + SO2Cl2

Cl 5.

CH3

IUPAC name of

Br

Cl is

CH3

5.

C2H5

Br

(A)4-Bromo-6-chloro-2-ethyl-1-

;kSfxd dk IUPAC uke gS&

C2H5

(A)4-czkseks-6-Dyksjks-2-,fFky-1-esfFkylkbDyksgsDl-1-

methylcyclohex-1-ene

bZu

(B)5-Bromo-1-chloro-3-ethyl-2-

(B)5-czkseks-1-Dyksjks-3-,fFky-2-esfFkylkbDyksgsDl-2-


bZu

(C)5-Bromo-3-chloro-1-ethyl-2-

(C)5-czkseks-3-Dyksjks-1-,fFky-2-esfFkylkbDyksgsDl-1-


bZu

(D)1-Bromo-5-chloro-3-ethyl-4-

(D)1-czkseks-5-Dyksjks-3-,fFky-4-esfFkylkbDyksgsDl-3-


bZu

6.

The oxidation number of Oxygen in Na2O2 is: (A) + 1 (B) + 2 (C) – 2 (D) – 1

6.

Na2O2 esavkWDlhtudkvkWDlhdj.kvadgS% (A) + 1 (B) + 2 (C) – 2 (D) – 1

7.

Select the correct order (I) –OH > – NH2 > –Cl , – I effect (II) –NH2 > –OH > –Cl , + M effect (III) – Cl > –OH > –OR > –NH2 – I effect (A) I, II & III (B) II & III (C) II only (D) I & III

7.

lgh Øe dk p;u fdft,s

Calculate the oxidation number of Co un derlined ele ment s in the following compounds : K[Co(C 2O4)2(NH 3)2] (A) +3 (B) +4 (C) + 5 (D) +6

8.

8.

(I) –OH > – NH2 > –Cl ,– I effect (II) –NH2 > –OH > –Cl ,+ M effect (III) – Cl > –OH > –OR > –NH2 – I effect (A) I, II & III (B) II & III (C) II only (D) I & III

fuEu;kSfxdksaesafpfUgrrRoCodkvkWDlhdj.kvadKkr djksA K[Co(C 2O4)2(NH 3)2] (A) +3 (B) +4 (C) + 5 (D) +6


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Page # 18

In which of the following compounds + M effect is present

OH (I)

N=O (II)

9.

fuEufyf[kresalsdksuls;kSfxdesa+MizHkkomifLFkr gksxkA

CH3

OH

(III)

(I)

N=O (II)

(III)

O

CHO (IV)


CH3

O

CHO

O–C–R (V)

(IV)


O–C–R (V)



In FeCr2O4, the oxidation numbers of Fe and Cr are : (A) + 2 and + 3 (B) 0 and + 2 (C) + 2 and + 6 (D) + 3 and + 6

10.

11.

Which statement are correct about Inductive effect (I.E.) and mesomeric(m) effect. (A) I.E. is temporary and mesomeric is permanent effect (B) I. E. is permanent and works only on  electron. (C) Mesomeric effect is permanent and works only on  electron but less dominating on I.E. (D) Both have partial displacement of bonded electron.

11.

fuEueslsdkSulkdFkuizsjf.kdizHkko(I.E.)oehtkseSfVd izHkko(m)dsfy,lghgSA (A)I.E.izsj.khdizHkkovLFkkbZgSvkSjehtkseSfVdizHkko LFkkbZgSA (B)I. E.izsj.khdizHkkoLFkkbZgSvkSjbysDVªksuij dk;ZjrgSA (C)ehtksesfjdizHkkoLFkkbZgSvkSj;gbysDVªksuij dk;ZjrgSaijUrqI.E.defunsZ'khgSA (D)caf/krbysDVªksudkvkaf'kdizfrLFkkiunksuks(IEoM) esagksrkgSA

12.

The oxidation number of Phosphorus in

12.

Mg2P2O7esaQkWLQksjldkvkWDlhdj.kvadgS%

10.

Mg2P2O7 is : (A) + 3 (C) + 5

FeCr2O4esa]Fe rFkkCrdsvkWDlhdj.kvad gaS% (A) + 2 rFkk + 3 (B) 0 rFkk + 2 (C) + 2 rFkk + 6

(B) + 2 (D) – 3

(A) + 3

(B) + 2

(C) + 5

(D) – 3


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(D) + 3 rFkk + 6



Page # 19

13.

14.

Which of the following is least stable reso-







(A) RS-1

(B) RS-2

(C) RS-3

(D) RS-1=RS-2

(A) RS-1 (C) RS-3

The oxidation states of Sulphur in the anions (A) (B) (C) (D)

, S2O42– and S2O62– < S2O42 S2O42– < SO32– SO32– < S2O42– S2O42 < S2O62–

S2O62– follow < SO32– < S2O62– < S2O62– < SO32–

14.

the order:

R.S 1

H2N

H2N

OH

OH

R.S 2

H2N

R.S 3

(C) R.S 3

(D) R.S 4

rFkk S2O62– esa lYQj dh vkWDlhdj.k voLFkkdkØegS%

(C) SO32– < S2O42– < S2O62– (D) S2O42 < S2O62– < SO32– 15.

fuEuesalslclsvf/kdLFkk;hvuquknhljapukcrkb;sA H2N

OH

OH

(B) R.S 2

SO32– , S2O42–

(B) S2O42– < SO32– < S2O62–

R.S 1

H2N

R.S 4

(A) R.S 1

(B) RS-2 (D) RS-1=RS-2

(A) S2O62– < S2O42 < SO32–

Identify the most stable Resonating structure. H2N

16.

fuEufyf[kresalsdkSulhvuquknhlajpuklclsdeLFkk;h g SA

SO32–

15.

13.

nating structure.

OH

H2N

OH

H2N

R.S 3

(A) R.S 1 (C) R.S 3 16.

R.S 2

OH

OH R.S 4

(B) R.S 2 (D) R.S 4

elements in the following compounds :

fuEu;kSfxdksaesafpfUgrrRoPdkvkWDlhdj.kvadKkr djksA

(A) +3

(B) +4

(A) +3

(B) +4

(C) + 5

(D) +6

(C) + 5

(D) +6

Calculate the oxidation number of P underlined


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Page # 20

Identify the odd species out Which of the species among the following is different from others ?

(A)

17.

fuEuLih'khtksaesalsmlLih'khtdhigpkudhft,tks lcls vyx gS \

(A)

(B) O

(C)

NH

(B) O

(C)

(D)

NH

(D) N

N

18.

18 .

Correct order of stability of the following carbanion is

fuEufyf[krdkcZ_.kk;udsLFkkf;RodklghaØegS&

(I) (I)

19.

(II)

(II)

(III)

(III)

(A) I > II > III

(B) I > III > II

(C) III > II > I

(D) III > I > II 19.

Among the following molecules, the correct order of C - C bond length is

(A) I > II > III

(B) I > III > II

(C) III > II > I

(D) III > I > II

fuEuv.kqvksaesaC-CdhcU/kyEckbZdklghØegS\ (A) C2H6 > C2H4 > C6H6 > C2H2

(A) C2H6 > C2H4 > C6H6 > C2H2

(B) C2H6 > C6H6 > C2H4 > C2H2 (C6H6 csathu esa)

(B) C2H6 > C6H6 > C2H4 > C2H2 (C6H6 is benzene)

(C) C2H4 > C2H6 > C2H2 > C6H6

(C) C2H4 > C2H6 > C2H2 > C6H6

(D) C2H6 > C2H4 > C2H2 > C6H6

(D) C2H6 > C2H4 > C2H2 > C6H6 20.

20.

The oxidation number of Oxygen in Na2O2 is: (A) + 1 (B) + 2 (C) – 2 (D) – 1

Na2O2 esavkWDlhtudkvkWDlhdj.kvadgS% (A) + 1 (C) – 2


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(B) + 2 (D) – 1


Page # 21

21.

Find out the hybridisation state of carbon atoms in given compounds from left to right.

21.

CH3 – CH = CH – CH = C = CH – C  C – CH3

22.

CH3 – CH = CH – CH = C = CH – C  C – CH3

(A) sp3 sp2 sp2 sp2 sp sp2 sp sp sp3

(A) sp3 sp2 sp2 sp2 sp sp2 sp sp sp3

(B) sp3 sp2 sp2 sp sp sp sp sp sp3

(B) sp3 sp2 sp2 sp sp sp sp sp sp3

(C) sp3 sp2 sp2 sp2 sp2 sp2 sp sp sp3

(C) sp3 sp2 sp2 sp2 sp2 sp2 sp sp sp3

(D) sp3 sp sp sp2 sp sp2 sp sp sp3

(D) sp3 sp sp sp2 sp sp2 sp sp sp3

In FeCr2O4, the oxidation numbers of Fe and Cr are : (A) + 2 and + 3 (C) + 2 and + 6

22.

(A)

N = N (B)

(C)

N

blv.kqesalclsvf/kdLFkk;hdsuksfudylajpukgS\

N=N

N (D) All are equally stable

The oxidation number of Phosphorus in Mg2P2O7 is : (A) + 3 (B) + 2 (C) + 5 (D) – 3

24.

Total number of  and -bonds are in naphthalene is

25.

(A) 5 and 18 (C) 5 and 19

(B) 6 and 19  (D) 7 and 26

(A)

N = N (B)

(C)

N = N (D)lHkhLFkk;hgSA

|

N=N

Mg2P2O7esaQkWLQksjldkvkWDlhdj.kvadgS% (A) + 3

(B) + 2

(C) + 5

(D) – 3

us¶FkyhuesavkSj-cU/kksadhdqyla[;kgS\ (A) 5 rFkk 18

(B) 6 rFkk 19 

(C) 5 rFkk 19

(D) 7 rFkk 26


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(B) 0 rFkk + 2 (D) + 3 rFkk + 6

N=N

23.

The most stable canonical structure of this molecule is

25.

FeCr2O4esa]Fe rFkkCrdsvkWDlhdj.kvad gaS% (A) + 2 rFkk + 3 (C) + 2 rFkk + 6

(B) 0 and + 2 (D) + 3 and + 6

N=N

23.

24.

fn;sx;s;kSfxdksaesacka;s]lsnka;sdkcZuijek.kqvksadhladj.k voLFkkdksKkrdhft;s\



27.

28.

Page # 22

The order of stability of the following carbanion is

26.

fuEudkcZ,suk;udsLFkkf;RodkØegS\

(I) CH3CH2

(II)

(I) CH3CH2

(II)

(III)

(IV)

(III)

(IV)

(A) I > II > III > IV (C) IV > III > II > I

(B) I > III > II > IV (D) III > IV > I > II

(A) I > II > III > IV (C) IV > III > II > I

(B) I > III > II > IV (D) III > IV > I > II

Consider the following reaction: 3Br2 + 6CO32 – + 3H2O  5Br – + BrO3– + 6 HCO3– Which of the following statements is true regarding this reaction: (A) Bromine is oxidized and the carbonate radical is reduced. (B) Bromine is reduced and the carbonate radical is oxidized. (C) Bromine is neither reduced nor oxidized. (D) Bromine is both reduced and oxidized.

27.

(I) CH2 = CH – CH = CH2

28.

fuEu vfHkfØ;k dk voyksdu dhft, % 3Br2 + 6CO32 – + 3H2O BrO3– + 6 HCO3–



5Br

–

+

blvfHkfØ;kdslUnHkZesadkSulkdFkulR;gS% (A)czksehuvkWDlhd`rgksrkgSrFkkdkcksZusVewydvipf;r gksrkgSA (B)czksehuvipf;rgksrkgSrFkkdkcksZusVewydvkWDlhd`r gksrkgSA (C)czksehuurksvipf;rgksrkgSvkSjughvkWDlhd`rgksrkgSA (D)czksehuvipf;rovkWDlhd`rnksuksagksrkgSA

(II) CH2 – CH = CH – CH2

(I) CH2 = CH – CH = CH2 (II) CH2 – CH = CH – CH2



Amo ng, thes e, w hich are structures of bufa-1,3 diene

can onic al

buesalsdkSulhdsuksfudylajpuk,sC;wVk1<3 MkbZbZudh

(A) I and II

(B) I and III

g SA

(C) II and III

(D) all

(A) I rFkk II

(B) I rFkk III

(C) II rFkk III

(D)lHkh


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Page # 23

29.

Statement-1 : In the reaction, MnO4–

+

29.

5Fe2+ + 8H+

2+ + 5Fe3+ +  Mn 4H2O, acts as oxidising agent. Statement-2 : : In the above reaction, Fe2+ is converted to Fe3+. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.

 Mn2+ + 5Fe3+ + 4H2O esa MnO4–

MnO4–

vkWDlhdkjd dh rjg O;ogkj djrk gSA oDrO;&2:mi;qZDrvfHkfØ;kesaFe2+,Fe3+esaifjofrZr gksrkgSA (A) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA (B) oDrO;-1 lR; gS, oDrO;-2 lR; gS ; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO;-1 lR; gS, oDrO;-2 vlR; gSA (D) oDrO;-1 vlR; gS, oDrO;-2 lR; gSA 30.

30.

oDrO;&1 : vfHkfØ;k MnO4– + 5Fe2+ + 8H+

Oxidat ion numb er of iron in + Na2[Fe(CN)5(NO )] is : (A) + 2 (B) + 3 (C) +8/3 (D) none of the three

Na2[Fe(CN)5(NO+)] esa vk;ju dh vkWDlhdj.k la[;k

fuEugS% (A) + 2

(B) + 3

(C) +8/3

(D)rhuksaesalsdksbZugha


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2016 SAMPLE PAPER

SAMPLE PAPER (1)

Avail Free Video Solutions of Experienced Faculty www.etoosindia.com

IIT JEE SAMPLE PAPER 2016 Duration : 3 Hours Max Mark : 360

In each part of the paper, Section-A contains 30 questions. Total number of pages are 28. Please ensure that the Questions paper you have received contains ALL THE QUESTIONS in each section and PAGES. SECTION - A 1.



Use only blue/black pen (avoid gel pen) for darkening the bubble.

2.


3.

The Answer sheet will be checked through computer hence, the answer of the question must be marked by shading the circles against the question by dark blue/black pen.

4.


Batch

Roll Number

10+1 10+1 10+2 10+3 Crash

0

0

0

0

0

0

0

0

0

1

1 1

1

1

1

1

1

1

2

2 2 3 3

2 3

2 3

2 3

2 2 3 3

2 3

4 4 5 5

4

4

4

4

4

4

3 4

Paper

1

5 6

5 6

5 6

5

6 6

6

5 6

5 5 6 6

Paper 1

7

7 7

7

7

7

7

7

7

8

8 8 9 9

8

8

8

8

9

8 9

8

9

9

9

9

Paper 2

Name

9

For example if only 'A' choice is correct then, the correct method for filling the bubble is A B C D E For example if only 'A & C' choices are correct then, the correct method for filling the bubble is A B C D E



L A S T N A ME

D D MMY Y


For example If Correct match for (A) is P; for (B) is R, S; for (C) is Q; for (D) is Q, S, T then the correct method for filling the bubble is P

Q

R S T

Ensure that all columns are filled . Answers, having blank column will be treated as incorrect. Insert leading zero(s) if required : '6' should be filled as 0006


0 0 0 0 1 1 1 1 2 2 2 2

0 0 0 0 1 1 1 1 2 2 2 2

3 3 3 3 4 4 4 4

3 3 3 3 4 4 4 4

5 5 5 5

5 5 5 5

6 6 6 6 7 7 7 7

6 6 6 6 7 7 7 7

5 5 5 5 6 6 6 6 7 7 7 7

8 8 8 8 9 9 9 9

8 8 8 8 9 9 9 9

8 8 8 8 9 9 9 9

0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4

VELOCITY (XII) - JEE MAIN

Page # 2

PART - I [MATHEMATICS] SECTION - A [STRAIGHT OBJECTIVE TYPE] Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct 1.

SECTION - A [ oLr q fu"B i zd kj dsi z'u ] i z-.1 l si z-.30 r d pkj fodYi (A), (B), (C), (D) fn; sgS ft uesa^^dsoy , d^^ l ghgS A

The domain of 1

4

[cos 1 x 4 ] | [ x  2 tan 1 x ] |  sin(nx )

1

[cos x ] | [ x  2 tan x ] |  sin(nx ) {3x 2  7}  a

sin x  3 cos x

 1  n cos   2  x

1.

{3x 2  7}  a

   

2)

(C) (–1, 1)

2.

(A) (–2,

(B) (0, 1)

x b and a  x 

2)

   

dki zkUr ]

(B) (0, 1)

(C) (–1, 1)

(D) None of these

If a, b > 0 and f(x) = xlim 0 

 1  n cos   2  x

t gk¡ [.] r Fkk{.}; x dsegÙke i w . kk± d r Fkk fHkUukRed Hkkx dksi znf' kZ r dj r sgS a ] gksxk-

where [.] and {.} denotes greatest integer and fractional part of x, is (A) (–2,

sin x  3 cos x

2.

(D)

buesal sdksbZugha

x b

; fn a, b > 0 r Fkk f(x) = xlim   r Fkk 0  a x  

x b

g(x) = xlim   then 0   a  x

x b

g(x) = xlim   0   a  x

(A) f(x) = g(x) (B) f(x) > g(x) (C) f(x) = g(x) only if a = b (D) f(x) < g (x)

(A) f(x) = g(x) (B) f(x) > g(x) (C) f(x) = g(x)

gS ] rc -

dsoy ; fn a = b

(D) f(x) < g (x) (SPACE FOR ROUGH WORK)

TSS

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Education is the most powerful weapon which you can use to change the world


Page # 3

3.

The shortest distance (S.D.) between lines

x 3 y 8 z3 x3 y7 z6     and is3 1 2 4 1 3

4.

(A) 3 30

(B)

(C) 2 30

(D) 4 30

3.

x3 y7 z6   dse/; y ?kq r e nw j h (S.D.) gS 3 2 4

30

A football match may be either won, draw or

4.

lost by the host country’s team. so , there are three ways of forecasting the result of any one match, one correct and two incorrect. Find the probability of forecasting at least three correct results for four matches.

5.

(A) 1/9

(B) 2/9

(C) 2.7

(D) None 5.

The root of the equation 2(1 + i)x2 – 4(2 – i)x – 5 – 3i = 0 where i =

x 3 y 8 z3   r Fkk 3 1 1

j s[kkvksa

(A) 3 30

(B)

(C) 2 30

(D) 4 30

30

, d Qq VckW y esp est cku ns'kdhVhe } kj kt hr k, MkW ; kgkj k t kl dr kgS Afdl h, d esp dsi fj .kke dsckj sesai q okZ uq eku y xkusdsr hu r j hdsgS ] , d l ghr Fkk nksxy r A r c pkj espksadsi q okZ uq ekuksaesal sde l sde r hu i q oku Zq eku l ghgksus dhi zkf; dr k gS(A) 1/9

(B) 2/9

(C) 2.7

(D) dks bZugha

l ehdj .k 2(1 + i)x2 – 4(2 – i)x – 5 – 3i = 0 t gk¡ i =  1 , dk ew y ft l dk eki ka d vf/kd gS , gS

 1 , which has greater modulus, is

 3  5i    2 

(B) 

3i    2 

(D) 

 3  5i   (A)   2 

 5  3i   (B)   2 

(A) 

3i   (C)   2 

 1  3i   (D)   2 

(C) 

 5  3i    2   1  3i    2 


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cM+k lkspks vkSj gkjus ls er MjksA

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VELOCITY (XII) - JEE MAIN 6.

Page # 4





Let   a1î  a 2ˆj  a 3kˆ ,   b1î  b 2ˆj  b3kˆ a nd

6.

     c1î  c2 ˆj  c3kˆ , |  | = 2,  makes angle /6 

       c1î  c2 ˆj  c3kˆ , |  | = 2,  ;  , oa dsl er y





with the plane of  and  and angle between

a1   b  and  is /4, then 1 c1

a2 b2 c2

a3 b3 c3

 ekuk   a1î  a 2ˆj  a 3kˆ ,   b1î  b 2ˆj  b3kˆ r Fkk



dsl kFk/6 dks.kcukr kgSr Fkk  , oa dse/; dks.k/ a1 4 gS , r c b1 c1

n

is equal to

a2 b2 c2

a3 b3 c3

n

dk eku gS(t gk¡ n l e i zkÑr

la [ ; kgS )

(where n is even natural number)

  n  |  ||  |  2

(A) 



2

  (|  ||  |) n (B) 2n / 2

  

  (|  ||  |) n / 2 (C) 2n

7.

 r  r 1  If A(r) =  2 r  

  n  |  ||  |  2  (A)    2    (|  ||  |) n / 2 (C) 2n

(D) None of these

r 1 r 2 r  2

r  2   r 1  , where  is com2 r  4 

plex cube root of unity. Then (A) A(r) is singular only if r is even (B) A(r) is singular only if r is odd (C) A(r) is singular (D) A(r) is non-singular

  (|  ||  |) n (B) 2n / 2

7.

 r  r 1  ; fn A(r) =  2 r  

(D)

r 1 r

 2 r  2

buesal sdksbZugha

r  2   r 1  , t gk¡  bdkbZdk 2 r  4 

l fEeJ ?kuew y gS ] rc (A) A(r) vO ;q RØe.kh; gksxk dsoy ; fn r l e gS (B) A(r) vO ;q RØe.kh; gksxk dsoy ; fn r fo"ke gS (C) A(r) vO ;q RØe.kh; gksxk (D) A(r) O ;q RØe.kh; gksxk


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Page # 5

n Pn If f(x) = n Cn n

8.

n 1 Pn 1 n 1 C n 1 n 1

n2 Pn  2 . Then f(x) is n 2 Cn2

n 8.

9.

(A) n2 + n

(B) n  1

(A) n2 + n

n2

(D) None

(C)

 1   1   cos 1 x  – tan   cos 1 x  = 4 2 4 2    

9.

tan 

2 1 x2 (A) x

 2 1 x2 (B) x

10.

Area bounded by the curve y = f(x), y = x and the lines x = 1, x = t is (t +

1 t 2 –

2

–1) sq. units for all t > 1. If f(x) satisfying f (x) > x for all x > 1, then f (x) is equal to -

x 1 x 2

n 2

Pn  2 , r ksf(x) fuEu Cn2

(D) dks bZugha

 1   1   cos 1 x  – tan   cos 1 x  = 4 2  4 2 

2 1 x2 x

(B)

x

(A) x + 1 +

n2

(B) n  1

n2

(C)

(D) None of these

1  x2

(C) 1 +

Pn 1 n 1 C n 1

tan 

(A)

x (C)

n2

n 1

; fn f(x) = n Pn Cn } kj kfoHkkft r gksxk-

divisible by -

(C)

n 1

n

n2

x 1 x

2

(D) bues sal

1  x2

1 x 2

i fj c) {ks=kQy l Hkht > 1 dsfy , (t + 1  t 2 – 2 – 1) oxZbdkbZgS A ; fn f(x) l Hkh x > 1 ds fy , f (x) > x dksl a rq "V dj r k gS ,r c f(x) cj kcj gSx (A) x + 1 +

1 x

x 1 x 2

sdksbZugha

oØ y = f(x), y = x r Fkk j s[kkvksax = 1, x = t l s

x

(B) x + (D)

10.

 2 1 x2 x

x 2

(B) x +

x (C) 1+

1 x

1 x 2 x

2

(D)

1 x 2



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iwjs fo'okl ds lkFk vius liuks dh rjQ cM+k]s ogh ftanxh ft;ks ftldh dYiuk vkius dh gSA


11.

1  Given f(x) = 4 –   x  2  

 tan [ x ]  ; g(x) =  x  1 ;

x0 x0

Page # 6

3/ 2

;

11.

1  fn; k gSfd f(x) = 4 –   x  2  

 tan [ x ]  ; x  1 ;

g(x) = 

; h(x) = {x},

 f (t)dt

(A) f, g, h (B) h, k (C) f, g (D) g, h, k t gk¡[x] r Fkk{x} egÙke i w . kkZ a d Qy u r FkkfHkUukRed

Hkkx Qy u dksn' kkZ r sgS A x

and f(x) satisfies the fol-

12.

ekukg(x) =  f ( t )dt r Fkk f(x) fuEufy f[ kr kr Z a

a

lowing condition f(x + y) = f(x) + f(y) + 2xy

f(x + y) = f(x) + f(y) + 2xy – 1,  x, y  R

– 1,  x, y  R and f (0) =

] rc dksl Ur q "B dj r k gSr Fkk f (0) = 3  a  a 2 gS x dsekuks adk l a iw . kZl eq Pp; gS ] t gk¡ g(x) o/kZ eku gS-

3  a  a 2 , then

the exhaustive set of values of x where g(x) increases is -

3  (A)   ,  2  (C) (0, ) 13.

; h(x) = {x},

x0

gS ] r c fdl dsfy , [0, 1] esa ya xzkt e/; eku i zes; y kxqughagS-

x

Let g(x) =

;

k (x) = 5log2 ( x 3) ,

k (x) = 5log2 ( x 3) , then in [0, 1] Lagranges Mean value theorem is NOT applicable to (A) f, g, h (B) h, k (C) f, g (D) g, h, k where [x] & {x} denotes G.I.F. and fractional part function respectively.

12.

x0

3/ 2



 3  (B)   ,0   2  (D) (– , )

cos x  x sin x dx is equal to x ( x  cos x )

13.

x   c x  cos x  

(A) log{x(x+cosx)}+c (B) log

 x  cos x    c (D) None of these x  

(C) log

3  (A)   ,  2 

 3  (B)   ,0   2 

(C) (0, )

(D) (– , )



cos x  x sin x dx cj kcj gS & x ( x  cos x )

x   c  x  cos x 

(A) log{x(x+cosx)}+c (B) log

 x  cos x    c (D) buesl sdks bZugha x  

(C) log


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Page # 7

14.

If p and q are order and degree of differen2

14.

2

2

d y  dy     + x2y2 = tial equation y  2  + 3x  dx  dx   

 d2 y   dy   + x2y2 = sin x dhdks fV y  2  + 3x   dx   dx 

2

sin x, then : (A) p > q (C) p = q

15.

 (1  x  x

1

1

(C)  xe x  x  c 16.

r Fkk ?kkr gS , rc: (A) p > q (C) p = q

1

(A) ( x  1)e x  x  c 1

2

(B) p/q = 1/2 (D) p < q

)e x  x dx is equal to -

15. 1

(A)

(B)

(C) (D)

1

)e x  x dx dk eku gS-

1

1

(D) xe x x  c

Statement-I : If |z1| = 1, |z2| = 2, |z3| = 3 and |z1 + 2z2 + 3z3| = 6. Then the value of |z2z3 + 8z3z1 + 27z1z2| is 36. Statement-II : |z1+ z2 + z3|  |z1| + |z2| + |z3| Both Statement-I and Statement-II are true but Statement-II is not correct explanation of the Statement-I. Both Statement-I and Statement-II are true and the Statement-II is correct explanation of the Statement-I. Statement-I is true but the Statement-II is false Statement-I is false but Statement-II is true

1

(A) ( x  1)e x  x  c (C)  xe x  x  c 16.

1

(B) ( x  1)e x  x  c 1

(D) xe x x  c

l ehdj .k ax2 + bx + c = 0, t gk¡ a, b, c f=kHkq t ABC dhH kq t k, ¡gS ar Fkkl

ehdj .kx2 + 2 x + 1 = 0 ds ew y l eku gSr ksC dk eki gksxk-

x2 + 2 x + 1 = 0 have a common root. The measure of C is(A) 90° (B) 45° (C) 60° (D) None of these 17.

 (1  x  x

(B) p/q = 1/2 (D) p < q

1

(B) ( x  1)e x  x  c

The equation ax2 + bx + c = 0, where a, b, c are the sides of a ABC and the equation

; fn p r Fkkq Øe' k%vody l ehdj .k

(A) 90° (C) 60° 17.

(A) (B) (C) (D)

(B) 45° (D) bues al

sdksbZugha

dFku-I : ; fn |z1| = 1, |z2| = 2, |z3| = 3 r Fkk |z1 + 2z2 + 3z3| = 6 r c |z2z3 + 8z3z1 + 27z1z2| dk eku 36 gS A dFku-II : |z1+ z2 + z3|  |z1| + |z2| + |z3| dFku-I vkS j dFku-II nksuka sl ghgSi j Ur qdFku-II, dFkuI dk l ghLi "Vhdj .k ughagS dFku-I vkS j dFku-II nksuka sl ghgSr FkkdFku-II, dFku-I dk l ghLi "Vhdj .k gS dFku-I l ghgSi j Ur qdFku-II xy r gS dFku-I xy r gSi j Ur qdFku-II l ghgS



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eqf'dys oks phtsa gksrh gS] tks gesa rc fn[krh gS tc gekjk /;ku y{; ij ugha gksrkA


Page # 8

2 1 18.

3 4 

If three matrices A = 4 1 , B = 2 3 and    

2 1

18.

 3  4  . Then  2 3 

3

C= 

C =  2 

 A(BC) 2  ABC   +tr  tr(A)+tr  4  2  

+…+  = (A) 6 (C) 12 19.

3    +tr  A(BC)   8  

(B) 9 (D) None

For x1, x2, y1, y2  R, if 0 < x1 < x2, y1 = y2 and z1 = x1 + iy1, z2 = x2 + iy2 and z3 =

19.

1 (z 2 1

3    +tr  A(BC)   8  

+…+  = (A) 6

(B) 9

(C) 12

(D) dks bZugha

   

x1, x2, y1, y2  R dsfy , ] ; fn 0 < x1 < x2, y1 = y2

, oaz1 = x1 + iy1, z2 = x2 + iy2 r Fkkz3 =

1 (z + 2 1

z2), r ksz1, z2, z3 fuEu dksl a rq "V dj r sgS(A) | z1 | = | z2 | = | z3 | (B) | z1 | < | z2 | < | z3 | (C) | z1 | > | z2 | > | z3 | (D) | z1 | < | z3 | < | z2 |

+ z2), then z1, z2, z3 satisfy (A) | z1 | = | z2 | = | z3 | (B) | z1 | < | z2 | < | z3 | (C) | z1 | > | z2 | > | z3 | (D) | z1 | < | z3 | < | z2 | 20.

 4 ] rc 3  gS

 A(BC) 2  ABC  tr(A)+tr  2  +tr  4   

   

3 4 

; fn r hu vkO ;w g A = 4 1 , B = 2 3 r Fkk    

The equation of the plane passing through the point (–1, 3, 2) and perpendicular to planes x + 2y + 3z = 5 and 3x + 3y + z = 9 is(A) 7x – 8y – 3z + 25 = 0 (B) 7x – 8y + 3z + 25 = 0 (C) 7x – 8y + 3z – 25 = 0 (D) 7x + 8y + 3z – 25 = 0

20.

fcUnq(–1, 3, 2) l sxq t j usoky sr Fkk l er y ksa x + 2y + 3z = 5 o 3x + 3y + z = 9 dsy Ecor ~ l er y dhl ehdj .k gS(A) 7x – 8y – 3z + 25 = 0 (B) 7x – 8y + 3z + 25 = 0 (C) 7x – 8y + 3z – 25 = 0 (D) 7x + 8y + 3z – 25 = 0


Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.


Page # 9

21.

The length of the perpendicular from the ori-

 gin to the plane passing through the point a  and containing the line r  b   c is -

21.

l er y i j ew y fcUnql sMky sx; sy Ec dhy EckbZgS  [a b c ] [a b c ] (A)       (B)     [a  b  b  c  c  a ] [a  b  b  c ]

  [a b c ] [a b c ] (A)       (B)     [a  b  b  c  c  a ] [a  b  b  c ]  [a b c ]     (C) [b  c  c  a ] 2

22.



 [a b c ] (C)     [b  c  c  a ]

 [a b c ] (D)     [c  a  a  b]

2

1 1  cos ec101 x  dx  x x 

22.

23.

24.

(B) 1 (D) 101/2

If the function f (x) = x3 + 3 (a – 7)x2 + 3(a2 – 9) x – 1 has a positive point of maximum, then (A) a  (3, )  (–, – 3) (B) a  (– ,–3)  (3, 29/7) (C) (– , 7) (D) (–, 29/7)

23.

If the normal to the curve y = f(x) at x = 0 be given by the equation 3x – y + 3 = 0, then the value of

24.

Lim x 2 {f ( x 2 )  5f ( 4x 2 )  4f (7 x 2 )}–1 , isx 0 (A) 1/3 (C) –1/5

1

(B) – 1/3 (D) 1/4

 [a b c ] (D)     [c  a  a  b]

1  x  dx  x 

101

 x cos ec

1/ 2

1/ 2

(A) 1/4 (C) 0



fcUnqa l sxq t j usokysr Fkkj s[kk r  b   c dksj [ kusokys

(A) 1/4

(B) 1

(C) 0

(D) 101/2

; fn Qy u f (x) = x3 + 3 (a – 7)x2 + 3(a2 – 9) x – 1 kukRed mfPp"B fcUnqj [ kr k gSr c (A) a  (3, )  (–, – 3) (B) a  (– ,–3)  (3, 29/7) (C) (– , 7) (D) (–, 29/7)

,d /

; fn oØ y = f(x) dsx = 0 i j vfHky Ec] l ehdj .k 3x – y + 3 = 0 ds} kj k fn; k t kr k gS , rc Lim x 2 {f ( x 2 )  5f ( 4x 2 )  4f (7 x 2 )}–1 dk eku gS x 0 (A) 1/3 (C) –1/5


(B) – 1/3 (D) 1/4


Page # 10

Suppose that f is a differentiable function

25.

with the property that f(x + y) = f(x) + f(y)

ekukfd f vody uh; Qy u gSr Fkk fuEu xq . k/keZj [ kr k gS f(x + y) = f(x) + f(y) + xy r Fkk

1 f(h) = 3, thenh

lim

1 f(h) = 3 r c h

(A) f is a linear function

(A) f

j s[kh; Qy u gS

(B) f(x) = 3x + x2

(B) f(x) = 3x + x2

+ xy and lim

h0

h0

2

(C) f(x) = 3x +

x 2

(C) f(x) = 3x +

(D) none of these

26.

27.

lim n 

(D)

1 1 2 3n    .....    is equal to n  n 1 n  2 4n 

26.

x2 2

buesal sdksbZugha

lim n 

1 1 2 3n    .....    cj kcj gSn  n 1 n  2 4n 

(A) log 4

(B) – log 4

(A) log 4

(B) – log 4

(C) 1 – log 4

(D) None of these

(C) 1 – log 4

(D)

A class has 21 students. The class teacher

27.

buesal sdksbZugha

, d d{kkesa21 fo| kFkhZgSd{kkv/; ki d dksr fo| kfFkZ ; ksads

has been asked to make n groups of r stu-

nl

dents each and go to zoo taking one group

dksfpfM+ ; k?kj y st kr sgSl ew g esafdr usy ksx vk; sxsa(r dk

at a time. The size of group (i.e. The value of r) for which the teacher goes to the zoo maximum number of times is – (No group can

ew g cukusdksdgkt kr kgSvkS j , d l e; i j , d l ew g

eku) ft l dsfy ; sv/; ki d fpfM+ ; k?kj vf/kd l svf/kd ckj t kr kgks]gS(dksbZHkhl ew g nksckj fpfM;+k?kj ughat kl dr k) (A) 9

go to the zoo twice) (A) 9 or 10

(B) 10 or 11

(C) 11 or 12

(D) 12 or 13

; k10

(C) 11


; k12

(B) 10

; k11

(D) 12

; k13


Page # 11

n

28.

The value of nlim  (A)  (C) /4

29.

30.

 cot r 1

1

1  3 r r  r  2   is    

n

28.

(B) /2 (D) – /2

The function f : R  R is defined by f(x) = cos2 x + sin4 x for x  R, then range of f (x) is

lim n

 cot r 1

(A)  (C) /4 29.

(B) /2 (D) – /2

Qyu f : R  R ; f(x) = cos2 x + sin4 x ; x  R } kj k i fj Hkkf"kr gS ] r ksf (x) dki fj l j gS 3 

 

3



3  (A)  , 1 4 

3  (B)  , 1 4 

(A)  , 1 4

3  (C)  , 1 4 

3  (D)  , 1 4 

(C)  , 1 4 

If f (x) be a continuous function defined for 1  x 3, f (x) Q  x [1, 3], f (B) = 10, then f (1.8) is (where Q is a set of all rational numbers) (A) 1 (B) 5 (C) 10 (D) 20

30.

1

1  3 r r  r  2   dk eku gS   

3



3 4

 

(B)  , 1 4  (D)  , 1

; fn f (x) l a r r ~Qy u gS ] t ks 1  x  3 ds fy , f (x)  Q  x  [1, 3], } kj k i fj H kkf"kr gSr Fkk f (B) = 10 gS ] r ksf (1.8) gS a (t gk¡Q l H khi fj es; l a [ ; kvksdk l eq Pp; gS ) (A) 1 (C) 10


(B) 5 (D) 20


Page # 12


SECTION - A

[ oLr q fu"B i


zd kj dsi z'u ]

out of which ONLY ONE is correct.

i z-.1 l si z-.30 r d pkj fodYi (A), (B), (C), (D) fn; sgS ft uesa^^dsoy , d^^ l ghgS A

1.

1.

Q.1 to Q.30 has four choices (A), (B), (C), (D)

2.

An infinitely long rod lies along the axis of a concave mirror of focal length f . The near end of the rod is at a distance u > f from the mirror. Its image will have a length (A)

f2 u–f

(B)

uf u–f

(C)

f2 uf

(D)

uf uf

Two equal positive point charges q are held at a fixed distance a apart. A point test charge is located in a plane that is normal to the line joining these charges and midway between them. What is the radius r of the circle in this plane for which the force on the test particle has maximum value ?

q

2.

, d vuUr : i l sy Ech NM+, d f Qksd l y EckbZds, d vor y ni Z . kdhv{kdsvuq fn' kj [ khgS ANM+dkfudV fl j k ni Z . k l s, d nw j hu > f i j fLFkr gS A bl dsçfr fcEc dh y EckbZgksxh(A)

f2 u–f

(B)

uf u–f

(C)

f2 uf

(D)

uf uf

nkscj kcj /kukRed fcUnqvkos'kksaq dksa nw j hi j fLFkj j [ kk x; k gS A , d fcUnqi fj {k.k vkos'k dksml r y esaj [ kk x; k gSt ksmu vkos'kksadkst ksMu +sokyhj s[kkdsyEcor ~r Fkkmuds e/; esagS Abl r y esao`Ùkdhf=kT; kr D; kgksxhft l dsfy; s i fj {k.k d.k i j cy vf/kdr e eku j [ ksxk?

q

q

a 2

q

a 2

a 2

(A) a/ 2

(B)

a/2 2

(C) a/2

(D)

2a

a 2

(A) a/ 2

(B) a/2 2

(C) a/2

(D)


2a


Page # 13

3.

Four identical charged particles, each +Q, are fixed at the corners of a square of side ‘a’. A free point charge of mass ‘m’ and charge ‘–q’ is placed at the centre of square. It is slightly displaced along a line passing through the centre of square and perpendicular to the plane of square and then released. Assuming that displacement of –q along the said line is quite small as compared to ‘a’ ‘–q’ will -

3.

(A)

1 3 2qQ (A) execute S.H.M. of frequency 2 m 0a 3

3 2qQ

dj r k gSr Fkk oxZdk dsUæ ml dh LFkkbZl kE; fLFkfr esagksxkA (B)

1 2 2qQ (B) execute S.H.M. of frequency 2 m 0a 3

1

2 2qQ

oxZds dsUæ ds l ki s{k 2 vko`fÙk l s m 0a 3

dj r k gSr Fkk oxZdk dsUæml dh vLFkkbZl kE; fLFkfr esagksxkA S.H.M.

about the centre of square and centre of square will be its unstable equilibrium position

1 2 2qQ (C) execute S.H.M. of frequency 2 m 0a 3

(C)

1

2 2qQ

oxZds dsUæ ds l ki s{k 2 m a 3 vko`fÙk l s 0

dj r k gSr Fkk oxZdk dsUæ ml dh LFkkbZl kE; fLFkfr esagksxkA S.H.M.

about the centre of square and centre of square will be its stable equilibrium position

1 3 2qQ (D) execute S.H.M. of frequency 2 m 0a 3

(D)

1

3 2qQ


dj r k gSr Fkk oxZdk dsUæml dh vLFkkbZl kE; fLFkfr esagksxkA S.H.M.

about the centre of square and centre of square will be its unstable equilibrium position In the fusion reaction, 1H2 + 1H2  2He3 + 1 0n The masses of deutrons, helium and neutron expressed in amu are 2.015, 3.017 and 1.009 respectively. If 1 kg of deuterium undergoes complete fusion. Find the amount of total energy release, 1 amu = 931 MeV/C2 (A) 6 × 1013 J (B) 5.6 × 1013 J (C) 9 × 1013 J (D) 0.9 × 1013 J

1


S.H.M.

about the centre of square and centre of square will be its stable equilibrium position

4.

pkj l e: i vkosf' kr d.kksa ] çR; sd +Q dksa Hkq t kds, d oxZ ds dksuksa i j fLFkj j [ kk x; k gS A m æO ; eku r Fkk –q vkos ' kds, d Lor a =kfcUnqvkos'kdksoxZdsdsUæi j j [ kk x; kgS Aml soxZdsdsUæl sr FkkoxZdsr y dsy Ecor ~j s[kk dsvuq fn' kfoLFkkfi r fd; kt kr kgSr FkkfQj eq ä fd; kt kr k gS A; g ekfu; sfd –q dkfoLFkki u nhxbZj s[kkdsvuq fn' k a dh r q y uk esacgq r de gS ] r ks–q -

4.

la y ; u vfHkfØ; kesa1H2 + 1H2  2He3 + 0n1 M~ ;w VªkW u] ghfy ; o U;w VªkW u dsamu esaO ; Dr fd; sx; snzO ; eku Øe' k% 2.015, 3.017 o 1.009 gS A ; fn 1 kg M~ ;w Vhfj ; e dk iw . kZl a y ; u gksrkgS Aeq Dr dq y Åt kZdhek=kkKkr dhft , (1 amu = 931 MeV/C2) (A) 6 × 1013 J (B) 5.6 × 1013 J 13 (C) 9 × 10 J (D) 0.9 × 1013 J



Page # 14

Figure shows the intensity - wavelength relations of X-rays coming from two different

5.

nksfofHkUu dw fy t ufy dkvksal svkusoky hX- fdj .kksadh r hozrk o r j a xnS /; Zdse/; xzkQ fp=kkuq l kj gS a A l r r ~oØ

coolidge tubes. The solid curve represents the relation for the tube A in which the

uy hA dsfy ; sft l esavkj ksfi r foHkokUr j VA r Fkk y {;

potential difference between the target and the filament is VA and the atomic number of

/kkr qdk i j ek.kqØeka d ZA] nw l j huy hdsfy ; sbudkeku

the target material is ZA. These quantities

VB

o ZB gS a A r ks& Intensity

Intensity

are VB and ZB for the other tube. Then

K K B A

K K B A

Wavelength

Wavelength (A) VA > VB, ZA > ZB (C) VA < VB, ZA > ZB

(A) VA > VB, ZA > ZB (B) VA > VB , ZA < ZB (C) VA < VB, ZA > ZB (D) VA < VB, ZA < ZB 6.

Four masses are fixed on a massless rod as

0.2 m

0.2 m

2 kg

0.2 m

6.

0.2 m

P

5 kg

2 kg

2 kg

shown in the figure. The moment of inertia about the axis P is about (A) 2 kg × metre2 (B) 1 kg × metre2 (C) 0.5 kg × metre2 (D) 0.3 kg × metre2 7.

pkj nzO ; eku fp=kkuq l kj nzO ; ekughu NM+i j fp=kkuq l kj fQDl gS A 0.2 m

0.2 m 5 kg

Some equipotential plane parallel surface are shown in the figure. The plane are inclined to x-axis by 45º and distance from one plane to another plane along x-axis is 20 cm. The electric field is –

(B) VA > VB , ZA < ZB (D) VA < VB, ZA < ZB

5 kg

0.2 m

P

0.2 m

5 kg

2 kg

P v{k dsl ki s {k t M+ Ro vk?kw . kZgSy xHkx (A) 2 kg × metre2 (B) 1 kg × metre2 2 (C) 0.5 kg × metre (D) 0.3 kg × metre2

7.

dq N l efoHko l ekUr j l r gsauhpsfp=k esan' kkZ bZxbZgS A ; g r y x-v{k l s45º vki fr r gSr Fk , d r y dh nw l j sr y l sx-v{k dsvuq fn' k nw j h 20 cm gS A r ks fo| q r {ks=k gS –



Page # 15

Y

Y

45º 20cm

45º 20cm

X

40cm 60cm 80cm 150V 125V 100V

150V 125V 100V

75V

ij ij (C) 177 V, x-v{k l s45º i j (D) 125 V, x-v{k l s135º i j (B) 125 V, x-v{k l s45º

Three atomic states of a hydrogenic atom are shown in the figure. The transition from C to B yields a photon of wavelength 364.6 nm and the transition from B to A yields a photon of wavelength 121.5 nm. Then the transition from C to A will yield a photon of wavelength

8.

fp=kesagkbMªkst u i j ek.kqdhr hu voLFkk, ¡n' kkb ZZxbZgS A aC l sB rd l a Øe.kl s, d QksVkW u mRl ft rZ gksrkgSft l dhr j a xnS /; Z 364.6 nm gSr FkkB l sA es al Ø a e.k dsnkS j ku , d QksVku W mRl ft Z r gksrkgSft l dhr j a xnS /; 121.5 Z nm gS ] r ksC l sA r d la Øe.kdsnkS j ku mRl ft Z r QksVkW u dhr j a xnS /; Zgksxh C

C

364.6 nm

364.6 nm

B

B 121.5 nm

121.5 nm

A

A (A) 91.2 nm (C) 486.1 nm 9.

(C)

3 2

1 3

(A) 91.2 nm (C) 486.1 nm

(B) 243.1 nm (D) none of above

Light incident at angle 60º on a medium. Reflected light is found perfectly plane polarised. Find refractive index of that medium (A)

75V

(A) 177 V, x-v{k l s135º

(A) 177 V at 135º with x-axis (B) 125 V at 45º with x-axis (C) 177 V at 45º with x-axis (D) 125 V at 135º with x-axis 8.

X

40cm 60cm 80cm

(B)

3

9.

i zd k' k , d ek/; e i j 60º dks.k i j vki fr r gksrk gS i j kofr Z r i zd k' ki .wkZ : i l sl er y /kzq for i k; kt kr kgS Abl ek/; e dk vi or Z uka d Kkr dhft , (A)

(D) None of these

(B) 243.1 nm (D) bues al sdksbZugha

(C)

3 2 1


3

(B) (D)

3

buesal sdksbZugha


The visibility of fringes for intensity ratio 3 in YDSE is -

4 5

(A)

;a x f} fNnzi z;ksx esar hozrk vuq i kr 3 dsfy , fÝUt ksadh n`' ; r k gksxh4 5

(A)

3 2

(D)

3

10.

5 4

(B)

2

(C)

11.

Page # 16

2

(C)

A conducting bar is moving along two parallel conducting bars, connected across a resistor R. The velocity of the conducting bar is constant in magnetic field B. The conducting bar oscillates between positions B and C,

11.

5 4

(B)

3 2

(D)

3

, d pky d NM+, d i zfr j ks/kR l st q M+ snksl ekUr j pky d NM+ ksdsvuq fn' kxfr ' khy gS Apky d NM+dkosx pq Ecdh; {ks=kB esafu; r gSpky d NM+cká ; cy ksadsdkj .kfLFkfr A l s kq : gksd j fLFkfr ; kB r FkkC dse/; nksy u dj r h

starting from position A due to some external

gS A l e; (t) dsfo: ) i zsfj r /kkj k (I) dsi fj ek. k dk

forces. Plot the magnitude of induced current

xzkQ gksxk :

(I) against time (t):

  L

L0

R C

L

B

L0 A

I

R C

B

L0 A

B

I

(A)

L0

I

B

I

t

(B)

(A)

t

t

(B)

t

I (C)

I

t

(D)

I t (C)


I t

(D)

t


Page # 17

12.

For two circuits as shown, find th e co rre ct graph between current (i) and time (t), if L1 < L2 : (The steady state current is same for both circuits)

L1

circuit-1 v L2

R2 circuit-2

2 1

(B)

1 (B)

2

t

t

t

2

i

1

i 1

2

(D)

t

(C)

2

1

t

t

Find the reading of voltmeter across the resistance and ammeter in the circuit :

13.

100

V V V 250V 250V 500V, 50Hz A

~

~

(A) 500 V, 5 A (C) 250 V, 10 A

(A) 500 V, 5 A (B) 250 V, 5 A (C) 250 V, 10 A (D) 500 V, 2.5 A An alternating current is given by ( 3 sin t + cos t). The rms current is : (B) 2

(C) 2 2

(D) 4

t

i fj i Fkesai zfr j ks/kdsfl j ksi j oksYVehVj dk i kB; ka d r Fkk vehVj i kB~ ; ka d Kkr dhft , :

V V V 250V 250V 500V, 50Hz A

(A) 2

2

(D)

100

14.

R2 circuit-2

1

2

i 1

L2

v i

(A)

t i (C)

v

2 1

R1 circuit-1

i i

(A)

L1

n' kkZ , x, nksi fj i Fkksa ds fy , /kkj k (i) r Fkk l e; (t) d s e/; l gh xzkQ gksxk L1 < L2 : (LFkk; h voLFkk/kkjknksuksai fj i Fkksa dsfy, l eku gS )

v

i

13.

12.

R1

14.

(B) 250 V, 5 A (D) 500 V, 2.5 A

, d i zR; kor hZ/kkj k ( 3 sin t + cos t) } kj k nhxbZ gS A oxZ ek/; ew y /kkj k gksxh: (A) 2

(B)

(C) 2 2

(D) 4


2


16.

A chain of mass M is placed on a smooth table with 1/n of its length L hanging over the edge. The work done in pulling the hanging portion of the chain back to the surface of the table is (A) MgL/n (B) MgL/2n (C) MgL/n2 (D) MgL/2n2

15.

Three masses 700 gm, 500 gm and 400 gm are suspended at the end of a spring as shown and are in equilibrium. 700 g When the 700 gm mass 500 g is removed, the system 400 g oscillates with a period of 3 seconds; when the 500 gm mass is also removed, it will oscillate with a period of : (A) 1 sec (B) 2 sec

16.

(C) 3 sec 17.

18.

Page # 18

(D)

M nz O ; eku

dh, d psu , d fpduhest i j ml dhy EckbZ

L ds1/n H kkx

l s, d fdukj sl sy Vdhgq bZgSpsu dsy Vds

gq , sHkkx dksoki l est dhl r g i j [ kha pusesafd; k x; k dk; Zgksxk (A) MgL/n

(B) MgL/2n

(C) MgL/n2

(D) MgL/2n2

fp=kkuq l kj , d fLi za x dsfl j ksa i j r hu nzO ; eku 700 xzke, 500 xz ke r Fkk400 xzke yVds

gS avkS j l kE; koLFkk esagS a At c 700

700 g

xzke oky snzO ; eku dks

500 g

gVk; kt kr kgS ]rc ; gl a d k; 3

400 g

l sd . M d snksy u dky l s

nksy u dj r k gS ( t c 500 xzke dksHkhgVk; k t kr k gS ] rc bl dk nksy u dky gksxk:

12 sec 5

A projectile is thrown in the upward direction making an angle of 60º with the horizontal direction with a velocity of 147 ms–1. Then the time after which its inclination with the horizontal is 45º, is (A) 15 s (B) 10.98 s (C) 5.49 s (D) 2.745 s

17.

A projectile is fired vertically upwards from the surface of the earth with a velocity kve, where ve is the escape velocity and k < 1. If R is the radius of the earth, the maximum height to which it will rise measured from the centre of earth will be : (Neglect air resistance)

18.

(A) 1 sec

(B) 2 sec

(C) 3 sec

(D)

12 sec 5

, d i z{ksI; dksÅ/okZ /kj fn' kkesa{kS fr t l s60º dkdks.kcukr s –1 gq ; s147 ms dsosx l sQS a d kt kr kgS a Aog l e; ft l ds ckn ml dk {kS fr t l svkur dks.k45º gS ] gksxk(A) 15 s (C) 5.49 s

(B) 10.98 s (D) 2.745 s

, d i z{ksi kve osx l si `Fohdhl r g l sÅ/okZ /kj Åi j dhvksj nkxkx; kgS , t gk¡ve i y k; u os x gSr Fkkk < 1 gS A; fn R i `Fohdhf=kT; k gS , r ksvf/kdr e Å¡pkbZt gk¡r d ; g i `Foh dsdsUnzl sc<+ r k gq v k i gq ¡p i k; sxk : (ok; qi zfr j ks/k ux.; fyft , )



Page # 19

1 k2 R

(B)

(C) R(1 – k2)

(D)

(A)

19.

R 1 k

1 k2 R

(B)

1 k2

(C) R(1 – k2)

(D)

1 k2

(A)

2

R 1 k2

The figure shows two NAND gates followed

19.

by a NOR gate. The system is equivalent to

R

R

fp=k n' kkZ r k gSfd nksNAND } kj , d NOR } kj r d vuq l fj r gS A fudk; fuEu r kfdZ d } kj dsr q Y; gS–

the following logic gate – A A

X

X B

B

A

20.

Z

Z Y

A

Y

(A) OR

(B) AND

(C) NAND

(D) None of these

Charge ‘q’ is uniformly distributed over the surface of an annula, non-conducting disc of inner radius R1 and outer radius R2. The

20.

(A) OR

(B) AND

(C) NAND

(D)

buesal sdksbZugha

vkUr fj d f=kT; k R1 o cká f=kT; k R2 dh , d oy ; kdkj vpky d pdr h dh l r g i j vkos'k ‘q' , dl eku : i l s

disc is made to rotate about an axis passing

for fj r gS A; g pdr h] vi usdsUnzl sxq t j usoky sv{kr Fkk

through its centre and perpendicular to its

bl dsr y dsy Ecor fu; r vko`fr v l s?kw . kZ u dj usdsfy ,

plane with a constant frequency v (rotations

fufeZ r dhxbZgS(?kw . kZ u i zfr l sd .M) pdr hdk pq Ecdh;

per second). Magnetic moment of the disc can be expressed as :

vk?kw . kZi znf' kZ r gS:

qv( R 22  R 12 ) (A) 2

qv( R 22  R 12 ) (B) 2

(A)

qv( R 22  R 12 ) 2

(B)

qv( R 22  R 12 ) 2

qvR 22 (C) 4

qv( R 22  R 12 ) (D) 4

(C)

qvR 22 4

(D)

qv( R 22  R 12 ) 4



Page # 20

In the diagram shown × × × i = 2A a semicircular wire loop case 'a' × 1m is placed in a uniform × × magnetic field B = 1 T. × × The plane of the loop × is perpendicular to the × × × i = 2A magnetic field. Current B × i = 2A flown in the loop 1m ×A × case 'b' C in the direction shown. D 2A Fin d th e ma gnet ic × × × force in both the cases a and b. Radius of the loop is 1m : (A) 0, 4 N (B) 4 N, 0 (C) 0, 8 N

(D) 8 N, 0

Passage Based Question (Q.22 to 24) We know that a simple harmonic motion has the following properties. (i) It is oscillatory (ii) It is periodic (iii) Acceleration of the particle is directly proportional to displacement from mean position and also always directed towards mean position There are many situations such that a charge may be oscillating. However, the motion may or may not be simple harmonic. Consider a system of two equal point charges, each Q = 8µC, which are fixed at points (2m, 0) and (–2m, 0). Another charge q is held at a point (0,0.1m) on the Yaxis. Mass of the charge q is 91 mg. At t = 0, q is released from rest and it is observed to oscillate along Y-axis in a simple harmonic manner. It is also observed that, at t = 0, force experienced by q is 9 × 10–3 N.

21.

j s[kkfp=kesan' kkZ , , d v) Z o`Ùkkdkj

×

r kj dkyw i , dl eku pq Ecdh; {k= skB = 1 T es afLFkr gS a A yw i

×

dkr y pq Ecdh; {ks=kdsyEcor ~ × gS ayw i esa/kkj k i = 2A n' kkZ ;h

×

fn' kk esai zokfgr gksrhgS A nksuksa ×A

fLFkfr ; ksaa r Fkkb esapq Ecdh; cy Kkr dj ksA yw i dh f=kT; k

×

× ×

i = 2A

× case 'a'

1m

×

×

×

×

i = 2A

×

1m

D 2A

×

× B

C

× case 'b' ×

1m gS: (A) 0, 4 N

(B) 4 N, 0

(C) 0, 8 N

(D) 8 N, 0

x| ka ' k i j v k/kkfj r ç' u (Q.22 l s24) ge t kur sgSfd l j y vkor Zxfr fuEu xq . k j [ kr hgS A (i) og nks y uhgksrhgS (ii) og vkor hZgks r hgS (iii) d.kdkR oj .k ek/; fLFkfr l sfoLFkki u dsl ekuq i kr h gksrkgSr Fkk ges'kkek/; fLFkfr dhvksj funsZ f' kr gksrkgS A , sl hcgq r l hfLFkfr ; kagSft uesavkos'knksy u dj l dr kgS A gk¡y kfd] xfr l j y vkor hZgks; k ughagksl dr hgS A çR; sd Q = 8µC dsnkscj kcj fcU nqvkos'kt ksfcUnq v ksa(2m, 0) o (–2m, 0) i j fLFkj gS ] dsfudk; i j fopkj dhft ; sA vU; vkos'k q dksY-v{k i j fcUnq(0,0.1m) i j j [ kk gS A vkos'kq dkæO ; eku 91 mg gS At = 0 i j q dksfoj ke l s NksM+ kt kr kgSr Fkkog Y-v{kdsvuq fn' kl j y vkor hZ: i l snksy u dj r k çsf{kr fd; k t kr k gS A ; g Hkh çsf{kr fd; k t kr kgSfd t = 0 i j q } kj kvuq Hko cy 9 × 10–3 N gS A fuEu ç' uksadsmÙkj nhft ; sA



Page # 21

22.

23.

24.

Answer the following questions Charge q is (A) – 8 µC (B) – 6.5 µC (C) –5 µC (D) +6.5 µC

22.

Amplitude of motion is (A) 10 cm (B) 20 cm (C) 30 cm (D) 40 cm

23.

Frequency of oscillations is (A) 8 (B) 10 (C) 5 (D) 2

24.

(A) – 8 µC (C) –5 µC

C

(B) 20 cm (D) 40 cm

nksy uksadhvko`fÙk gS(B) 10 (D) 2

x| ka ' k i j v k/kkfj r (Q.25 l s26) xksy h; foi Fku (aberration) xksy h; ni Z . kdk, d , sl knk"sk gS At ksQksd l yEckbZf dhvki r u dks.k‘' i j fuHkjZr kds dkj .kgksrkgS ] uhpsn' kk;Zsfp=kkuq l kj A l EcU/kfuEu çdkj gS f=R–

R sec  2

t gk¡R ni Z . k dh oØr k f=kT; k gSr Fkk  vki r u dks.k gS A fdj .ksat kseq [ ; v{kdsl ehi gSmi k{kh; (paraxial) fdj .ksa dgy kr h gSr Fkk eq [ ; v{k l snw j oky h fdj .ksal heka r (marginal) fdj .ks adgy kr hgS Ami j ksDr fuHkZ j r kdsdkj .k fdj .ksaQksd l i j fofHkUu fcUnq v ksai j y kbZt kr hgSr Fkk, d fcUnqfcEc dkçfr fcEc , d fcUnqughagksrkA

where R is radius of curvature of mirror and  is the angle of incidence. The rays which are closed to principal axis are called paraxial rays and the rays far away from principal axis are called marginal rays. As a result of above dependence rays are brought to focus at different points and the image of a point object is not a point.

Principalaxis

xfr dk vk; ke gS(A) 10 cm (C) 30 cm

R sec  2

 

(B) – 6.5 µC (D) +6.5 µC

(A) 8 (C) 5

Passage Based Question (Q.25 to 26) Spherical aberration in spherical mirrors is a defect which is due to dependence of focal length ‘f ’ on angle of incidence ‘’ as shown in figure is given by f=R–

vkos'k q gS-

Principalaxis

C

Pole(P)

 

Pole(P)

F

F f f

25.

If fp and fm represent the focal length of paraxial and marginal rays respectively, then correct relationship is (A) fp = fm (B) fp > fm (C) fp < fm (D) None

25.

; fn fp o fm Øe' k%mi k{kh; r Fkkl heka r fdj .kksadhQksd l y Eckb; ksadhn' kkZ r sgSr ksbudse/; l ghl EcU/k gS(A) fp = fm (C) fp < fm


(B) fp > fm (D) dks bZugha


Page # 22

26.

The total deviation suffered by the ray falling on mirror at an angle of incidence equal to 60º is (A) 180º (B) 90º (C) Can’t be determined (D) None Passage : (Q.27 to Q.29) A tungsten target (z = 74) is bombarded by electrons in an X-ray tube. The K, L and M atomic X-rays energy levels for tungsten are 69.5, 11.3 and 2.3 keV 27.

The minimum value of the accelerating potential that will permit the production of the characteristic k and klines of tungsten– (A) 69.5 keV (B) 58.2 keV (C) 67.2 keV (D) 11.3 keV

28.

For the same accelerating potential, what is min (A) 16.9 pm (B) 17.9 pm (C) 18.9 pm (D) 19.9 pm

29.

What is k wavelength ? (A) 16.5 pm (B) 17.5 pm (C) 18.5 pm (D) 21.5 pm

30.

Statement-I : As shown, the  B straight conductor when moves through a magnetic V field with a constant velocity (length, magnetic field and velocity are mutually perpendicular) then the rod will experience a force. Statement-II : The electrons will accumulate at the lower and of the rod and an electric field is produced inside the conductor. (A)If both Statement- I and Statement- II are true, and Statement - II is the correct explanation of Statement– I. (B)If both Statement - I and Statement-II are true but Statement - II is not the correct explanation of Statement – I. (C) If Statement-I is true but Statement-II is false. (D) If Statement-I is false but Statement-II is true.

26.

ni Z . k i j , d 60º dks.k i j vki fr r gksusoky hfdj .kksaesa mRiUu dq y fopy u gS(A) 180º

(B) 90º

Kkr ughagksl dr k (D) dksbZugha x| ka ' k:(Q.27 l sQ.29) X-fdj .kufydkes aVa xLVu(z = 74) dsy{; i j bysDVªkW uksadh ceckj hdht kr hgS AVa xLVu dsfy, X-fdj .kksadsK, L o M ds i j ekf.o; Åt kZLr j Øe' k%69.5, 11.3 o 2.3 keV gS A 27. Rofj r foHko dkU;u wr e eku t ksVa xLVu dsvfHky k{kf.kd k o kj s[kkvksadsmRiknu dhvuq efr nsrk gS ] gksxk – (C)

(A) 69.5 keV (C) 67.2 keV 28.

29.

30.

(B) 58.2 keV (D) 11.3 keV

l eku Rofj r foHko dsfy , ,min D; k gS? (A) 16.9 pm (C) 18.9 pm

(B) 17.9 pm (D) 19.9 pm

k r j a xnS /; ZD; k gS? (A) 16.5 pm (C) 18.5 pm

(B) 17.5 pm (D) 21.5 pm

d Fku -I : , d l h/kk pky d , d  B pq Ecdh; {ks=k esa, d fu; r osx ds V l kFk¼ y EckbZ ] pq Ecdh; {ks=kr Fkkosx i j Li j y Ecor gS ½xfr dj r kgksr c NM+, d cy vuq Hko dj sxhA dFku -II : by sDVªkW u NM+dsfupy sfl j si j Rofj r gksrsgS r Fkk fo| q r {ks=k pky d dsvUnj mRiUu gksrk gS A (A) dFku-1 l R ; gS ] dFku-2 l R; gS A dFku-2, dFku-1 dk l ghLi "Vhdj .k gS A (B) dFku-1 l R ; gS ] dFku-2 l R; gS A dFku-2, dFku-1 dk l ghLi "Vhdj .k ughagS A (C) dFku-1 l R ; gS ] dFku-2 vl R; gS A (D) dFku-1 vl R ; gS ] dFku-2 l R; gS A



Page # 23

PART - III [CHEMISTRY] SECTION - A [STRAIGHT OBJECTIVE TYPE] Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct

SECTION - A [ oLr q fu"B i zd kj dsi z'u ] i z-.1 l si z-.30 r d pkj fodYi (A), (B), (C), (D) fn; sgS ft uesa^^dsoy , d^^ l ghgS A l kE; koLFkkfLFkj ka d fuEu esal sfdl l si zHkkfor gksrkgS A (A) l kU nzrkmRikn (B) mR i zsjd (C) fØ; kdkj d l kU nzrkdki nkFkZ (D) r ki eku fuEu esal sdkS u j kl k; fud vf/k' kks"k.kdsfy , ekU; ughagS A (A) vf/k' kks "k.k dhm"ek mPp gksrhgS A (B) ; g mPp r ki i j gks r hgS A (C) ; g mR Øe.kh; gS a (D) ; g , dy vk.kfod i j r ds: i es agksrhgS A

1.

The numerical value of the equilibrium constant is affected by changing the (A) concentration of product (B) catalyst (C) concentration of reacting substance (D) temperature

1.

2.

Which one of the following is not applicable to chemisorption ? (A) Its heat of adsorption is high (B) It takes place at high temperature (C) It is reversible (D) It forms mono-molecular layers

2.

3.

pH of an aqueous solution of NaCl at 85°C should be (A) 7 (B) > 7 (C) < 7 (D) 0

3.

4.

The vapour pressure of the solution of two liquids A (P° = 80 mm) and B(P° = 120 mm) is found to be 100 mm when xA = 0.4. The result shows that (A) Solution exhibits ideal behavior (B) Solution shows positive deviation (C) Solution shows negative deviation (D) None

4.

nksnzoksaA (P° = 80 mm) r Fkk B(P° = 120 mm) ds foy ; u dk ok i nkc 100 mm i k; k t kr k gS Atc xA = 0.4 gS A i fj .kke n' kkZ r sgS afd & (A) foy ; u vkn' kZO ; ogkj n' kkZ r k gS (B) foy ; u /kukR ed fopy u n' kkZ r k gS (C) foy ; u _ .kkR ed fopy u n' kkZ r k gS (D) dks bZugha

5.

The volume of atoms present in a face centred cubic unit cell of a metal (r is atomic radius) is: (A) 20/3 r3 (B) 24/3 r3 3 (C) 12/3 r (D) 16/3 r3

5.

, d /kkr q(i j ek.kqf=kT; k r gS ) dhQyd ds fUær ?kuh; bdkbZ dksf"Bdkesami fLFkr i j ek.kq v ksadkvk; r u gS-

NaCl

dst y h; foy ; u dk 85ºC i j pH D; k gksxk \

(A) 7 (C) < 7

(A) 20/3 r3 (C) 12/3 r3


(B) > 7 (D) 0

(B) 24/3 r3 (D) 16/3 r3


7.

Page # 24

Order of no. of revolution/sec 1, 2 3 and 4 for I, II, III and IV orbits is (A) 1 > 2 > 3 > 4 (B) 4 > 3 > 2 > 1 (C) 1 > 2 > 4 > 3 (D) 2 > 3 > 4 > 1

6.

When ethyne is passed through a red hot tube, then formation of benzene takes place:

7.

9.

H fº (C 2 H 2 )(g ) = 230 kJ mol–1

H fº (C 6 H 6 )( g ) = 85 kJ mol–1

H fº (C 6 H 6 )( g ) = 85 kJ mol–1

, Fkkbu l scS a t hu dsf=ky dhdj .kdhekud Å"ekdhx.kuk dhft , A (A) 205 kJ mol–1 (C) – 605 kJ mol–1 8.

In the balanced chemical reaction,

9.

+

aI–

+

bH+

11.

(B) 605 kJ mol–1 (D) – 205 kJ mol–1

; fn Ca3 (PO4)2 vkS j H3PO3 esa‘P’ i j ek.kqdh l eku la [ ; kmi fLFkr gksr c bu ; kS fxdksaesavkW Dl ht u i j ek.kqdk vuq i kr Øe' k%gS(A) 8/3 (C) 3

(B) 2/3 (D) 4/3

fuEu l a rq fy r j kl k; fud vfHkfØ; kesa , IO3– + aI– + bH+  cH2O + dI2

 cH2O + dI2

a, b, c and d respectively correspond to (A) 5, 6, 3, 3 (B) 5, 3, 6, 3 (C) 3, 5, 3, 6 (D) 5, 6, 5, 5 10.

t c bFkkbu j Dr r Ir uy hl sxq t kj ht kr hgSr kscS a t hu dk fuekZ . k i k; k t kr k gS -

H fº (C 2 H 2 )(g ) = 230 kJ mol–1

If Ca3 (PO4)2 and H3PO3 contain same number of ‘P’ atom then the ratio of oxygen atoms in these compounds respectively is (A) 8/3 (B) 2/3 (C) 3 (D) 4/3

IO3–

dks’k’ds’fy ; s’1, 2 3 o 4 ?kw . kZ u@l S - dh la [ ; k dk Øe gksxk (A) 1 > 2 > 3 > 4 (B) 4 > 3 > 2 > 1 (C) 1 > 2 > 4 > 3 (D) 2 > 3 > 4 > 1

Calculate the standard heat of trimerisation of ethyne to benzene. (A) 205 kJ mol–1 (B) 605 kJ mol–1 (C) – 605 kJ mol–1 (D) – 205 kJ mol–1 8.

I, II, III o IV

a, b, c o d Øe' k%r q Y; (A) 5, 6, 3, 3 (C) 3, 5, 3, 6

In a reaction A2B3(g)  A2(g) + 3/2B2(g), the pressure changes from 60 torr to 75 torr in 2.5 minutes. The rate of disappearance of A2B3 is(A) 8 torr min–1 (B) 18 torr min–1 –1 (C) 4 torr min (D) 10 torr min–1

10.

The following compounds are to be arranged in order of their increasing thermal stabilities. Identify the correct order : K2CO3(I), MgCO3 (II), CaCO3 (III), BeCO3(IV) (A) I < II < III < IV (B) IV < II < III < I (C) IV < II < I < III (D) II < IV < III < I

11.

gksxsa(B) 5, 3, 6, 3 (D) 5, 6, 5, 5

vfHkfØ; kA2B3(g)  A2(g) + 3/2B2(g), esankc 2.5 feuV ea s60 VkW j l s75 VkW j esacny r kgS AA B dsfoy q Ir 2 3 gksusdh nj gS(A) 8 torr min–1 (C) 4 torr min–1

(B) 18 torr min–1 (D) 10 torr min–1

fuEu ; kS fxd dksmudsr ki h; LFkkf; Ro dsc<+ r sØe esa O ; ofLFkr dhft ; s] r ksl ghØe gS: K2CO3(I), MgCO3 (II), CaCO3 (III), BeCO3(IV) (A) I < II < III < IV (B) IV < II < III < I (C) IV < II < I < III


(D) II < IV < III < I


Page # 25

Loa ; vi p; u fof/kdki z;ksx fdl dsfu"d"kZ . kesafd; kt kr kgS

Auto reduction process is used for the extraction of : (A) Hg (B) Pb (C) Cu (D) All of these

12.

Which of the following does/do not exhibit optical isomerism (A) tetrahedral complexes (B) square planar complexes (C) octahedral complexes (D) polynuclear complexes

13.

A colourless solid (A) on heating evolve CO2 and give a white residue (B) soluble in water. (B) also give CO2 when treated with dilute acid. (A) is : (A) Na2CO3 (B) NaHCO3 (C) CaCO3 (D) Ca(HCO3)2

14.

Among the following substituted silanes the one which will give rise to cross linked silicone polymer on hydrolysis is : (A) R3SiCl (B) R4Si (C) RSiCl3 (D) R2SiCl2

15.

The basic strengths of the hybrides group 15 elements decreases in the order : (A) SbH3 > PH3 > AsH3 > NH3 (B) NH3 > SbH3 > PH3 > AsH3 (C) NH3 > PH3 > AsH3 > SbH3 (D) SbH3 > AsH3 > PH3 > NH3

16.

17.

Acidic solution of S2O32– is converted to ....... in presence of I2. (A) S4O62– + I– (B) SO42– + I– (C) SO3 + I– (D) S4O62– + I3–

17.

S2O32– dkvEyh; foy; u I2 dhmi fLFkfr es acny (A) S4O62– + I– (B) SO42– + I– (C) SO3 + I– (D) S4O62– + I3–

18.

Favourable conditions to get good yield of SO3 in contact process is /are : (A) Low temperature, high pressure (B) Low temperature, low pressure (C) High temperature, high pressure (D) High temperature, low pressure

18.

l Ei dZfof/k l sSO3 dksvPNhek=kk esai zkIr dj usdsfy , mi ; q Dr fof/k gS: (A) fuEu r ki ] mPp nkc (B) fuEu r ki ] fuEu nkc (C) mPp r ki ] mPp nkc (D) mPp r ki ] fuEu nkc

12.

13.

14.

15.

16.

(A) Hg (C) Cu

(B) Pb (D)mi j ks Dr

l Hkh

fuEu esal sdkS u i zd kf' kd l eko; or ki znf' kZ r ughadj r sgS A (A) pr q "Qy dh; l a dq y (B) o. kZl er y h; l a dq y (C) (D)

v"VQy dh; l a dq y i ksy hU; w fDy ; j l a dq y

ja xghu Bksl gS A t ksxeZdj usi j CO2 xS l nsrk gSr Fkk vo' ks"k (B) nsrkgS At ksfd t y esa?kq y u' khy gSr Fkk(B) vEy l sfØ; k dj dsCO2 xS l Hkhnsrk gS ] r ks(A) gS:

A

(A) Na2CO3 (C) CaCO3

(B) NaHCO3 (D) Ca(HCO3)2

fdl dst y vi ?kVu l scross linked (Network) t ky la j puk cgq y d cur k gS: (A) R3SiCl (C) RSiCl3

(B) R4Si (D) R2SiCl2

l ew g 15 dsr RoksadsgkbMªkbMds{kkj h; l keF; Z r kdk?kVr k Øe gS: (A) SbH3 > PH3 > AsH3 > NH3 (B) NH3 > SbH3 > PH3 > AsH3 (C) NH3 > PH3 > AsH3 > SbH3 (D) SbH3 > AsH3 > PH3 > NH3


t kr kgS A


20.

Page # 26

Which one of the following has an optical isomer ? (A) [CO(en)3]3+ (B) [Co(H2O)4(en)]3+ (C) [Zn (en)2]2+ (D) [Zn(en)(NH3)2]2+ Consider the following complexes – (i) K2PtCl6 (ii) PtCl4.2NH 3 (iii) PtCl4.3NH 3

19.

fuEu esal sdkS ul k , d i zd k' kh; l eko; or knsrk gS? (A) [CO(en)3]3+ (C) [Zn (en)2]2+

20.

fuEu l a dq y ksai j fopkj (i) K2PtCl 6 (iii) PtCl4.3NH3

(iv) PtCl4.5NH 3

(B) [Co(H2O)4(en)]3+ (D) [Zn(en)(NH3)2]2+ dhft , – (ii) PtCl4.2NH3 (iv) PtCl4.5NH3

Their electrical conductances in aq. solutions are

budh t y h; foy ; u esafo| q r pky dr k fuEugS&

(A) 256, 0, 97, 404 (B) 404, 0, 97, 256

(A) 256, 0, 97, 404 (B) 404, 0, 97, 256

(C) 256, 97, 0, 404 (D) 404, 97, 256, 0

(C) 256, 97, 0, 404 (D) 404, 97, 256, 0

21.

Which of the following compound will show geometrical isomerism ? (A) Cyclohexene (B) 2-Hexene (C) 3-Hexyne (D) 1,1-Diphenyl ethylene

21.

fuEu esal sdkS ul k; kS fxd T; kfer h; l eko; or kn' kkZ r kgS? (A) l kbDy ksgs Dl su (B) 2-gs Dl su (C) 3-gs Dl su (D) 1,1-MkbZ Qsfuy bFkby hu

22.

The absolute configuration of the following :

22.

fuEu dkl ghvfHkfoU; kl gS:

is

is (A) 2S, 3R (C) 2R, 3S 23.

(A) 2S, 3R (C) 2R, 3S

(B) 2S, 3S (D) 2R, 3R

The major product formed in the reaction is

23.

fuEu vfHkfØ; k dk i zeq [ k mRikn gS



NO 2

NO 2

?

?

NO2

NO2

(B)

(A)

(B) 2S, 3S (D) 2R, 3R

(B)

(A) NO2

NO2

NO2

(D)

(C)

NO2

(C)

(D) NO2

NO2



Page # 27

24.

24.

Identify the struucture of the major product 'X' : (A)

(B)

(C)

(D)

eq [ ; mRikn 'X' dhl a j puk cr kb; s: (A)

(B)

(C)

(D)

25.

In order to get propane gas, which of the following should be subjected to sodalime decarboxylation ? (A) Sodium formate (B) Mixture of sodium acetate and sodium ethanote (C) Sodium butyrate (D) Sodium propionate

25.

i zksisu xS l i zkIr dj usdkØe gS AfuEu esal sdkS ul hl ksMk y kbe dsMh&dkcksZ fDl y hdj .kds} kj k? (A) l ks fM; e QkesZ V (B) l ks fM; e , fl VsV vkS j l ksfM; e , sFksuksLV dkfeJ.k (C) l ks fM; e C;w j sV (D) l ks fM; e i zksfi ; ksuks,V

26.

CH3COOH is reacted with HCCH in presence of Hg2+, the product is :

26.

CH3COOH

27.

t c HCCH l sHg2+dhmi fLFkfr esafØ; k dj r k gS ] r ksmRikn gS:

(A)

(A)

(B) CH3CH2(OOC–CH3) (C) CH3CH(OOC–CH3)2 (D) None of these

(B) CH3CH2(OOC–CH3) (C) CH3CH(OOC–CH3)2 (D) bues al sdksbZugha

3-phenyl propene on reaction with HBr gives (as a major product) : (A) C6H5CH2CH(Br)CH3 (B) C6H5CH(Br)CH2CH3 (C) C6H5CH3CH2CH2Br (D) C6H5CH(Br)CH=CH2

27.

3-Qs fuy

i zksihu t c HBr l sfØ; k dj r k gS ] r ksnsrk gS A (eq [ ; mRikn) : (A) C6H5CH2CH(Br)CH3 (B) C6H5CH(Br)CH2CH3 (C) C6H5CH3CH2CH2Br (D) C6H5CH(Br)CH=CH2



Page # 28

In which of the following pairs both members on heating with alc. KOH result in the formation of same alkene as major product ? (A) (CH3)3C—Br, (CH3)2CHCH2Br (B) n-Propyl iodide, isopropyl iodide

(C)

28.

(A) (CH3)3C—Br, (CH3)2CHCH2Br (B) n-i z ksfi y vk; ksMkbM, vkbl ksizksihy vk; ksMkbM

,

(C)

(D) All of the above

29.

fuEu esal sdkS ul k; q Xe , Ydksgyhd KOH dsl kFkxeZdj us i j , d l eku , Ydhu dk fuekZ . k gksrk gS A eq [ ; mRikn ds : i esa?

(D) mi j ks Dr

Which of the following compounds is the most likely to undergo a bimolecular nucleophilic substitution reaction with aqueous NaOH ?

29.

30.

(B)

(B)

(C) (C)

l Hkh

fuEu esal sdkS ul k ; kS fxd vR; f/kd v.kq d ukfHkdLusgh i zfr LFkki u vfHkfØ; knsxkt y h; NaOH dsl kFk?

(A) (A)

,

(D)

(D)

To distinguish between 2-pentanone and 3pentanone which reagent can be used ? (A) NaOH/I2 (B) SeO2 (C) K2Cr2O7/H+ (D) Zn-Hg, HCl

30.

2-i s UVsukW u vkS j 3-i sUVsukW u esafoHksn dj usdsfy , vfHkdeZ d dki z;ksx dj sxsa? (A) NaOH/I2 (B) SeO2 (C) K2Cr2O7/H+ (D) Zn-Hg, HCl


dkS ul s

2016 SAMPLE PAPER

SAMPLE PAPER (2)







2.


3.


4.


Batch

Roll Number

10+1 10+1 10+2 10+3 Crash

0

0

0

0

0

0

0

0

0

1

1 1

1

1

1

1

1

1

2

2 2 3 3

2 3

2 3

2 3

2 2 3 3

2 3

4 4 5 5

4

4

4

4

4

4

3 4

Paper

1

5 6

5 6

5 6

5

6 6

6

5 6

5 5 6 6

Paper 1

7

7 7

7

7

7

7

7

7

8

8 8 9 9

8

8

8

8

9

8 9

8

9

9

9

9

Paper 2

Name

9




L A S T N A ME

D D MMY Y



Q

R S T



0 0 0 0 1 1 1 1 2 2 2 2

0 0 0 0 1 1 1 1 2 2 2 2

3 3 3 3 4 4 4 4

3 3 3 3 4 4 4 4

5 5 5 5

5 5 5 5

6 6 6 6 7 7 7 7

6 6 6 6 7 7 7 7

5 5 5 5 6 6 6 6 7 7 7 7

8 8 8 8 9 9 9 9

8 8 8 8 9 9 9 9

8 8 8 8 9 9 9 9

0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4


Page # 2

PART - I [MATHEMATICS] SECTION - A [ oLr q fu"B i zd kj dsi z'u ]

SECTION - A [STRAIGHT OBJECTIVE TYPE] Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct


1.

1.

If f ( x) 

x  1  x  24  10 x  1 , 1 < x < 26

is a real valued function, then for 1 < x < 26, f (x) is equal to (A) 0

(B)

(C) 2 x  1  5

2.

 

 

(C) 3.

43 – 2x 7

x 1

(D) None of these

Find cot–1 cot  2x 

(A) 3 –  2x 

1

  7

 22  , if 3 < x < 7 7

(A) 0

(B)

(C) 2 x  1  5

(D)

 

cot–1 cot  2x 

 

41 7

(A) 3 –  2x 

(D) 2x –

45 7

(C)

1 1  i  2 2 1 i 1 2  1 equals i i   1 1 (B) 1 (D) 

3.

43 – 2x 7

1 x 1

buesal sdksbZugha

 22  Kkr dj s a ] ; fn 3 < x < gS 7 7

(B) 2x –

If w ( 1) is a cube root of unity, then

(A) 0 (C) i

2.

; fn f ( x)  x  1  x  24  10 x  1 , 1 < x < 26 okLr fod eku Qy u gS , r ks1 < x < 26 dsfy , f (x) dk eku gS

  7

(B) 2x –

41 7

(D) 2x –

45 7

; fn w ( 1) bdkbZdk , d ?kuew y gS ] rc 1 1  i  2 2 1 i 1 2  1 cj kcj gSi i   1 1 (A) 0 (C) i


(B) 1 (D) 


Page # 3

4.

   If a , b and c are non-coplanar unit vectors

4.

  (b  c ) 3      .k gS , r c a , oab dse/; dks a × (b×c ) = 2 gS: (A) /6 (B) 5/6 (C) /3 (D) 2/3

    (b  c ) 3  , then the such that a × ( b × c ) = 2   angle between a and b is : (A) /6 (B) 5/6 (C) /3 (D) 2/3 5.

Let f be a real valued function defined by f(x) =

e x  e | x | , then range of f is : e x  e| x |

(A) R (C) [0, 1) 6.

5.

6.

a1 a n 1 a 2 n 1 (A) 2 (C) 0

a2 a n2 a 2n  2

a3 a n 3 is equal to a 2n 3 (B) 23 (D) None

(B) [0, 1] (D) [0, 1/2)

[sin(sin 1 x )] =  , oa ekuk xlim 1 1 lim [sin 1 (sin x )] =  gS , t gk¡[.] = egÙke i w . kkZ d 2

x  / 2

Qy u, r c-

Then(A) 1 = 1, 2 does not exist (B) 1 = 0, 2 = 1 (C) 1 does not exist, 2 = 1 (D) both 1 and 2 do not exist If 2 a1 , 2 a 2 , 2 a 3 , …, 2 a n are in G.P., then

e x  e | x | } kj k i fj Hkkf"kr , d okLr fod e x  e| x |

(A) R (C) [0, 1)

lim [sin 1 (sin x )] =  , where [.] = G.I.F.. 2 x  / 2

7.

ekukf ; f(x) =

eku Qy u gS ] r c f dk i fj l j gS:

(B) [0, 1] (D) [0, 1/2)

[sin(sin 1 x )] =  and Let xlim 1 1

 ; fn a , b , oac vl er y h; bdkbZl fn' kbl i zd kj gSfd

(A) 1 = 1, 2 fo| eku ughagS (B) 1 = 0, 2 = 1 (C) 1 fo| eku ughagS , 2 = 1 (D) 1 , oa2 nks uksafo| eku ughagS 7.

. kksÙkj Js<+ h esa gS ] ; fn 2 a1 , 2 a 2 , 2 a 3 , …, 2 a n xq a1

r c a n 1 a 2 n 1 (A) 2 (C) 0


a2 a n2 a 2n  2

a3 a n 3 cj kcj gksxk a 2n 3 (B) 23 (D) dks bZugha


9.

Page # 4

f (x) = 1+x (sin x) [cos x], 0 < x  /2 ([.] denotes the greatest integer function) (A) is continuous in (0, /2) (B) is strictly decreasing in (0, /2) (C) is strictly increasing in (0, /2) (D) has global maximum value 2 The function f (x) = 2x (sin x  tan x ) (x 

 x  21  2   41   

10

8.

f (x) = 1+x (sin x) [cos x], 0 < x  /2 ([.]

egÙke i w . kkZ d Qy u dksi znf' kZ r dj r k gS ) (A) (0, /2) esl r r ~gS (B) (0, /2) es afuj Ur j o/kZ eku gS (C) (0, /2) es afuj Ur j âkl eku gS (D) O ; ki d mPPr e eku 2 j [ kr k gS 9.

Qy u f (x) = 2x (sin x  tan x ) gS A(x,  dki w . kkZ a fd;  x  21  2   41   

integarl multiple of ) is ( [.] G.I.F.) : (A) even (B) odd (C) neither even nor odd (D) cannot be determined

xq . ku ughagS ) ( [.]egÙke i w . kkZ a d Qy u) : (A) l e (B) fo"ke (C) u r ksl e u ghfo"ke (D) Kkr ughadj l dr s

 2 0  3 4 3 1  is expressed as the sum If A =   5 7 2 

 2 0  3   ; fn A =  4 3 1  , , d l efer r Fkkfo"ke l efer  5 7 2 

10.

of a symmetric and skew-symmetric matrix, then the symmetric matrix is -

vkO ;w g ds; ksx ds: i esaO ; Dr gS ] r c l efer vkO ;w g gS-

 2 2  4  2 3 4   (A)   4 4 2 

 2 2  4  2 3 4   (A)   4 4 2 

 2 4  5 0 3 7  (B)   3 1 2 

 4 4  8 4 6 8  (C)   8 8 4 

(D)

 4 4  8 4 6 8  (C)   8 8 4 

 2 4  5 0 3 7  (B)   3 1 2 

(D) none of these


buesal sdksbZugha


Page # 5

11.

cot 1 x , | x | 1  Let f (x) =  1 | x |    1 , | x | 1 , then num 4 2 2

11.

dhl a [ ; k ft udksf (x) dk i zkUr xzg.k ughdj r k gS ] gS ,d (B) nks r hu (D) bues al sdksbZugha

ber of points which domain of f (x) does not contain is(A) one (B) two (C) three (D) none of these 12.

13.

14.

(A) (C)

The equation of the normal to the curve x + y = xy, where it cuts x-axis is(A) y = x (B) y = x + 1 (C) y = x – 1 (D) x + y = 1

12.

Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn+1 – Tn = 21, then n equals (A) 5 (B) 7 (C) 6 (D) None

13.

A region in the x-y plane is bounded by the

14.

curve y =

15.

(B) y = x + 1 (D) x + y = 1

ekuk Tn mu f=kHkq t ksadhl a [ ; k dksi znf' kZ r dj r k gSt ksn Hkq t kvksaoky s, d l ecgq Hkq t ds' kh"kksZ a} kj kcuk; st kl dr sgS A ; fn Tn+1 – Tn = 21 gS ] r c n cj kcj gS (A) 5 (C) 6

25  x 2 and the line y = 0. If the

The total number of three-letter words that can be formed from the letters of the word ‘SAHARANPUR’ is equal to (A) 210 (B) 237 (C) 247 (D) 227

oØ x + y = xy dsvfHky Ec dh l ehdj .k, t gk¡ ; g x-v{k dksdkVr k gS ] gS(A) y = x (C) y = x – 1

point (a, a + 1) lies in the interior of the region, then (A) a  (–4, 3) (B) a  (–, –1)  (3, ) (C) a  (–1, 3) (D) None of these 15.

cot 1 x , | x | 1  ekuk f (x) =  1 | x |    1 , | x | 1 gS , r c fcU nq v ksa  4 2 2

(B) 7 (D) dks bZugha

x-y l

eer y esa, d {ks=koØ y = 25  x 2 r Fkkj s[kky = 0 ds} kj k i fj c) gS A ; fn fcUnq(a, a + 1) {ks=k ds vUnj gS ] rc (A) a  (–4, 3)

(B) a  (–, –1)  (3, )

(C) a  (–1, 3)

(D)

buesal sdksbZugha

' kCn 'SAHARANPUR' dsv{kj ksal sfufeZ r fd; st kl dus oky sr hu v{kj h; ' kCnksadhdq y la [ ; k gksxh(A) 210 (C) 247


(B) 237 (D) 227


Page # 6

Let the function f(x) be defined as below,

sin 1   x 2 , f (x)   2x,

16.

0  x 1

sin 1   x 2 , f (x)   2x,

x 1

f(x) can have a minimum at x = 1 then value of  is (A) 1 (B) –1 (C) 0 (D) none of these

17.

If f (x) =

tan [( 2x  3)3 ] ([.] denotes the 1  [ 2x  3]2

f(x) , x = 1 (A) 1 (C) 0

17.

19.

; fn f (x) =

0  x 1 x 1

i j fufEu"B j [ kr k gSr c  dk eku gS(B) –1 (D) bues al

sdksbZugha

tan [( 2x  3)3 ] ([.] egÙke i w . kkZ a d Qy u 1  [ 2x  3]2

dksi znf' kZ r dj r k gS ) r ks

greatest integer function), then (A) f (x) is continuous in R (B) f (x) is continuous in R but not differentiable in R (C) f (x) exists everywhere but f (x) does not exist at some x R (D) None of these 18.

ekuk Qy u f(x) fuEu i zd kj i fj Hkkf"kr gS

esal a r r ~gS a r r ~gSfdUr qR esavody uh; ugha xg fo| eku gSfdUr qf (x) dq N x R i j fo| eku ugh (D) bues al sdksbZugh (A) f (x), R (B) f (x), R es al (C) f(x) l H kh t

A is known to tell the truth in 5 cases out of 6 and he states that a white ball was drawn from a bag containing 8 black and 1 white ball. The probability that the white ball was drawn, is (A) 7/13 (B) 5/13 (C) 9/13 (D) none of these

18.

The curve x + y – ln (x + y) = 2x + 5 has a vertical tangent at the point (). Then   is equal to (A) –1 (B) 1 (C) 2 (D) –2

19.

A dsl

R; cksy usdh6 esal s5 fLFkfr ; k¡gSr Fkkog dgr k gSfd , d FkS y sesa8 dky ho 1 l Qsn xsa n j [ khgSft l esal s , d l Qsn xsa n dksfudky kt kr k gSr ksfudky hxbZxsa n ds l Qsn gksusdhi zkf; dr k gksxh(A) 7/13 (C) 9/13

(B) 5/13 (D) bues al

sdksbZugha \

, d oØ x + y – ln (x + y) = 2x + 5 fcUnq() i j m/oZLi ' kZj s[kk j [ kr k gS A r c  cj kcj gS(A) –1 (C) 2


(B) 1 (D) –2


Page # 7

20.

The points of extremum of the function F(x) =



x

e

1

t2 / 2

2

(1  t ) dt are -

(A) ± 1 (C) ± 1/2

21.

20.



21.

23.

/2

(1  t 2 ) dt dspj e fcU nqgksxsa (B) 0 (D) ± 2

(B) – 3/8 (D) bues al sdksbZugha

2

22.

 x  c x4

x (B) e 

 x2 c x4

(D)  x  4   c  

x (C) e 

2

(A) 3/4 (C) – 3/4

(B) – 3/8 (D) None of these

x2 x   e dx is equal to x4

x (A) e 

e t

1

   1  cos 8  i sin 8   arg  1 – cos   i sin   cj kcj gS 8 8 

2

22.



x

(A) ± 1 (C) ± 1/2

(B) 0 (D) ± 2

   1  cos 8  i sin 8   arg  1 – cos   i sin   is equal to 8 8   (A) 3/4 (C) – 3/4

Qy u F(x) =

x2 c x4

range of ‘a’ so that f (x) has maxima at x = – 2, is (A) |a|  1 (B) |a| < 1 (C) a > 1 (D) a < 1

 x  c x4

x (B) e 

x2 c x4

 x2 c x4

(D) 

x (A) e 

 2xe x 

2 | x 2  5 x  6 | x  2 If f(x) =  , t hen the  a2 1 x  2



x2 x   e dx dk eku gSx4

 2xe x  c   x4

x (C) e 

23.

2 | x 2  5 x  6 | x  2

; fn f(x) =  

a 2 1

x  2

gS ] r c 'a' dk

i fj l j r kfd f (x), x = –2 i j mfPp"B j [ kr k gS ] gS (A) |a|  1 (C) a > 1


(B) |a| < 1 (D) a < 1


24.

Page # 8

The solution of the equation (x + 2y3)

dy –y dx

24.

  [ x  1] 

0 

e

x

 dx (where [.] denotes greatest 

25.

integer function) is equal to (A) 0 (B) 1 (C)  (D) None of these

26.



( 2x12  5x 9 ) dx is equal to ( x 5  x 3  1) 3

x 2  2x c (A) ( x 5  x 3  1) 2

27.

dy – y = 0 dk gy gS dx

(t gk¡A dks bZLosPN fu; r ka d gS ) (A) y(1 – xy) = Ax (B) y3 – x = Ay (C) x (1 – xy) = Ay (D) x (1 + xy) = Ay

= 0 is : (where A is any arbitrary constant) (A) y(1 – xy) = Ax (B) y3 – x = Ay (C) x (1 – xy) = Ay (D) x (1 + xy) = Ay

25.

vody l ehdj .k (x + 2y3)

  [ x  1] 

0 

. kkd ± Qy u dksi znf' kZ r  dx (t gk¡[.] egÙke i w ex 

dj r k gS ) cj kcj gS(A) 0 (C) 

26.

x10 c (B) 2( x 5  x 3  1) 2



(B) 1 (D) bues al

sdksbZugha

( 2x12  5x 9 ) dx cj kcj gS( x 5  x 3  1) 3

(A)

x 2  2x c ( x 5  x 3  1) 2

(B)

x10 c 2( x 5  x 3  1) 2

(C) log | x 5  x 3  1 |  (2 x 7  5x 4 )  c

(C) log | x 5  x 3  1 |  (2 x 7  5x 4 )  c

(D) none of these

(D)

x2 y2  =1 50 20 from which pair of perpendicular tangents are

Number of point on the ellipse

2

2

27.

buesal sdksbZugha

nh?kZ o`Ùk 2

x2 y2  = 1 50 20

i j fcUnq v ksadh l a [ ; k] ft l l s

2

x y  =1ij 16 9

l ekUr j Li ' kZj s[kkvksadsl ew g [ kha psx; s

x y  drawn to the ellipse = 1 is 16 9 (A) 0 (B) 2

gS ] gksxh& (A) 0

(B) 2

(C) 1

(C) 1

(D) 4

(D) 4



Page # 9

28.

Let LL' be the latus rectum through the 2

2

x y  = 1 and A' be a2 b2 the farther vertex. If A'LL is equilateral, then the eccentricity of the hyperboal is (axes are coordinate axes) (A) 3 (B) 3  1

28.

focus of the hyperbola

3 1

(C)

2

(A)

3 1

(C)

2

3

Co-ordinates of N are

30.

N

 52 78  156  , ,   49 49 49 

(C) 

Equation of PQ is

x y z   2 3 6

gS A fcUnq

dsfunZ s'kka d gksa xs-

(A) 

(D) 

3

y Ec PN Mky kt kr kgS ] t gk¡N yEc d k i kn gS A , d l j y j s[ kk PQ l er y 3x + 4y + 5z = 0 dsl ekU r j [ kha pht kr hgSt ksAB dks Q i j fey r h gS Ar c

  52 78 156  , ,   49 49 49 

 52  78 156  , ,   49 49 49 

(C) 

3 1

P(1, 2, 5) l sAB i j

29. (B) 

3 1

(D)

ekuk AB , d l j y j s[ kk

the point P(1, 2, 5) perpendicular PN is drawn to AB, where N is foot of perpendicular. A straight line PQ is drawn parallel to plane 3x + 4y + 5z = 0 to meet AB in Q. Then

 52 78 156  , ,   49 49 49 

dhukfHk l sgksd j

x| ka ' k(Q.29 r Fkk30)

x y z   . From Let AB be the straight line 2 3 6

(A) 

(B)

3

Passage(Q.29 & 30)

29.

x2 y2  =1 a2 b2

xq t j usokykukfHkyEc gSr FkkA' fudVr e ' kh"kZgS ] ; fn f=kHktq A'LL l eckgqgS ] r c vfr i j oy ; (v{kfunsZ ' khv{kgS ) dh mRdsUnzrkgksxh&

3 1

(D)

ekfu; sLL' vfr i j oy ;

30.

PQ

  52 78 156  , ,   49 49 49 

 52 78 156  , ,   49 49 49 

(B) 

 52  78 156  , ,   49 49 49 

(D) 

 52 78  156  , ,   49 49 49 

dk l ehdj .k gS-

(A)

x 1 2  y z  5 x 1 y  2 z  5     (B) 4 13 8 4 13 8

(A)

x 1 2  y z  5 x 1 y  2 z  5     (B) 4 13 8 4 13 8

(C)

x 1 y  2 5  z x 1 y  2 z  5     (D) 4 13 8 4 13 8

(C)

x 1 y  2 5  z x 1 y  2 z  5     (D) 4 13 8 4 13 8



Page # 10


SECTION - A

[ oLr q fu"B i


zd kj dsi z'u ]

out of which ONLY ONE is correct.


1.

A ray of light incident at an angle  on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5º and the prism is made of a material of refractive index 1.5, the angle of incidence is : (A) 7.5º (B) 5º (C) 15º (D) 2.5º

1.

A potentiometer wire of length 1 m has a resistance of 10 ohm. It is connected in series with a resistance R and a cell of emf 3V and negligible internal resistance. A source of emf 10 mV is balanced a length of 60 cm of the potentiometer wire. Find the value of R (A) 10 (B) 1790 (C) 100 (D) 1970

2.

A square of side a is kept at distance d from a long wire as shown. The mutual inductance between them will be :

3.

Q.1 to Q.30 has four choices (A), (B), (C), (D)

2.

3.

µ0a  d  a  (A) 2 n a    (B)

(C)

µ0a d n 2 a µ0a  d  a  n  2  d 

, d i zd k' kfdj .k , d fi zTe dhi j kor Z d l r g i j  dks.k i j vki fr r gksd j nw l j hl r g l svfHky Ec : i l sfuxZ r gks t kr hgS A; fn fi zTe dks.k5º gSr Fkkfi zTe 1.5 vi or Z uka d dsi nkFkZl scuk gks]r ksvki r u dks.k gS: (A) 7.5º (C) 15º

(B) 5º (D) 2.5º

y EckbZdk , d foHkoeki h r kj 10 ohm dki zfr j ks/k j [ kr k gS A bl sR i zfr j ks/k r Fkk 3V fo-ok-cy r Fkkux.; vkUrfj d i zfr j ks/kdsl sy dsl kFkJs.khØe esat ksM+ kt kr kgS A 10 mV fo-ok-cy dkL=kks r foHkoeki hr kj dh60 cm dh y EckbZi j l Ur q fy r gksrk gS A R dk eku Kkr dhft ; s

1m

(A) 10 (C) 100

(B) 1790 (D) 1970

aH kq t kdsox oxZdksn' kkZ ; svuq l kj

, d yEcsr kj l sd nw jh i j j [ kk x; k gS A mudse/; vU; ksU; i zsjdRo gksxk : (A)

µ0a  d  a  n  2  a 

(B)

µ0a d n 2 a

(C)

µ0a  d  a  n  2  d 

(D) dks bZugha

(D) None (SPACE FOR ROUGH WORK)


Page # 11

4.

An inductor of inductance L = 400 mH and resistors of resistances R1 = 2 and R2 = 2 are connected to a batery of emf 12V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is : (A) 6e–5tV (B)

4.

12 –3t e V t (C) 6(1 – 9–t/0.2)V (D) 12 e–5tV

(B)

5.

; fn cká dq . My hB esai zofkgr /kkj ki = 2t } kj knht kr h gS ] r ksvkUr fj d dq . My hA esai zsfj r /kkj kgksxh: (A) okekor Z (B) nf{k.kkor Z (C) nf{k.kkor Z ; k okekor Zgksl dr hgS (D) dks bZ/kkj ki zsfj r ughagksrhgS L-C-R

5.

If current flowing in outer coil B is given as i = 2t, then induced current in inner coil A will be (A) Anticlockwise (B) Clockwise (C) May be clockwise or Anticlockwise (D) No induced current

6.

An L-C-R circuit contains resistance of 100 ohm and a supply of 200 volt and 300 radian angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by 60º. If on the other hand, only inductor is taken out the current leads by 60º with the applied voltage. The current flowing in the circuit is : (A) 1 amp (B) 1.5 amp (C) 2 amp (D) 2.5 amp

6.

For a transistor a change of 8mA in the emitter current results in a change of 7.9 mA in the collector current then the change in base current will be : (A) 1µA (B) 10µA (C) 100µA (D) 1000µA

7.

7.

i szjd r FkkR1 = 2 r Fkk R2 = 2 i z fr j ks/kksadksfp=kkuq l kj 12V fo-ok-cy dhcS Vªhds l kFkt ksM+ kx; kgS AcS VªhdkvkUrfj d i zfr j ks/kux.; gS AfLop S ] t = 0 i j ca n gS A l e; dsQu ds: i esaL dsfl j ksa dk foHko i r u gS: (A) 6e–5tV

12 –3t e V t

(C) 6(1 – 9–t/0.2)V (D) 12 e–5tV

L = 400 mH i z sjdRo ds, d

i fj i Fk esa100 vkse dk i zfr j ks/k vkS j 200 oksYV] 300 j s fM; u dks.kh; vko`fÙkdhfo| q r l IykbZgS A; fn i fj i Fk l sdsoy l a /kkfj =kdksckgj fudky dj ' ks"ki fj i Fkdksi q u% t ksM+ kt kr kgS ] r c /kkj koksYVst l s60º i ' pxkehgksrhgsA ; fn i zsjd dksckgj fudky fn; kt kr kgS ] r c /kkj kvkj skfi r oksYVst l s60º vkxsgksrhgS ] r c /kkj k dk eku gS: (A) 1 amp (C) 2 amp

(B) 1.5 amp (D) 2.5 amp

, d VªkfW t LVj esamRl t Z d /kkj kesa8mA dki fj or Z ul a xzkgd /kkj k esa7.9 mA dk i fj or Z u dj r hgS ] r ksvk/kkj /kkj k esa i fj or Z u gksxk: (A) 1µA (C) 100µA


(B) 10µA (D) 1000µA


Page # 12

For a common emitter npn transistor amplifier (with a = 0.99) the input resistance is 10 and output resistance is 10k then voltage gain will be : (A) 9.9 (B) 99 (C) 990 (D) 9900

8.

mHk; fu"B mRl t Z d npn i zo/kZ d esa(a = 0.99) gS A ; fn fuos'khi zfr j ks/k 10 gS ] r ksoksYVr ky kHkgksxk: (A) 9.9 (C) 990

(B) 99 (D) 9900

Passage for (Q.9 to Q.10) Kirchoffs law suggest that (i) i = 0 at a junction (ii) V = 0 in a closed loop. Using these laws, answer the following questions about the given D-C circuit.

Passage for (Q.9 to Q.10)

9.

The potential at D is : (A) 4V (B) 5V (C) 6V (D) None

9.

The energy stored in the capacitor is : (A) 2 × 10–6 J (B) 1.5 × 10–5 J (C) 2 × 10–5 J (D) 144 µJ

10.

The figure shows the variation of photocurrent with anode potential for a photosensitive surface for three different radiations. Let Ia, Ib and Ic be the intensities and fa, fb and fc be the frequencies for the curves a, b and c respectively :

11.

10.

11.

fdj pkW Q dsfu; e dsvuq l kj (i) l a f/ki = 0 (ii) cUn y w i esaV = 0 gksrk gS A bu fu; eksadsmi ; ksx l sfn; sx; s D-C i fj i Fkdsl ki s {kfuEu i z'uksadsmÙkj nhft ; sA

D i j foH ko gS: (A) 4V (C) 6V

(B) 5V (D) dks bZugha

la /kkfj =kesal a fpr Åt kZgS: (A) 2 × 10–6 J (C) 2 × 10–5 J

(B) 1.5 × 10–5 J (D) 144 µJ

r hu fHkUu-fHkUu fofdj .kksadsfy , , d i zd k' kl a osnhl r g ds fy , i zd k' k/kkj kr Fkk, uksMfoHko dki fj or Z u fp=kesan' kkZ ;k x; k gSekuk fd a, b r Fkk c oØksdsfy , Ia, Ib r FkkIc ‘gks‘r c‘: r hozrk r Fkk fa, fb r Fkkfc vko‘fr Z I

I

C C

B A

B A V

V

(A) fa = fb and Ia  Ib (B) fa = fc and Ia = Ic (C) fa = fb and Ia = Ib (D) fb = fc and Ib = Ic

r FkkIa  Ib (B) fa = fc r FkkIa = Ic (C) fa = fb r FkkIa = Ib (D) fb = fc r FkkIb = Ic

(A) fa = fb



Page # 13

12.

Which of the following curve gives correct graphical representation between refractive index  of glass and wavelength  of light -



O

3 NC–1

(C) 5 NC–1

O

2cm

6 2 × 10 m/s

3 NC–1

(C) 5 NC–1

(D) 20 NC–1

u

3

E=10 V/m

45°

(B) 2 2 × 106 m/s (D)

(B) 4 NC–1

by sDVªkW u dscj kcj vkos'kr Fkk1.6 × 10–30 kg nzO ; eku dk, d d.kfp=kkuq l kj l ekUrj Iy sV l /akkfj =kdhuhpsoky h Iy sV ds{kS fr t l s45° dks.ki j i zkj fEHkd pky u l si z{ksfi r gS AIy sVsai ; kZ Ir y EckbZdhr Fkk 2 cm dhi Fkdr kj [ kr h gS Ad.kdsosx dkvf/kdr e eku ft l l sfd ; g Åi j oky h Iy sV l su Vdj k; s] gS (Iys Vksadse/; fo| q r {ks=k= 103V/m Åi j dhvksj gS ) (g = 0) 2cm

3

45°

(C)

q r {ks=kE gS A; fn x-v{kdsl kFk60° dks.k cukj ghj s[kkdsvuq fn' k2m dhnw j hl sxq t j usesa0.2 C vkos'kdhxfr esafo| q r {ks=k} kj kfd; kx; kdk; Z4J gSr c E dk eku D; k gS?

(D) 20 NC–1 14.



x-fn' kkes afo|

(A)

E=10 V/m

(A) 2 × 106 m/s

13.

O



(B) 4 NC–1

A particle having charge that of an electron and mass 1.6 × 10–30 kg is projected with an initial speed u at an angle 45° to the horizontal from the lower plate of a parallel plate capacitor as shown in figure. The plates are sufficiently long and have separation 2 cm. The maximum value of velocity of particle for it not to hit the upper plate. (Electric field between plates = 10 3 V/m directed upward) (g = 0) -

u

(D)



There is an electric field E in x-direction. If the work done by electric field in moving a charge of 0.2 C through a distance of 2 m along a line making an angle 60° with x-axis is 4J, then what is the value of E ?





(C)

O



O





(D)

O

(B)

O





(C)



(A)

O





14.



(B)

(A)

fuEu esal sdku Sl k oØ dk¡p dsvi or Z ukd a  r Fkki zd k' k ds r j xanS /; Z dse/; l ghxzkÝh; i zn' kZ u nsrkgS-



(A)

13.

12.

1 2

× 106 m/s

(A) 2 × 106 m/s (C)


6 2 × 10 m/s

(B) 2 2 × 106 m/s (D)

1 2

× 106 m/s


Page # 14

Passage (Q. No. 15 & 16) In YDSE, when dichromatic source of light is used we get interference pattern which is obtained by combination of interference pattern due to each wavelength. A black line is obtained on screen where a minima due to each wavelength coincides. Similarly, when polychromatic source is taken say white light source, resulting interference pattern is superimposition of interference pattern from each wavelength as a result at which we get white central bright fringe surrounded by few coloured fringes on both sides of central bright fringe and then uniform illumination of screen. In a typical YDSE set up, distance between slits (d) is 1 mm and screen is placed at a distance (D) of 1 m from slits. The source of light used is dichromatic emitting waves of wavelength 1 = 500 nm and 2 = 700 nm. 15. Fringe width due to 1 is (A) 5 × 10–4 m (B) 4 × 10–5 m –6 (C) 4 × 10 m (D) None

x| ka ' k(Q. No. 15 , oa16) YDSE es at

c i zd k' kdsf} o.khZL=kksr dkmi ; ksx fd; kt kr kgS

ge O ; fr dj .ki zk: i i zkIr dj r sgSt ksi zR;sd r j xanS /; Zdsdkj .k O ; fr dj .k i zk: i dsl ;akst u } kj k i zkIr gksrk gS A i nsZi j , d dkyhj s[kk i zkIr gksrhgSt gk¡ i zR;sd r j a xnS /; Zdsdkj .k , d fufEu"B l Eikr hgS A bl hi zd kj ] t c cgq o.khZL=kksr t lSs osr i zd k' kL=kksr fy; kt kr kgS ] r ksi fj .kkehO ; fr dj .ki zk: i ] i zR;sd r j xanS /; Zl sO ; fr dj .ki zk: i dkv/; kj ksi.kgS ] i fj .kkeLo: i ml i j ge dsUnzh; pedhyhfÝU t dsnksuksavksj i j j a xhu fÝUt ksa } kj kf?kj h osr dsUnzh; pedhy hfÝUt i zkIr dj r sgSr FkkfQj i nsZi j , dl eku i znhi u gksrkgS A, d YDSE O ; oLFkkesa ] fLyVksa dse/; nw j h(d) 1 mm r Fkki nsZdksfLyVksal s1 m dhnw jh (D) i j

j [ kkx; kgS Ai zd k'kdkL=kksr 1 = 500 nm r Fkk2

= 700 nm r j x ans/; Zdkf} o.ku Zrja xsamRl ft 15.

1

Z r dj r kgS A

dsdkj .k fÝUt pkS M+ kbZgS-

(A) 5 × 10–4 m

(B) 4 × 10–5 m

(C) 4 × 10–6 m

(D)

dksbZugha

16.

If dichromatic source is replaced by white light source, then fringe nearest to central bright fringe is (A) Violet (B) Yellow (C) Red (D) None

16.

; fn f} o.kZ d L=kksr dks osr i zd k' kL=kksr } kj ki zfr LFkkfi r dj fn; kt k; sr ksdsUnzh; pedhy hfÝUt dhl ehi LFkfÝUt gS(A) cS a xuh (B) i hy h (C) y ky (D) dks bZugha

17.

A nucleus X, initially at rest, undergoes alphadecay according to the equation. A 228 +  92X  ZY The alpha particle produced in the above process is found to move in a circular track of radius 0.11 m in a uniform magnetic field of 3 Tesla. Find the energy (in MeV) released during the process (A) 2.223 MeV (B) 5.336 MeV, (C) 4.167 MeV (D) None of these

17.

, d ukfHkd X, i zkj fEHkd : i l sfoj ke esagS ] ; g {k; ds v/khu gS ] fuEu l ehdj .kdsvuq l kj A 92X

 ZY228 +  mi j ksDr i zØe esamRiUud.kksadkl e: i pq Ecdh; {ks=k3 Vsl yk esao Ùkh; i Fkft l dhf=kT; k0.11 m gS ] esaxfr dj uki k; kt kr k gS Ai zfØ; kdsnkS j ku eq Dr Åt kZ(MeV es )aKkr dhft ; s (A) 2.223 MeV (B) 5.336 MeV, (C) 4.167 MeV (D) bues al sdksbZugha



Page # 15

18.

The transistor circuit shown here has a current gain of 100. What should be the bias resistance RB so that the voltage across the collector and emitter terminals VCE is 5 volts ? (Neglect voltage across base-emitter VBE)–

; gk¡i znf' kZ r Vªka ft LVj 100 dk/kkj ky kHkj [ kr kgS Ack; l i zfr j ks/kRB D; kgksxkft l l sfd l a xzkgd o mRl t Z d VfeZ uyksa i j oksYVr k VCE, 5 volts gS? (vk/kkj -mRl t Z d ij oksYVr k ux.; y hft , ) –

1k

RB

1k

RB

18.

+ C

10 V

C

+ 10 V –

– VCE

VCE E E

The potential energy of magnet is : (A) U = –

MB 2

MB (B) U =  3 2 (C) U =

20.

19.

Dipole Moment = M S 30°

B

N

(C) U =

(D) None of the above Focal length of the shown plano-convex lens is 15 cm. Plane surface of the lens is silvered. An object is kept on the principal axis of the lens at a distance 20 cm. Image of the object will form (A) 60 cm, left (B) 60 cm, right (C) 12 cm, left (D) 30 cm, right

MB 2

(C) U =  3

MB 2

Object

20cm

(B) 2 k (D) 200 k

pq Ecd dhfLFkfr t Åt kZgS: (A) U = –

////// ///// //// // /

19.

(A) 1 k (C) 100 k

(B) 2 k (D) 200 k

Dipole Moment = M

MB 2

S

30° B

N

MB 2

buesal sdksbZugha fp=k ea si znf' kZ r l er y ksÙky (plano-convex) y S a l dh Qksd l nw j h15 cm gS Ay S a l dsl er y Hkkx i j pk¡nhdk y siu (silvered) dj fn; kt kr k gS Ay S a l l s20 cm dh nw j hi j eq [ ; v{k(principal axis) i j , d oLr qj [ knh t kr hgS A oLr qdk i zfr fcEc cusxk (A) 60 cm, ck; s a Object (C) 60 cm, nk; s a (C) 12 cm, ck; s a 20cm (D) 30 cm, nk; s a (D)

20.


////// ///// //// // /

(A) 1 k (C) 100 k


22.

23.

Page # 16

n resistors each of resistance R are joined with a capacitor of capacity C (each) and a battery of e.m.f. E as shown in the figure. In steady condition, ratio of charges stored in the first and last capacitor is (A) n : 1 (B) (n – 1) : R E R C R C C R (C) (n2+1):(n2–1) (D) 1 : 1

21.

The B-H curve for a B certain specimen is schematically shown H by the given diagram. Wh ich one of t he following is the correct magnetic nature of the specimen ? (A) Diamagnetic and not ferromagnetic paramagnetic (B) Ferromagnetic and not diamagnetic paramagnetic (C) Paramagnetic and not diamagnetic ferromganetic (D) Applicable to all the three types magnetism mentioned above

22.

fdl h, d uew usdsfy , B-H B oØ fn, x, vkj s[k} kj kn' kkZ ;k x; kgS afuEu esal sdkS ul h, d H uew usdhl ghpq Ecdh; i zo`fRr gS\ (A) i z fr pq Ecdh; gSr Fkky kS gpq Ecdh; ; kvuq pE qcdh; ughagS A (B) ykS gpq Ecdh; gSr Fkki zfr pq Ecdh; ; kvuq pq Ecdh; ughgS A (C) vuq pq Ecdh; gSr Fkki zfr pq Ecdh; ; kykg Spq Ecdh; ughagS A (D) mi j ks Dr o.khZ r l Hkhr huksai zd kj dhpq EcdRo i zo`fRr ds fy ; sy kxqgS A

23.

tS l k fd fp=k esafn[ kk; k x; k gS ] P r Fkk Q nksl ek{kh; pky d y w i ga St ksdq N nw j h i j fLFkr ga S A t c fLop S dks pky wdj r sgS ] r ksP esanf{k.kkor hZ/kkj k Ip i zokfgr gksrhgS

or or of

and an induced current IQ1 flows in Q. The switch remains closed for a long time. When S is opened, a current IQ2 flows in Q. Then the directions of IQ1 and IQ2 (as seen by E) are P

/kkfj r kdsl a /kkfj =ko E fo- okcy okyhcS Vj hl sfp=kkuq l kj t ksM+ sx; sgS ALFkkbZvoLFkkesa ] i zFke r Fkkva fr e l /akkfj =kesal a xzfgr vkos'kksadkvuq i kr gS

(A) n : 1 (C) (n – 1) : R E (C) (n2+1):(n2–1) (D) 1 : 1

or

As shown in the figure, P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed, a clockwise current Ip flows in P (as seen by E)

R eku dsi z R; sd n i zfr j ks/kC

C R

A sj [ skusi j ) , oaQ esai zsfj r /kkj k IQ1 i zokfgr gksrhgS fLop cgq r y Ecsl e; r d pky wj gr k gS at c fLop S dks

(E l

] r c E } kj k [ kksy r sga S ] r ksQ esaIQ2 d/kkj ki zokfgr gksrhgS IQ1 r Fkk IQ2 dhi z sf{kr fn' kk; sagS& P

Q

Q E

E S

S Battery

R C R C

Battery



Page # 17

Øe' k%nf{k.kkor Z, oaookekor Z (B) nks uksanf{k.kkor Z (C) nks uksaokekor Z (D) Øe' k%okekor Z, oanf{k.kkor Z

(A) Respectively clockwise and anticlockwise (B) Both Clockwise (C) Both anticlockwise (D) Respectively anticlockwise and clockwise 24.

25.

(A)

Work done in increasing voltage across a capacitor from 5 V to 10 V is W. The work done in increasing the voltage from 10 V to 15 V will be (A) W (B) 4/3W (C) 5/3 W (D) 2W

24.

A current i varies with time in a coil of resistance R as shown in the graph. Find the heat dissipated in the resistor (A) i02 RT

25.

(A) W (C) 5/3 W

(B)

i RT 3

(C)

5 2 i RT 6 0

(D) 26.

i0

i zfr j ks/k dh, d dq . M+ y hesal e; dsl kFk/kkj k i xzkQ esa n' kkZ ; svuq l kj i fj ofr Z r gksrhgS ai zfr j ks/kesaO ; f; r Å"ekKkr dhft ; s&

R

(C)

i20 RT

T

2

l

(D)

t

Two conducting rails are connected to a source of emf and form an incline as shown. A bar of mass m slides without friction down the incline through a vertical magnetic field B. If the length of the bar is l and current i is provided by battery, then the value of B, which will make the bar slide at a constant velocity is –

(B) 4/3W (D) 2W

(A) i02 RT i2 RT (B) 0 3

i

2 0

,d l a /kkfj =kdsfl j ksai j oksYVst 5 V l s10 V r d c<+ kus esafd; k x; k dk; ZW gS A oksYVst dks10 V l s15 V r d c<+ kusesafd; k x; k dk; Zgksxk &

26.

i i0

5 2 i RT 6 0 i20 RT T

2

t

nkspky d i V~ fj ; ksa, d fo-ok-cy L=kksr l st q M+ hgSr Fkkuhps n' kkZ , svuq l kj , d vkur fufeZ r dj r hgS A, d m nzO ; eku dhNM+bl vkur l suhpsdhvksj ?k"kZ . kghu : i l sfQl y j ghgSr Fkkl kFkgh, d Å/oZpq Ecdh; {ks=kB fo| eku gS ANM+ dhy EckbZl gSr FkkcS Vj h} kj ki znku dht kj gh/kkj ki gSr ks B dk og eku D; k gSr kfd NM+, d fu; r os x dsl kFk fQl y l ds– l

B

B v

v







27.

28.

Page # 18

(A)

mg sin  il

(B)

mg cos  il

(A)

mg sin  il

(B)

mg cos  il

(C)

mg tan  il

(D)

mg il cosθ

(C)

mg tan  il

(D)

mg il cosθ

When an electron in the hydrogen atom in ground state absorbs a photon of energy 12.1 eV, its angular momentum : (A) decreases by 2.11 × 10–34 J-s (B) decreases by 1.055 × 10–34 J-s (C) increases by 2.11 × 10–34 J-s (D) increases by 1.055 × 10–34 J-s In the given figure shows a potentiometer circuit (AB is the potentiometer wire) for comparison of two resistances R and X. A student fails to get the balance point.

A B

27.

28.

t c gkbMªkst u i j ek.kqesavk?kvoLFkkesabysDVªkW u 12.1 eV Åt kZdk, d QksVku Wvo' kksf"kr dj r kgSbl dkdks.kh; l a osx : (A) 2.11 × 10–34 J-s } kj k ?kV t k, xk (B) 1.055 × 10–34 J-s } kj k ?kV t k, xk (C) 2.11 × 10–34 J-s } kj k c<+t k, xk (D) 1.055 × 10–34 J-s } kj k c<+t k, xk n' kkZ ; sfp=kesa, d foHkoeki hi fj i Fk(AB foHkoeki hr kj gS ) nksi zfr j ks/kksaR o X dhr q y ukdsfy ; smi ; ksx esafy ; kt kr k gS A, d Nk=kvfo{ksi fcUnqi zkIr dj usesavl Qy gkst kr kgS A A B

R

X

R

X

B

B E

E Consider the following statements . (a ) Po tent ial differe nce acro ss

t he

potentiometer wire is greater than that across R and X if one fails to get the balance point (b ) Po tent ial differe nce acro ss t he potentiometer wire is less than that across R and X if one fails to get the balance point (c) A series resistor is used in the external circuit to get the balance point (d) A shunt is used in the external circuit to get the balance point Which of the statements given above are correct (A) a and c

(B) b and c

(C) a and d

(D) b and d

fuEu dFkuksai j fopkj dhft ; s(a) foH koeki hr kj dsfl j ksai j foHkokUrj R o X dsfl j ksai j foHkokUrj l svf/kd gksxk; fn dksbZl Urq y u fcUnqi zkIr dj us esavl Qy gkst kr k gS (b) foH koeki hr kj dsfl j ksai j foHkokUrj R o X dsfl j ksai j foHkokUrj l sde gksxk; fn dksbZl Urq y u fcUnqi zkIr dj usesa vl Qy gkst kr k gS (c) l U rq y u fcUnqi zkIr dj usdsfy ; scká i fj i Fk esa, d Js.khi zfr j ks/kdkmi ; ksx fd; kt kr kgS (d) l U rq y u fcUnqi zkIr dj usdsfy ; scká i fj i Fk esa, d "ka V dk mi ; ksx fd; k t kr k gS fuEu esal sdkS ul sdFku l ghgS? (A) a r Fkk c (B) b r Fkkc (C) a r Fkkd (D) b r Fkkd



Page # 19

29.

A galvanometer of 50 ohm resistance has

29.

25 divisions. A current of 4 × 10–4 ampere

xsYosuksehVj 25 [ kkusj [ kr kgS A

4 × 10–4 , s fEi; j

range of 25 volts, it should be connected with

dh/kkj k, d [ kkusdkfo{ksi nsrhgS Abl xsYosuksehVj dks25 oksYV dh i j kl oky soksYVehVj esa cny usdsfy ; si zfr j ks/kt ksM+ ukgksxk–

a resistance of –

(A) 2500 

‘ka V‘ds‘: Ik‘esa

(A) 2500  as a shunt

(B) 2450 

‘ka V‘ds‘: Ik‘esa

gives a deflection of one division. To convert this galvanometer into a voltmeter having a

(B) 2450  as a shunt

Js.khØe esa (D) 2450  Js . khØe esa (C) 2550 

(C) 2550  in series (D) 2450  in series 30.

50 ohm i z fr j ks/kdk, d

In figure, AB is a non-conducting rod, Equal charges of magnitude q are fixed at various points on the rod as shown. The rod is rotated uniformly about an axis passing through O and perpendicular to its length such that linear speed of the end A or B is the rod is 3 m/s. Magnetic induction at O is -

q A

q

q

q q

O 1m 1m 2m 3m

30.

uhpsn' kkZ , sfp=kesa ] AB , d vpky d NM+gS ] l eku vkos'kksa ] i zr; sd dki fj ek.kq gSNM+dsfHkUu fcUnq v ksai j uhpsn' kkZ ;s vuq l kj fQDl fd; sx; sgS a AO l sxq t j usoky hr Fkkbl dh yEckbZdsyEcor ~, d v{kdsi fj r %NM+bl i zd kj ?kw . kZ u dj j ghgSfd NM+dsfl j sA ; k B dhj s[kh; pky 3 m/s gS A O i j pq Ecdh; {ks=k gS-

q

q

B

A

2m 3m

q

q

q q

O 1m 1m 2m 3m

q B

2m 3m

(A)

11 0 q 12

(B)

3 0 q 7

(A)

11 0 q 12

(B)

3 0 q 7

(C)

0q 2

(D)

6 0 q 13

(C)

0q 2

(D)

6 0 q 13



Page # 20

PART - II [CHEMISTRY] 1.

The reaction : 3O2 2O3,H=+ 69,000 calories favourable conditions to produce O3 are :

1.

mPp r ki eku r Fkk fuEu nkc (B) mPp r ki eku r Fkk mPp nkc (A)

(A) high temperature and low pressure (B) high temperature and high pressure

fuEu r ki eku r Fkkvf/kd nkc (D) fuEu r ki eku r FkkfuEu nkc (C)

(C) low temperature and high pressure (D) low temperature and low pressure 2.

3.

Given below are a few electrolytes, indicates which one among them will bring about the coagulation of a gold sol. quickest and in the least of concentration ? (A) NaCl (B) MgSO4 (C) Al2(SO4)3 (D) K4[Fe(CN)6]

2.

The pH of an aqueous solution of 1.0 M solution of a weak monoprotic acid which is 1% ionised is:

3.

(A) 1 (C) 3 4.

5.

j kl k; fud vfHkfØ; k 3O2 2O3, H= +69,000 dS y ksjhO3 dsfuekZ . kdhl gk; d fLFkfr ; k¡gS%

dq N vi ?kV~ ; fn, x, gS ] fuEu esl sdkS u l ksusdsfoy ; u dkU; w uRe l kUnzrkesaLdUnu dj r kgS a A (A) NaCl (C) Al2(SO4)3

1 M , dy i z ksVh; nq cZ y vEy dk t y h; 1% vk; uhd`r gq v kgS , dhpH gks xh\ (A) 1 (C) 3

(B) 2 (D) 11

Osmotic pressure of 30% solution of glucose is 1.20 atm and that of 3.42 % solution of cane sugar is 2.5 atm. The osmotic pressure of the mixture containing equal volumes of the two solutions will be – (A) 2.5 atm (B) 3.7 atm (C) 1.85 atm (D) 1.3 atm

4.

In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be: (A) X4Y3 (B) X2Y3 (C) X2Y (D) X3 Y4

5.

(B) MgSO4 (D) K4[Fe(CN)6]

foy ; u t ksfd

(B) 2 (D) 11

30% Xy w d kst dsfoy ; u dk i j kl j .k nkc 1.20 atm r Fkk3.42 % ‘' kdZ j k‘ds‘foy ; u‘dk‘i j kl j .k‘nkc‘2.5 atm gS AfeJ.kft l esanksuksafoy ; uksadsl eku vk; r u gS ]

dki j kl j .knkc gksxk& (A) 2.5 atm (C) 1.85 atm

(B) 3.7 atm (D) 1.3 atm

; kS fxd esar Ro Y dsi j ek.kqccp t ky d cukr sgSr Fkkr Ro X ds2/3rd pr q "Qy dh; fNædksHkj r sgS; kS fxd dkl w =k gksxk (A) X4Y3 (C) X2Y


(B) X2Y3 (D) X3 Y4


Page # 21

6.

If the radius of first Bohr orbit of hydrogen atom is ‘x’ then de Broglie wavelength of electron in 3rd orbit is nearly (A) 2x (B) 6x (C) 9x

7.

8.

9.

6.

x (D) 3

The enthalpy changes for the following processes are listed below : Cl2(g) 2Cl(g) ; H = 242.3 kJ mol–1 I2(g) 2I(g) ; H = 151.0 kJ mol–1 ICl(g)I(g) + Cl(g) ; H = 211.3 kJ mol–1 I2(s) I2(g) ; H = 62.76 kJ mol–1 Given that, the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICl (g) is : (A) –14.6 kJ mol–1 (B) –16.8 kJ mol–1 (C) +16.8 kJ mol–1 (D) +244.8 kJ mol–1

7.

Calculate the number of oxygen atoms required to combine with 7g of N2 to form N2O3 when 80 % of N2 is converted into N2O3 : (A) 2.3 × 1023 (B) 3.6 × 1023 (C) 1.8 × 1021 (D) 5.4 × 1021

8.

Which of the following groups of molecules act both as oxidising agent as well as reducing agent ? (A) KMnO4, O3, SO3 (B) HClO4, HNO2, H2O2 (C) HNO2, SO2, H2O2 (D) HNO3, SO2, H2SO4

9.

; fn H-i j ek.kqdhi zFke cksgj dks'kdhf=kT; k‘x’ gSr ksr hl j s dks'kesaby sDVªkW u dhMh&czksXy hr j a x nS /; ZfudV gksxh(A) 2x

(B) 6x

(C) 9x

(D)

x 3

fuEu i zØeksadsfy , , UFkS Yi hi fj or u Z uhpsl w phc) fd, x, gS a A Cl2(g) 2Cl(g) ; I2(g) 2I(g) ; ICl(g)I(g) + Cl(g) ; I2(s) I2(g) ;

H = 242.3 kJ mol–1 H = 151.0 kJ mol–1 H = 211.3 kJ mol–1 H = 62.76 kJ mol–1

fn; kx; kgSfd] vk; ksMhu o Dy ksjhu dsfy, ekud voLFkk, sa I2(s) o Cl2(g) gS a , ICl (g) dsfy , fuekZ . k dh ekud , UFkS Yi hgS (A) –14.6 kJ mol–1 (B) –16.8 kJ mol–1 (C) +16.8 kJ mol–1 (D) +244.8 kJ mol–1

N2O3 dsfuekZ . kesaN2 ds7g dsl

kFkt q Mu +sokysvkD W l ht u i j ek.kqdhl a [ ; kdhx.kukdj kst c N2 dk80 % N2O3 esai fj ofr Z r gksrk gS: (A) 2.3 × 1023

(B) 3.6 × 1023

(C) 1.8 × 1021

(D) 5.4 × 1021

fuEu esal sv.kq v ksadkdkS ul kl ew g vkD W l hdkjh, oavi p; udkjh nksuksadsl eku O ; ogkj dj r kgS ? (A) KMnO4, O3, SO3 (B) HClO4, HNO2, H2O2 (C) HNO2, SO2, H2O2 (D) HNO3, SO2, H2SO4



11.

12.

13.

Page # 22

The specific conductance of N/50 solution of HCl in a cell at 25ºC is 0.002765 mho cm–1. If the resistance of a cell containing this solution is 400 ohm, the cell constant is: (A) 1.106 cm (B) 1.106 cm–1 (C) 1 cm (D) 1 cm–1

10.

Match the column Column I I. Highest ionization Potential II. Lowest electronegativity III. Highest electron gain enthalpy IV. Second most electronegative element I II III (A) (iii) (ii) (iv) (B) (iii) (ii) (i) (C) (ii) (iii) (iv) (D) (iv) (ii) (i)

11.

i j ,d l S y esaHCl dsN/50 foy ; u dhfof' k"V pky dr k 0.002765 mho cm–1 gS A ; fn bl foy ; u ;q Dr , d l S y dh i zfr j ks/kdr k 400 ohm gS , r ksl S y fLFkj ka d gS 25ºC

(B) 1.106 cm–1 (D) 1 cm–1

(A) 1.106 cm (C) 1 cm

Column II (i) O (ii)Cs (iii) He (iv) Cl

Lr EHkdkfey ku dhft ; s dkW ye I I. mPp vk; uu foH ko II. fuEu fo| q r _ .kkRedr k III. mPp by s DVªkW u cU/kq rk IV. f} r h; fo| q r _ .kh; r Ro (A) (B) (C) (D)

IV (i) (iv) (i) (iii)

Auto reduction process is used for the extraction of : (A) Hg (B) Pb (C) Cu (D) All of these

12.

When H2O2 is added to an acidfied solution of K2Cr2O7 : (A) Solution turns green due to formation of Cr2O3 (B) Solution turns yellow due to formation of K2CrO4 (C) A deep blue-violet coloured compound CrO(O2)2 is formed (D) Solution gives gree ppt. of Cr(OH)3

13.

I (iii) (iii) (ii) (iv)

II (ii) (ii) (iii) (ii)

III (iv) (i) (iv) (i)

dkW y e II (i) O (ii)Cs (iii) He (iv) Cl IV (i) (iv) (i) (iii)

Loa ; vi p; u fof/kdki z;ksx fdl dsfu"d"kZ . kdsfy , fd; k t kr kgS: (A) Hg (C) Cu K2Cr2O7 dsvEyh;

gS ] r ks:

(B) Pb (D) mi j ks Dr

l Hkh

foy ; u esat c H2O2 dksfeyk; kt kr k

(A) Cr2O3dscukusdsdkj .kfoy ; u gj kgkst kr kgS A (B) K2CrO4 dscuusdsdkj .kfoy ; u fn; ki hykgkst kr k

gS A

(C) , d

xgj kuhy k&i j i y j a x i zkIr gksrkgS ACrO(O2)2 dscuusdsdkj .k (D) Cr(OH)3 foy ; u dkgj kvo{ks i i zkIr gksrkgS A



Page # 23

dk mi ; ksx gokbZt gkt i uMq Cch esafd; k t kr k gS ] D; ksa fd : (A) CO2 dksxz g.kdj r kgS ] O2 dhl kUnzrkdksc<+ kr kgS (B) uehdksxz g.kdj r kgS A (C) CO2 dksxz g.kdj r kgS A (D) vks t ksu mRiUu dj r kgS a A

14.

KO2 is used in space craft and submarines because is : (A) absorbs CO2 and increases O2 concentration (B) absorbs moisture (C) absorbs CO2 (D) produces ozone.

14.

KO2

15.

For the following oxoacids of phosphorus : H3PO2, H3PO3 and H3PO4 which of the following statements is true ? (A) All of these are tribasic acids (B) All of these are reducign in nature (C) All of theese have tetrahedral geometry of phosphorus (D) The order of acidity is H3PO4 > H3PO3 > H3PO2

15.

QkLQksjl dsfuEu vkW Dl hvEy H3PO2, H3PO3 r FkkH3PO4 dsfy , fuEu esal sdkS ul kdFku l R; gS? (A) mi j ks Dr l Hkhf=k{kkj h; vEy gS A (B) mi j ks Dr l Hkhvi pk; d i zd `fr dsgS A (C) mi j ks Dr l HkhesaQkLQksjl dhl a j pukpr q "Qy dh; gS A (D) vEy h; r kdkØe gSH3PO4 > H3PO3 > H3PO2

16.

The anhydride of ortho phosphoric acid is : (A) P2O3 (B) P2O5 (C) P3O5 (D) P4O8

16.

fut Z y h; vkFkksZQkLQksfj d vEy :

17.

Excess of PCl 5 reacts with conc. H 2 SO 4 giving : (A) Thionyl chloride (B) Sulphur monochloride (C) Sulphuryl chloride (D) Sulphur tetrachloride

17.

PCl5 dkvkf/kD; l kU nzH2SO4 (A) Fkk; ks fuy Dy ksjkbM (B) l YQj eks uksDy ksjkbM (C) l Y¶; w fj d Dy ksjkbM (D) l YQj Vs VªkDy ksjkbM

18.

Consider the following reaction

18.

fuEu vfHkfØ; kdsvk/kkj i j Cold Br2 + 2NaOH (l kU nz)  i kn   mR mi j ksDr vfHkfØ; kl smRiUu gS A

(A) P2O3 (C) P3O5

Cold Br2 + 2NaOH (dill.)    Products The products of the above reactions are (A) NaBrO + NaBrO3 + H2O (B) NaBrO + NaBr + H2O (C) NaBr + H2O (D) NaBrO +H2O

19.

Which of the following pairs represents linkage isomers ? (A) [Co(NH3)5NO3]SO4 and [Co(NH3)5SO4]NO3 (B) [PtCl2(NH3)4]Br2 and [Pt Br2(NH3)4]Cl2 (C) [Cu(NH3)4] [PtCl4] and [Pt(NH3)4] [CuCl4] (D) [Pd(PPh3)2(NCS)2] and [Pd(PPh3)2(SCN)2]

(B) P2O5 (D) P4O8

dsl kFknsrkgS:

(A) NaBrO + NaBrO3 + H2O (B) NaBrO + NaBr + H2O (C) NaBr + H2O (D) NaBrO +H2O 19.

fuEu esal sdkS ul k; q Xe fy a d st l eko; or kn' kkZ r kgS? (A) [Co(NH3)5NO3]SO4 r Fkk[Co(NH3)5SO4]NO3 (B) [PtCl2(NH3)4]Br2 r Fkk[Pt Br2(NH3)4]Cl2 (C) [Cu(NH3)4] [PtCl4] r Fkk[Pt(NH3)4] [CuCl4] (D) [Pd(PPh3)2(NCS)2] r Fkk[Pd(PPh3)2(SCN)2]



21.

22.

Page # 24

Which of the following has an optical isomer? (A) [Co(H2O)4 (en)]3+ (B) [Co(en)2(NH3)2]3+ (C) [Co(N3)3Cl]+ (D) [Co(en) (NH3)2]2+

20.

Match the List I with List II and select the correct answer from the codes given below the lists : List I (Reaction) (A) CH3CH=CHCHO  CH3CH=CHCOOH (B) CH3CH=CHCHO  CH3CH=CHCH2OH (C) R–COH  R —CH2OH (D) CH3CH2COCl  CH3CH2CHO List II (Reagent) (1) LiAlH4 (2) NaBH4, LiAlH4 (3) Ammoniacal AgNO3 (4) H2 ; Pd/BaSO4 Codes : A B C D (A) 4 3 2 1 (B) 1 2 3 4 (C) 3 1 2 4 (D) 2 3 1 4

21.

CH3COCl is used to detect the number of amino groups in the organic compounds. When a compound of molecular mass 180 is acetylated, gives a compound of molecular mass 390. The number of amino groups in the compound is : (A) 1 (B) 2 (C) 5 (D) 4

22.

fuEu esal sdkS ul ki zd kf' kd l eko; or kn' kkZ r kgS? (A) [Co(H2O)4 (en)]3+ (B) [Co(en)2(NH3)2]3+ (C) [Co(N3)3Cl]+ (D) [Co(en) (NH3)2]2+

Lr EHkI r FkkLr EHk II esal ghdw V dk fey ku dhft ; s: Lr EHkI (vfHkfØ; k) (A) CH3CH=CHCHO  CH3CH=CHCOOH (B) CH3CH=CHCHO  CH3CH=CHCH2OH (C) R–COH  R —CH2OH (D) CH3CH2COCl  CH3CH2CHO Lr EHkII (vfHkdeZ d) (1) LiAlH4 (2) NaBH4, LiAlH4 (3) veks fudy AgNO3 (4) H2 ; Pd/BaSO4 dw V: A B C D (A) 4 3 2 1 (B) 1 2 3 4 (C) 3 1 2 4 (D) 2 3 1 4

dkmi ; ksx dkcZ fud ; kS fxdksaesaNH2 l ew g dk i r kyxkusdsfy, fd; kt kr kgS A180 v.kqHkkj okys; kfSxd dk, l hVhy hdj .kdj usi j i zkIr ; kS fxd dkv.kq Hkkj 390 gks t kr kgS ] r ks; kS fxd esavehuksal ew g dhl a [ ; kgksxh: CH3COCl

(A) 1 (C) 5


(B) 2 (D) 4


Page # 25

23.

Which of the following amines will show carbylamine reaction ?

(A)

(C)

23.

fuEu esal sdkS ul k, ehu dkcksZ y , ehu vfHkfØ; knsrkgS?

(A)

(B)

(C)

(D)

(B)

(D)

24.

24.

which is the corret alternate ? (A) (i) CH3NH2,

dkS ul hfof/kmi ; q Dr gS? (A) (i) CH3NH2,

(ii)

(ii)

R1Schmidt reaction ;R2 HVZ reaction (B) (i) CH3CH2CONH2 (ii) CH3CH2COBr R1HVZ reaction ; R2 Schmidt reaction (C) (i) CH3CH2NH2, (ii) CH3CH2COBr R1HVZ reaction ; R2 Schmidt reaction (D) None is correct

R1f' eV vfH kfØ; k;R2 HVZ vfHkfØ; k (B) (i) CH3CH2CONH2 (ii) CH3CH2COBr R1HVZ vfH kfØ; k; R2 f' eV vfHkfØ; k (C) (i) CH3CH2NH2, (ii) CH3CH2COBr R1HVZ vfH kfØ; k; R2 f' eV vfHkfØ; k (D) bues al sdksbZugha



26.

Page # 26

What is the major product from the reaction?

(A)

(B)

(C)

(D)

What is the product of the reaction sequence below ?

(A)

(C)

(B)

(D)

25.

26.

vfHkfØ; kl seq [ ; mRikn i zkIr gksxk?

(A)

(B)

(C)

(D)

fuEu vfHkfØ; kØe esamRikn i zkIr gksxk?

(A)

(C)


(B)

(D)


Page # 27

27.

Choose the species that would be perdicted to be not produced in the following aldol reaction.

(A)

(C)

28.

27.

fuEu esal sdku Sl kmRikn , YMksy vfHkfØ; kl si zkIr ughagksxk

(A)

(B)

(C)

(D)

(B)

(D)

Samponification neutralization of O18 labelled ethyl acetate yields which of the following isotopically labelled products

28.

l kcq uhmnkl huhdj .kesaO18 bFkkbZ y , l hVsV esai zkIr gksrk gS ] r ksfuEu esal sO18 mRikn esai zkIr gksxk

(A) (A) (B)

(B)

(C) (D) (a)

(C) (D) approx equal amount of (a) and (c)


r Fkk (c) dscj kcj


What is the structure of the most stable enol form of the compound ?

(A)

(C)

30.

Page # 28

29.

fuEu ; kS fxd dhl okZ f/kd LFkk; hbZ ukW yla j pukgS?

(A)

(B)

(C)

(D)

(B)

(D)

Which is the Claisen condensation product of ethyl propanoate ?

30.

, fFky i zksfi vksusV dkDy st u l a ?kuu mRikn gksxk?

(A) (A)

(B) (B)

(C) (C)

(D) (D)


2016 SAMPLE PAPER

SAMPLE PAPER (3)







2.


3.


4.


Batch

Roll Number

10+1 10+1 10+2 10+3 Crash

0

0

0

0

0

0

0

0

0

1

1 1

1

1

1

1

1

1

2

2 2 3 3

2 3

2 3

2 3

2 2 3 3

2 3

4 4 5 5

4

4

4

4

4

4

3 4

Paper

1

5 6

5 6

5 6

5

6 6

6

5 6

5 5 6 6

Paper 1

7

7 7

7

7

7

7

7

7

8

8 8 9 9

8

8

8

8

9

8 9

8

9

9

9

9

Paper 2

Name

9




L A S T N A ME

D D MMY Y



Q

R S T



0 0 0 0 1 1 1 1 2 2 2 2

0 0 0 0 1 1 1 1 2 2 2 2

3 3 3 3 4 4 4 4

3 3 3 3 4 4 4 4

5 5 5 5

5 5 5 5

6 6 6 6 7 7 7 7

6 6 6 6 7 7 7 7

5 5 5 5 6 6 6 6 7 7 7 7

8 8 8 8 9 9 9 9

8 8 8 8 9 9 9 9

8 8 8 8 9 9 9 9

0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4

AITS _ 12th _ JEE MAIN

Page # 2

PART - I [MATHEMATICS] SECTION - A [ oLr q fu"B i zd kj dsi z'u ]

SECTION - A [STRAIGHT OBJECTIVE TYPE] Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct

1.

2.

   x  + cosx cos 3 

f (x) = sin 2x + sin 2 

i z-.1 l si z-.30 r d pkj fodYi (A), (B), (C), (D) fn; sgS ft uesa^^dsoy , d^^ l ghgS A 1.

    x  , g(x) = 2 then f (2013) – g (2013) = ? 3 

    x  , g(x) = 2 r c f (2013) – g (2013) = ? 3 

(A) 3/4

(B) –3/4

(A) 3/4

(B) –3/4

(C) 1/4

(D) None

(C) 1/4

(D) dks bZugha

Suppose that g(x) = 1 +

x & f(g(x)) = 3 +

2.

ekukfd g(x) = 1 + x o f(g(x)) = 3 + 2 x + x r ksf(x) gS -

2 x + x, then f(x) is

3.

   x  + cosx cos 3 

f (x) = sin 2x + sin 2 

(A) 1 + 2x2

(B) 2 + x2

(C) 1 + x

(D) 2 + x

The number of solutions of the equations af(x) + g(x) = 0, where a > 0, g(x)  0 and has minimum value 1/2, is (A) infinitely many

(B) only one

(C) two

(D) zero

3.

(A) 1 + 2x2

(B) 2 + x2

(C) 1 + x

(D) 2 + x

l ehdj .kaf(x) + g(x) = 0, t gk¡a > 0, g(x)  0 r Fkk U; w ur e eku 1/2 j [ kr hgS , dsgy ks adhl a [ ; k gksxh(A)

vuUr

(C) nks

(B)

dsoy , d

(D) ' kw U;


Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Education is the most powerful weapon which you can use to change the world


Page # 3

4.

f(x)= nlim 



then

x n  x n (x > 1) x n  xn

f ( x)n ( x  1  x 2 )

(B) x n |x +

x2 +c 2

f(x)= nlim 

rc

dx is -

1  x2

(A) n |x +

(C)

4.

x n  x n (x > 1) x n  xn

f ( x)n ( x  1  x 2 )



1  x2

(A) n |x +

1 x2 | + c

(B) x n |x +

1 x2 | + c (D) None

(C)

x2 +c 2

gksxk -

dx

1 x2 | + c 1 x2 | + c (D) dks bZugha

; fn A o B nks?kVuk; sabl i zd kj gSfd P(A  B)  3/4 r Fkk1/8  P(A  B)  3/8, r c -

If A and B are two events such that P(A  B)  3/4 and 1/8  P(A  B)  3/8, then (A) P(A) + P(B)  11/8 (B) P(A) . P(B)  3/8 (C) P(A) + P(B)  7/8 (D) none of these

5.

6.

The area bounded by the curve y = x + sin x and its inverse function between the ordinates x = 0 and x = 2, is (A) 8 sq unit (B) 4 sq unit (C) 8 sq unit (D) None of these

6.

dksfV; ksx = 0 r Fkkx = 2dse/; oØy = x + sin x r Fkk bl dsi zfr ykse Qyu } kj ki fj c) {ks=kdk{ks=kQy gS (A) 8 oxZbdkbZ (B) 4 oxZbdkbZ (C) 8 oxZbdkbZ (D) bues al sdksbZugha

7.

Let g(x) be a function defined on [–1, 1]. If the area of the equilateral triangle with two of its vertices at (0, 0) and (x, g(x)) is equal

7.

ekukg(x) , d Qy u gSt ksvUr j ky [–1, 1] esai fj Hkkf"kr gS A; fn nks' kh"kks±(0, 0) o (x, g(x)) oky sl eckgqf=kHkq t

5.

to

3 /4, then the function g(x) is -

(A) ± 1  x 2

(B)

(C) –

(D) None of these

1 x2

1 x2

(A) P(A) + P(B)  11/8 (B) P(A) . P(B)  3/8 (C) P(A) + P(B)  7/8 (D) bues al sdksbZugha

dk {kS =kQy 3 /4, gS ] r c Qy u g(x) gS(A) ± 1  x 2

(B)

(C) –

(D)

1 x2

1 x2

buesal sdksbZugha



cM+k lkspks vkSj gkjus ls er MjksA

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Page # 4

; fn l Hkh x  R dsfy ; sf (x) = x3 + x2 f (1) + x f(2) + f(3) gS ] r c f(x) =

If f (x) = x3 + x2 f (1) + x f(2) + f(3) for all x  R, then f(x) = (A) x3 + 2x2 – 5x + 6 (B) x3 + 2x2 + 6x – 5 (C) x3 – 5x2 + 6x + 2 (D) x3 – 5x2 + 2x + 6

8.

9.

If x  [0, 2] and g(x) = f(x) + f(2 – x). Also, f”(x) < 0, then g(x) (A) increases in [0, 2] (B) decreases in [0, 2] (C) decreases in [0, 1) and increases in (1, 2] (D) increases in [0, 1) and decreases in (1, 2]

9.

; fn x  [0, 2] r Fkk g(x) = f(x) + f(2 – x) gS A f"(x) < 0 H khgS ] r c g(x) (A) [0, 2] es ao/kZ eku gS (B) [0, 2] es aÐkl eku gS (C) [0, 1) es aÐkl eku r Fkk (1, 2] esao/kZ eku gS (D) [0, 1) es ao/kZ eku r Fkk (1, 2] esaÐkl eku gS

10.

If

10.

; fn

8.

dx

 x 2 ( x n  1) (n1) / n = – [f(x)]

1/n

is(A) (1 + xn) (C) xn + x–n 11.

n1/ 

n

1/n

(A) (1 + xn) (C) xn + x–n 11.

f ( x 2 / 4)[f ( x )  f ( x )] dx is g( x 2 / 4)[g( x )  g( x )]

(A) depend on  (C) zero

dx

 x 2 ( x n  1) (n1) / n = – [f(x)]

+c

gS , r ksf(x)

gS (B) 1 + x–n (D) None of these

If f(x) and g(x) are continuous functions, the



12.

+ c, then f(x)

(A) x3 + 2x2 – 5x + 6 (B) x3 + 2x2 + 6x – 5 (C) x3 – 5x2 + 6x + 2 (D) x3 – 5x2 + 2x + 6

; fn f(x) r Fkkg(x) l r r ~Qy u gS ] r ks



n1 / 

n

4 x  x 3  n (a 2  3a  3), 0  x  3 f(x) =  x  18, x 3  complete set of values of a such that f(x) as a local minima at x = 3 is (A) [–1, 2] (B) (–, 1)  (2, ) (C) [1, 2] (D) (–, –1)  (2, )

f ( x 2 / 4)[f ( x )  f ( x )] dx gS g( x 2 / 4)[g( x )  g( x )]

(A) i j (C) ' kw U;

(B) a non-zero constant (D) None of these

(B) 1 + x–n (D) bues al sdksbZugha

fuHkZ j

(B) , d v' kw U; vpj (D) bues al sdksbZugha

4 x  x 3  n (a 2  3a  3), 0  x  3

12.

f(x) = 



x  18,

x 3

dsekuksadk l Ei w . kZl eq Pp; r kfd f(x), x = 3 i j LFkkuh; fufEu"B gks

a

(A) [–1, 2] (C) [1, 2]

(B) (–, 1)  (2, ) (D) (–, –1)  (2, )


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Page # 5

13.

 

(A) e (C) 1 14.



n 

1/ n



is

14.

(3x  4y  7)2 is 4

dk eku gS

(B) e2 (D) e3

nh?kZ o`Ùk(5x – 10)2 + (5y + 15)2 = (A) 10 (C) 20/7

(B) 20/3 (D) 4

A natural number x is chosen at random from the first one hundred natural numbers. The

1/ n

15.

(B) 20/3 (D) 4

i zFke l kSi zkÑr l a [ ; kvksaesal s, d i zkÑr l a [ ; kx dk; kn`PN; k ( x  20)( x  40) < 0 gks us dh x  30

p; u fd; k t kr k gS ] r ks i zkf; dr k gksxh-

(A)

1 50

(B)

9 50

(A)

1 50

(B)

9 50

(C)

3 25

(D)

7 25

(C)

3 25

(D)

7 25

If the equation of a line and a plane be

x3 y4 z5   and 4x – 2y – z = 1 re2 3 2 spectively, then (A) the line is parallel to the plane (B) the line is perpendicular to the plane (C) the line lies in the plane (D) none of these

(3x  4y  7)2 4

dsnh?kZv{k dhy EckbZgS &

( x  20)( x  40) probability that < 0 is x  30

16.

 2 n   lim tan tan ... tan  n   2n 2n 2n   (A) e (C) 1

The length of the major axis of the ellipse

(A) 10 (C) 20/7

13.

(B) e2 (D) e3

(5x – 10)2 + (5y + 15)2 =

15.

2

tan ... tan  The value of lim tan n  2n 2n 2n

16.

; fn , d j s[kk r Fkk , d l er y dh l ehdj .k Øe' k% x3 y4 z5   r Fkk4x – 2y – z = 1 gS ] rc 2 3 2

j s[kk l er y dsl ekUr j gS j s[kk dsl er y dsy Ecor gS (C) j s [ kk l er y esafLFkr gS (D) bues al sdksbZugha (A)

(B)



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iwjs fo'okl ds lkFk vius liuks dh rjQ cM+k]s ogh ftanxh ft;ks ftldh dYiuk vkius dh gSA


17.

18.

19.

Page # 6

Passage (Q.17 to 18) : The number of ways in which n identical items can be divided into r groups so that no group contains less than m items and more than k (m < k), is Coe ffic ient of x n in th e expans ion of (xm + xm+1 +…+ xk)r The number of ways of distributing 5 identical balls into three boxes so that no box is empty (each box being large enough to accommodate all balls), is (A) 35 (B) 53 (C) 15 (D) None In how many ways can 20 identical toys be distributed among 4 children so that each one gets at least 3 toys ? (A) 11C2 (B) 12C3 11 (C) C3 (D) None   Statement–I : If u and v are unit vectors

inclined at an angle  and x is a unit vector bisecting the angle between them, then 



x| ka ' k(i z. 17 l s18) : mu r j hdksadhl a [ ; kft ul sn , d l eku oLr q v ksadksr l ew gksa esafoHkkft r fd; k t k l dr k gSr kfd fdl hHkhl ew g esam oLr vqksal sde r Fkkk (m < k) l svf/kd u gks](xm + xm+1 +…+ xk)r dsi z l kj esaxn dsxq . kka d dscj kcj gS A 17.

(A) 35 (C) 15 18.

19.



(A)

(B) (C) (D)

(B) 53 (D) dks bZugha

20 , d l eku f[ ky kS uksadks4 cPpksadse/; fdr usi zd kj l s for fj r fd; kt kl dr kgSr kfd i zR; sd cPpkde l sde 3 f[ ky kS usi zkIr dj sa? (A) (C)

x = ( u + v )/(2 sin(/2)). Statement-II : If ABC is an isosceles triangle with AB = AC = 1, then the vector representing the bisector of angle A is given by AD = ( AB + AC )/2. Both statement-I and statement-II are true but statement-II is not the correct explanation of statement-I. Both statement-I and statement- II are true, and statement-II is correct explanation of Statement-I. Statement-I is true but statement-II is false. Statement-I is false but statement-II is true

mu r j hdksadhl a [ ; kft ul s5 , d l eku xsa nksadksr hu cDl ksa esaj [ kkt kr kgSt cfd dksbZHkhcDl k[ kky hughaj gs(i zR; sd cDl kl Hkhxsa nksadksj [ kusft r ukcM+ kgS ), gks xh

11 11

(B) 12C3 (D) dks bZugha

C2 C3

dFku–I : ; fn u r Fkk v bdkbZl fn`k dks.ki j >q d sgq , gSr Fkk x bdkbZl fn`kgSt ksmudse/; dks.kdksl ef} Hkkft r dj r k gS ] rc 





x = ( u + v )/(2 sin(/2)). dFku-II : ; fn ABC , d l ef} ckgqf=kHktq gSt gk¡AB = AC = 1, r c dks . kA dsv/kZ d dksfu: fi r dj usoky kl fn`k (A) (B) (C) (D)

A AD = ( AB + AC )/2 ds} kj kfn; kt kr kgS dFku-I o dFku-II nksuksal R; gS ai j a r qdFku-II] dFku-I dk l ghLi "Vhdj .kughagS dFku-I o dFku-II nksuksal R; gS ar FkkdFku-II dFku-I dk l ghLi "Vhdj .k gS dFku-I l R; gSi j Ur qdFku-II vl R; gS dFku-I vl R; gSi j Ur qdFku-II l R; gS


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Page # 7

20.

21.

22.

If the system of linear equations x + 4ay + az = 0, x + 3by + bz = 0, x + 2cy + cz = 0 have a non-trivial solution, then a, b, c are in (A) H.P. (B) G.P. (C) A.P. (D) none of these

20.

Vertices of a variable triangle are (3, 4), (5 cos , 5 sin ) and (5 sin , –5 cos ), where   R. Locus of its orthocentre is (A) (x + y – 1)2 + (x – y – 7)2 = 100 (B) (x + y – 7)2 + (x – y – 1)2 = 100 (C) (x + y – 7)2 + (x + y – 1)2 = 100 (D) (x + y – 7)2 + (x – y + 1)2 = 100

21.

Equation of a circle which passes thought the point (2, 0) and whose centre is the limit of the point of intersection of the lines 3x + 5y = 1 and (2 + ) x + 52y = 1 as tends to 1 is (A) 25(x2 + y2) + 20x – 2y – 140 = 0 (B) 25(x2 + y2) – 20x + 2y – 60 = 0 (C) 9(x2 + y2) – 20x – 2y + 4 = 0

22.

; fn j S f[ kd l ehdj .kfudk; x + 4ay + az = 0, x + 3by + bz = 0, x + 2cy + cz = 0 , d vr q PN gy j [ kr h gS ] r c a, b, c gS (A) gj kR ed Js.khesa (B) xq . kksÙkj Js.khesa (C)

l ekUr j Js.khesa

(D)

buesal sdksbZugha

, d pj f=kHkq t ds' kh"kZ(3, 4), (5 cos , 5 sin ) r Fkk (5 sin , –5 cos ) gS ] t gk¡  R r c bl dsy EcdsUnz dk fcUnq i Fk gS (A) (x + y – 1)2 + (x – y – 7)2 = 100 (B) (x + y – 7)2 + (x – y – 1)2 = 100 (C) (x + y – 7)2 + (x + y – 1)2 = 100 (D) (x + y – 7)2 + (x – y + 1)2 = 100

ml o`Ùkdkl ehdj .kt ksfcUnq(2, 0) l sgksd j xq t j r kgSr Fkk ft l dkdsUnzj s[kkvksa3x + 5y = 1, oa(2 + ) x + 52y = 1 dsi z fr PNsn fcUnqdh 1 i j l hek gS , gks xk(A) 25(x2 + y2) + 20x – 2y – 140 = 0 (B) 25(x2 + y2) – 20x + 2y – 60 = 0 (C) 9(x2 + y2) – 20x – 2y + 4 = 0 (D) 9(x2 + y2) – 2x – 20y + 4 = 0

(D) 9(x2 + y2) – 2x – 20y + 4 = 0 23.

The equation of the circle touching the lines

y = x at a distance 2

2

2 units from the origin is

(A) x + y – 4x + 2 = 0 (B) x2 + y2 + 4x – 2 = 0 (C) x2 + y2 + 4x + 2 = 0 (D) None of these

23.

j s[kkvksa y = x dksew y fcUnql s 2 bdkbZnw j h i j Li ' kZ dj usoky so`Ùkdk l ehdj .k gksxk& (A) x2 + y2 – 4x + 2 = 0 (B) x2 + y2 + 4x – 2 = 0 (C) x2 + y2 + 4x + 2 = 0 (D) bues al sdksbZugha



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eqf'dys oks phtsa gksrh gS] tks gesa rc fn[krh gS tc gekjk /;ku y{; ij ugha gksrkA

AITS _ 12th _ JEE MAIN 24.

25.

Page # 8

ABCD is a square with side AB = 2. A point P moves such that its distance from A equals its distance from the line BD. The locus of P meets the line AC at T1 and the line through A parallel to BD at T2 and T3. The area of T1T2T3 is (A) 1/2 (B) 2/3 (C) 1 (D) 2

24.

2

a2

(A) 1/2 (C) 1



y

25.

2

b2

is the same and the is equal to

a  2b . 2

























( c × a ) × ( a × b )] ; (where a , b and c are non zero non-coplanar vector) is equal to   

(A) [ a b c ]2   

(C) [ a b c ]4

  

(B) [ a b c ]3   

(D) [ a b c ]

y2 b2

 1 i j nksfcU nqgS aft udhbl dsdsUnzl s a 2  2b 2 dscj kcj gS ] r ksnh?kZ o`Ùk dh 2

mRdsUnzrk gksxh(B) 1 / 2 (D) 1 / 3 2

(C) 1/3

[( a × b ) × ( b × c ) ( b × c ) × ( c × a ) 



(A) 1/2

The projection of the line 3x – y + 2z – 1 = 0 = x + 2y – z – 2 on the plane 3x + 2y + z = 0, is(A) 3x + 2y + z = 0 = 3x – 8y + 7z + 4 (B) 3x – 2y – z = 0 = 3x – 8y + 7z + 4 (C) 3x – 2y + z = 0 = 3x – 8y – 7z + 4 (D) None of these



a2

(B) 2/3 (D) 2

nw j h l eku r Fkk

2

Then the eccentricity of the ellipse is (A) 1/2 (B) 1 / 2 (C) 1/3 (D) 1 / 3 2



x2

nh?kZ o`Ùk

 1 whose distance from its centre 2

27.

oxZgSft l dhHktq kAB = 2 gS A, d fcUnqP bl i zd kj xfr eku gSfd A l sbl dhnw j hj s[kkBD l sbl dhnw jh dscj kcj gS AP dkfcUnq i Fkj s[kkAC dksT1 i j feyr kgSr Fkk BD dsl ekraj A l sgks d j t kusokyh j s[kk T2 , oaT3 i j feyr h gS , r c  T1T2T3 dk {ks =kQy gksxk -

There are exactly two points on the ellipse

x

26.

ABCD , d

26.

l er y 3x + 2y + z = 0 i j j s[kk 3x – y + 2z – 1 = 0 = x + 2y – z – 2 dki z {ksi gksxk(A) 3x + 2y + z = 0 = 3x – 8y + 7z + 4 (B) 3x – 2y – z = 0 = 3x – 8y + 7z + 4 (C) 3x – 2y + z = 0 = 3x – 8y – 7z + 4 (D) bues al sdksbZugha

27.

















[( a × b )× ( b × c )( b × c ) 







( c × a ) × ( a × b )] ; (t gk¡a , b vl er yh; l fn`kgS ), dkeku gks xk   

(A) [ a b c ]2   

(C) [ a b c ]4



×



o c v' kU w;

  

(B) [ a b c ]3   

(D) [ a b c ]


Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.



(c ×a )


Page # 9

28.

29.

If n1, n2, n3, ....... n100 are positive real number such that n1 + n2 + n3 + ............+ n100 = 20 and k = n1(n2 + n3 + n4)(n5 + n6 + ..... + n9) (n10 + ...... + n16) ..... (.... + n100) then k belongs to (A) (0, 100] (B) (0, 128] (C) [0, 144] (D) None of these

28.

Solution of the differential equation -

29.

(A) (0, 100] (C) [0, 144]

y(xy + 2x2y2) dx + x (xy – x2y2)dy = 0 is

1 (C) 2 log | x | + log | y | + xy = C 1 (D) 2 log | y | + log | x | + xy = C

x(1  y 2 ) , then the equation of 1 x2 the curve will be given that curve passes through the point (3,1) (A) 5(1+y2) = 1 + x2 (B) 1 + y2 = 1 + x2 (C) 1 + y2 = 5 (1 + x2) (D) none

dk gy

1 (A) 2 log | x | – log | y | – xy = C 1 (B) 2 log | y | – log | x | – xy = C

1 (B) 2 log | y | – log | x | – xy = C

at P. If NQ =

vody u l ehdj .k fn; k t k; sxk -

1 (A) 2 log | x | – log | y | – xy = C

A normal at P(x, y) on a curve meets the x-axis at Q & N is the foot of the ordinate

(B) (0, 128] (D) bues aesal sdksbZugha

y(xy + 2x2y2) dx + x (xy – x2y2)dy = 0

given by

30.

; fn /kukRed okLr fod l a [ ; k, ¡ n1, n2, n3, ....... n100 bl i zd kj gS afd n1 + n2 + n3 + ............+ n100 = 20 r Fkkk = n1(n2 + n3 + n4)(n5 + n6 + ..... + n9) (n10 + ...... + n16) ..... (.... + n100) gS ] r ksk fuEu l sl a ca f/kr gksxk-

1 (C) 2 log | x | + log | y | + xy = C 1 (D) 2 log | y | + log | x | + xy = C 30.

oØ dsfcUnqP(x, y) i j [ kha pkx; kvfHky Ec x-v{kdksQ i j feyr kgSr Fkk N fcU nqP dhdksfV dki kn gS; fn NQ = x(1  y 2 ) gS ] r ksoØ dkl ehdj .kD; kgks xkt cfd fn; kgSfd 1 x2 oØ fcUnq(3,1) l sgksd j xq t j r k gS (A) 5(1+y2) = 1 + x2 (B) 1 + y2 = 1 + x2 (C) 1 + y2 = 5 (1 + x2) (D) dks bZugha



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Page # 10

PART - II [PHYSICS] 1.

The energy levels of a certain atom for 1st, 2nd and 3rd levels are E, 4E/3 and 2E respectively. A photon of wavelength  is emitted for a transition 3  1. What will be the wavelength of emission for transition 2  1 (A) /3 (B) 4/3 (C) 3/4 (D) 3

2.

A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towards it. When its distance from the mirror is 20 cm its velocity is 4 cm/s. The velocity of the image in cm/s at that instant is (A) 6 towards the mirror (B) 6 away from the mirror (C) 9 away from the mirror (D) 9 towards the mirror

3.

A small source of light is 4 cm below the surface of a liquid of refractive index 5/3. In order to cut off all the light coming out of liquid surface, minimum diameter of the disc placed on the surface of liquid is (A) 3 m (B) 4 m (C) 6m (D) 

4.

A particle is moving with a real velocity V m/s along the straight line shown. An observer at the end of the same line is viewing the a particle. Which of the following graphs most appropriately represent the velocity of the particle as seen by the observer ? All the medium boundaries are in the state of rest -

1.

fdl hi j ek.kqds1st, 2nd r Fkk3rd Åt kZLr j ksadhÅt k,Z¡ Øe' k%E, 4E/3 r Fkk2E gS a A, d QksVkW u ft l dhr j a xnS /; Z gSl Ø a e.k3  1 gsrqmRl ft Z r gksrkgS Al a Øe.k2  1 gsrq mRl ft Z r rja xnS /; Zgksxh(A) /3 (C) 3/4

2.

, d i zfr nhIr fcUnqoLr q ] 12 cm Qksd l yEckbZds, d vor y ni .Zkdseq [ ; v{kl svuq fn' kbl dhvksj xfr ' khy gS At c ni Z .k l sbl dhnw j h20 cm gksr c bl dkosx 4 cm/s gksxkAbl {k.ki j i zfr fcEc dk osx cm/s esagksxk(A) ni Z . k dhvksj 6 cm/s (B) ni Z . k l snw j 6 cm/s (C) ni Z . k l snw j 9 cm/s (D) ni Z . k dhvksj 9 cm/s

3.

i zd k' k dk , d y ?kqL=kksr] 5/3vi or Z uka d ds, d nzo dh l r g l suhpsgS Anzo l r g l sckgj vkr si w j si zd k' kdksj ksd us dsfy , fdl U; w ur e O ; kl dh pdr h nzo dh l r g i j j [ kuh gksxh(A) 3 m (C) 6m

4.

Air

2

(B) 4 m (D) 

, d d.k] V m/s dsokLr fod osx dsl kFkn' kkZ bZxbZl h/kh j s[kkdsvuq fn' kxfr ' khy gS Al eku j s[kkdsfl j si j fLFkr , d i zs{kd d.kdksns[kj gkgS AfuEu esal sdku Sl kxzkQ i zs{kd } kj k ns[ksx, d.k dsosx dksl okZ f/kd mfpr r j hdsl si znf' kZ r dj r kgS A l Hkhek/; e l hek, safoj ke dhvoLFkk esagSAir



(B) 4/3 (D) 3



2

3 4

3 4


Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Optimism is the faith that leads to achievement. Nothing can be done without hope and confidence.

(D) Time

5.

100 (C) ohm 1.6  0.39

(B)

l e;

(D)

vkHkkl hosx

vkHkkl h osx

l e;

(C)

l e;

l e;

Time

Calculate the resistance of a semiconductor rod of length 10 cm and cross-section area 1 mm2, if it is doped with a total of 1015 donor atoms at room temperature. Given that the electron mobility = 0.39 m2/volt-sec and electron charge = 1.6 × 10–19 coulomb :

1 (A) ohm 1.60  0.39

6.

Time

Apparent Velocity

(C)

Apparent Velocity

Time

(A)

vkHkkl hosx

(B)

Apparent Velocity

Apparent Velocity

(A)

vkHkkl h osx


Page # 11

5.

1 (B) ohm 1.6  0.39  100 (D) 1.6×0.39×100 ohm

A p-n-p transistor is used as an amplifier in the common-base configuration. If the ratio of cha nge in c urre nt in e mitt er a nd corresponding change in current in collector is 1.25, then the value of current gain is : (A) 0.8 (B) 0.9 (C) 1.0 (D) 1.25

6.

10 cm yEckbZr Fkk1 mm2 {ks =kQy

dsvuq i zLFkdkV okyh, d v) Z pkyd NM+dki zfr j ks/kKkr dj ks]; fn ; g dej sdsr ki eku i j dq y 1015 nkr ki j ek.kql svi fefJr dht k; sAfn; kx; k gS: bysDVªkW u dh xfr ' khyr k = 0.39 m2/volt-sec o bysDVªkW u vkos'k = 1.6 × 10–19 dw y kW e: (A)

1 ohm 1.60  0.39

(B)

1 ohm 1.6  0.39  100

(C)

100 ohm 1.6  0.39

(D) 1.6×0.39×100 ohm

, d p-n-p Vªka ft LVj mHk; fu"B vk/kkj foU;kl esai zo/kd Z ds : i esai z;q Dr fd; kt kr kgS A; fn mRl t Z d esa/kkj kesai fj or Z u r Fkkbl dsl a xr l a xzkgd esa/kkj kesai fj or Z u dkvuq i kr 1.25 gS ] r c /kkj k ykHkdk eku gS: (A) 0.8 (C) 1.0

(B) 0.9 (D) 1.25



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In the circuit shown in figure all the diodes are ideal. The current drawn from the battery of 1.5 volts emf and 1internal resistance is :

1.5 A 10  (A)   1  3  1.5 A 10  (B)   1  2 

8.

(C)

1.5 A 11

(D)

1.5 A 10

(D) none of these

7.

1.5 A 10 (B)   1  2 

D2

10 10 1.5V – + 1

fp=k esan' kkZ , i fj i Fk esal HkhMk; ksMvkn' kZgS A 1.5 oksYV fo- ok- cy r Fkk1vkUr fj d i zfr j ks/kdhcS Vj hl si zokfgr /kkj k gS: 1.5 A (A)  10  1  3 

D1

10

D1

10

D2

10 10

D3 1.5 A (C) 11 (D)

In figure, a coil of single turn is wound on a sphere of radius r and mass m. The plane of the coil is parallel to the inclined plane and lies in the equatorial plane of the sphere. If the sphere is in rotational equilibrium, the value of B is :

mg (A) ir mg sin  (B) i mg sin  (C) ir

9.

Page # 12

8.

1.5V – + 1

D3

1.5 A 10

fp=k esa ] , dy Qsjsdh, d dq . My hr f=kT; ko m nzO ; eku dsxksy si j y i sVh xbZgS A dq . My h dk r y vkur r y ds l ekUrj gSr Fkkxksy sdsl er y esaj [ kkgS A; fn xksy k?k.wkZ u l kE; koLFkk esagS ] r ksB dkeku gS : mg ir mg sin  (B) i mg sin  (C) ir (A)

O B

 mg

A campound microscope has a magnification power of 100 when the image is formed at infinity. the objective has a focal length of 0.5 cm and the tube length is 6.5 cm. Find the focal length of eyepiece. (A) 1 cm (B) 2 cm (C) 20 cm (D) 40 cm

(D) 9.

buesal sdksbZugha

O B

 mg

,d l a ;q Dr l w {en' khZ100 vko/kZ u {ker k j [ kr k gSt c i zfr fcEc vuUr i j cur k gS A vfHkn`' ; d dh Qksd l nw jh 0.5 cm gS auS f=kdk dhQksd l nw j hKkr dhft ; s& (A) 1 cm (C) 20 cm

(B) 2 cm (D) 40 cm


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dk;Z gh lQyrk dh cqfu;kn gSA


Page # 13

Three charges +4q, Q and q are placed in a straight line of length  at points distances 0, /2 and  respectively. What should be Q in order to make the net force on q to be zero ? (A) –q (B) 4q (C) –

(A) –q (C) –

(D) –2q

A point source has been placed as shown in the

11.

q 2

(D) –2q

, d fcUnqL=kksr dksuhpsfp=kesan' kkZ ; svuq l kj fLFkr fd; k x; k gS A LØhu i j og y EckbZD; k gksxht gk¡ r d ni Z . kl s

will receive reflected light from the mirror ?

i j kofr Z r çdk' kçkIr fd; kt k; sxk?

(A) 2 H

(A) 2 H

(C) H

SOURCE H

(D) None H

(B) 3H

H

SOURCE

(C) H (D) dks bZugha

2H

Surface charge density on a ring of radius a

12.

H H

H

2H

, d oy ; dhf=kT; ka r FkkpkS M+ kbZd gS ] bl dki `"B vkos'k

and width d is . It rotates with frequency 

?kuRo gS A; g vi uhv{kdsi fj r %vko`fÙkdsl kFk?k.wkZ u

about its own axis. Assume that the charge

dj r hgS Aekukfd vkos'kdsoy cká l r g i j gS AdsUnzi j

is only on outer surface. The magnetic field

pq Ecdh; {ks=k gS-

induction at centre is -

13.

(B) 4q

figure. What is the length on the screen that

(B) 3H

12.

r hu vkos'kksa+4q, Q o q dks y EckbZdhl j y j s[kk i j Øe' k%0, /2 o  nw fj ; ksai j j [ kk x; k gS Aq i j usV cy ' kw U; gksusdsfy ; sQ D; k gksxk?

SCREEN

11.

q 2

10.

SCREEN

10.

(A) 0d

(B) 0d

(C) 20d

 d (D) 2 0

In single slit diffraction the linear position of

13.

(A) 0d

(B) 0d

(C) 20d

(D)

 d 2 0

, dy fNnzfoor Z u esadsUnzl sf} r h; fufEu"V dh j s[kh;

2nd minima from centre is y, D is screen distance,

fLFkfr y gS , D i ns Zdhnw j h]  r j a xnS /; Zr Fkkd fNnzpkS M+ kbZ

is wavelength, d is slit width than y is -

gks] r c y gksxk-

(A)

d D

(B)

2d D

(A)

d D

(B)

2d D

(C)

2D d

(D)

2Dd 

(C)

2D d

(D)

2Dd 



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Page # 14

A metal rod of mass m and length l is rotating

14.

with a constant angular velocity about an

dsl

axis passing through O and normal to its

kFkbl dhy EckbZdsvfHky Ec r FkkO l sxq t j r sv{k

Bl 2 gSx1 (x1 < x2) Kkr dhft , : (pq Ecdh; {ks=k B 6

Bl 2 rod is , Find x1 (x1 < x2) (Magnetic field 6

r y l sy Ecor uhpsdhvksj gS )

B is vertically downward) :

×

×

O×

×

×

×

×

O×

×

×

×

x1 ×

×

x2 ×

×

×

x1 ×

×

x2 ×

×

(A)

l 4

(B)

l 3

(C)

l 6

(D)

l 7

The height of the image formed by a converging

15.

lens on a screen is 8 cm. For the same position

(A)

l 4

(B)

l 3

(C)

l 6

(D)

l 7

, d vfHkl kj hy sUl } kj k i nsZi j fufeZ r i zfr fcEc dhÅ¡pkbZ 8 cm gS AoLr qr Fkki nsZdhl

of the object and screen again an image of

16.

/kkr qNM+fu; r dks.kh; osx

dsi fj r %?kw . kZ u dj j ghgS ANM+dsfl j ksadse/; foHkokUr j

length. Potential difference between ends of

15.

m nz O ; eku r Fkkl y EckbZdh, d

eku fLFkfr i j i q u% ] 12.5 cm

size 12.5 cm is formed on the screen by shifting

vkdkj dki zfr fcEc] y sUl dsfoLFkki u } kj kLØhu i j fufeZ r

the lens. The height of the object is -

gksrk gS A oLr qdhÅ¡pkbZgS-

(A) 15 cm

(B) 5 cm

(A) 15 cm

(B) 5 cm

(C) 10 cm

(D) None

(C) 10 cm

(D) dks bZugha

A square loop of length L and resistance R is pulled out of a magnetic field B. Which is perpendicular to the loop slowly and uniformly

16.

L y EckbZr FkkR i z fr j ks/kds, d (l

oxkZ d kj y iwdks/khj s-/khj st

sd .M) esa, dl eku : i l s, d pq Ecdh; {ks=k B t ksy w i

in t (sec). One edge of the loop was initially

dsr y l sy Ecor gS ] l sckgj fudky kt kr kgS Ay w i dk, d

placed at the boundary of the magnetic field.

fl j ki zkj fEHkd fLFkfr esapq Ecdh; {ks=kdhl heki j fLFkr FkkA

Find the work done in pulling the loop out :

yw i dksckgj [ kha pusesafd; k x; k dk; ZKkr dhft , :

(A) B2 L4 / Rt

(A) B2 L4 / Rt

(B) B2 L3 / Rt

(C) B2 L2 / Rt

(D) BL3 / Rt

(C)

B2

L2

/ Rt

(B) B2 L3 / Rt (D)

BL3

/ Rt



Å|e gh lQyrk dh daqth gSA


Page # 15

17.

The following figure shows a logic gate circuit

17.

with two inputs A and B and output C. The

, d r kfdZ d } kj i fj i Fk gS A A, B o C dsoksYVr k r j xka xz

voltage waveforms of A, B and C are as

uhpsf} r h; fp=kesai znf' kZ r gS

shown in second figure below

A B

Logic gate circuit

uhpsn' kkZ ; kx; kfp=knksfuos'khA o B r FkkfuxZ r C oky k

A B

C

Logic gate circuit Figure (i)

Figure (i) I

I A

A

t

t

I

I B

B

t

t

I

I C

C

t

t Figure (ii)

Figure (ii)

18.

C

The logic circuit gate is :

r kfdZ d i fj i Fk } kj gS:

(A) OR gate

(B) AND gate

(A) OR gate

(B) AND gate

(C) NAND gate

(D) NOR gate

(C) NAND gate

(D) NOR gate

If the input and output resistances in a

18.

; fn , d mHk; fu"B vk/kkj i zo/kZ d i fj i Fk esafuos'kh o

common-base amplifier circuit are 500 ohm

fuxZ r i zfr j ks/k Øe' k%500 ohm o 100 kilo-ohm gS ]

and 100 kilo-ohm respectively, what is the

oksYVr k i zo/kZ u D; kgS ] t c mRl t Z d /kkj k 5 mA o /kkj k

voltage amplification when the emitter current is 5 mA and current gain is 0.9 ? (A) 450

(B) 180

(C) 90

(D) 2.5

y kHk 0.9 gS? (A) 450

(B) 180

(C) 90

(D) 2.5



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Page # 16

A particle of charge ‘+q’ and mass ‘m’, at t = 0,

 enters a uniform magnetic field B  B 0 kˆ while  moving with a velocity V  V0 (î  kˆ ), then (A) it moves along a circular path of radius

(B) it moves along a helical path of pitch

(C) it moves along a helical path of pitch

19.

21.



d.k B  B0 kˆ 

, dl eku pq Ecdh; {ks=kesai zos'kdj r kgS ] t c V  V0 (î  kˆ ) osx l sxfr dj j gk gS ] rc

2mV0 qB 0

(A)

2mV0

; g qB 0

f=kT; k dso`Ùkkdkj i Fk dsvuq fn' k xfr

dj r kgS

2m V0 qB0

(B) ; g

2 2m V0 qB 0

2m V0 fi p dsl fi Z y kdkj i Fkdsvuq fn' kxfr qB0

dj r kgS (C)

(D) it moves along a straight line path along the z-axis 20.

t = 0 i j ‘+q' vkos ' ko ‘m' nzO ; eku dk, d

;g

2 2m V0 fi p dsl fi Z y kdkj i Fk dsvuq fn' k qB 0

xfr dj r k gS (D) ; g z-v{kdsvuq fn' kl j y j s[kh; i Fki j xfr dj r kgS

In the Boolean algebra ( A . B) . A equals to : 20.

(A) A  B

(B) A

(C) A . B

(D) A + B

In YDSE the ratio of intensity at path difference /4 and O is (A) 2 : 1 (B) 1 : 2 (C) 1 : 1 (D) 1 : 4

Passage: (Q.22 to 23) The potential difference across a 2H inductor as a function of time is shown in diagram. At time t = 0, current is zero.

21.

cw fy ; u cht xf.kr esa( A . B) . A cj kcj gS: (A) A  B

(B) A

(C) A . B

(D) A + B

;a x f} fNnzi z;ksx esa/4 r Fkk O i FkkUr j i j r hozrk dk vuq i kr gksxk(A) 2 : 1 (C) 1 : 1

x| ka ' k: (Q.22 l s23) , d 2H i zsjd dsfl j ksai j foHkokUrj l e; dsQy u ds: i esauhpsesan' kkZ , svuq l kj i fj ofr Z r gS At = 0 i j /kkj k' kw U; gS A

VL(volt)

V L(volt)

8

8

0

1

2

(B) 1 : 2 (D) 1 : 4

t(s)

0

1

2

t(s)



HkkX; lkglh dk lkFk nsrk gSA


Page # 17

22.

Area under VL-t graph gives -

22.

(A) Current

VL-t

xzkQ dsv/khu {ks=kQy nsrk gS-

(A) /kkj k

(B) Change in current

(B) /kkj kes ai fj or Z u

(C) Product of inductance and current

(C) i z sj.kr Fkk/kkj kdkxq . kuQy

(D) Product of inductance and change in current

(D) i z sj.kr Fkk/kkj kesai fj or Z u dkxq . kuQy 23..

24.

Current at t = 1S is (A) 1A

(B) 2A

(C) 4A

(D) 8A

23.

Two wavelengths of light 1 and 2 are sent

24.

through Young’s double slit apparatus

i j /kkj k gS-

(A) 1A

(B) 2A

(C) 4A

(D) 8A

1

o 2 nksr j a xnS /; ksZdsi zd k' k; a x f} j s[kkfNnzmi dj .k

l s, d l kFkxq t kj st kr sgS A 1 o 2 dsfy , D; kl ghgksuk

simultaneously. What must be true about

pkfg, ] ; fn r `r h; dksfV dh pedhy h fÝUt 1 dsl a xr

1and 2 if the third order bright fringe of 1

25.

t = 1S

coincides with fifth order dark fringe of 2 ?

i kpohadksfV dhvnhIr fÝUt 2 gS?

(A) 31 = 22

(B) 21 = 32

(A) 31 = 22

(B) 21 = 32

(C) 31 = 52

(D) 51 = 32

(C) 31 = 52

(D) 51 = 32

A ball of radius R carries a positive charge whose volume charge density depends only on separation r from the ball’s centre as

r   =  0 1 –  where  0 is a constant. R   Assuming the relative permittivities of the ball and the environment to be equal to unity. The electric field at distance r1from centre (For r1 < R) is -

0 r1  3r1   (A) 3 1 – 4R  0  0 r1  r1   (C) 2 1 – 2 R 0 

0 r1  r1  (B) 3 1 –  R 0  (D) None

25.

R f=kT; kdh, d

xsa n , d /kukRed vkos'kj [ kr hgSft l dk xsa n dsdsUnzl sr nw j h i j vk; r u vkos'k ?kuRo  = 0 r  1 –  R  

dsvuq l kj fuHkZ j dj r kgSt gk¡0 , d fu; r kd a gS A

; g ekfu; sfd xsa n r Fkk i fj os'k dhvki sf{kd fo| q r ' khy r k bdkbZds cj kcj gS A dsUnz l s r1 nw j h i j fo| q r {ks=k (r1 < R dsfy ; s ) gS0 r1  3r1   (A) 3 1 – 4R  0 

0 r1  r1  (C) 3 1 –  R 0 

0 r1  r1   (C) 2 1 – 2 R 0 

(D) dks bZugha



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Page # 18

Two concentric spherical shell of radii a,b(a
26.

has charge q and –q respectively. Electric field

a, b (a < b) f=kT; kvks adsnksl

a d sUnzh; xksy h; dks'kØe' k%q o –q vkos'kj [ kr hgS Afo| q r {ks=kmi fLFkr gksxk–

does exist –

b

b a

q –q

a

dsoy vkUr fj d xksy sdsvUnj (B) ds oy cká xksy sdsckgj (C) ds oy nksdks'kksadse/; (D) l H kht xg ' kw U; gS (A)

(A) Inside inner sphere only (B) Outside outer sphere only (C) between two shells only (D) Everywhere is zero 27.

A particle of mass m and carrying charge –q1

27.

is moving around a charge +q2 along a circular path of radius r. Find period of revolution of

nzO ; eku r Fkk–q1 vkos'k dk , d d.k+q2 vkos'k ds pkj ksavksj r f=kT; kdho Ùkh; d{kkdsvuq fn' kxfr dj r kgS A +q2 dsl ki s {k–q1 vkos'kdk?kw . kZ udky Kkr dhft ; s

m

the charge –q1 about +q2

16 3 0 mr 3 q1q 2

(A) (A)

16 3 0 mr 3 q1q 2

(B)

8 3 0 mr 3 q1q 2

q1q 2

(C) (C) 28.

q1q 2 16 3 0 mr 3

16 3 0 mr 3

of coordinates and extends along the positive x, y and z-axes. Suppose the electric field in this region is given by E = (a + by) ˆj .

28.

8 3 0 mr 3 q1q 2

(D) ' kw U;

H kq t kokys, d ?ku dk, d dksukfunsZ ' kka d ksadsew y fcUnqi j

gSr Fkk/kukRed x, y r Fkkz-v{kksadsvuq fn' kfoLr kfj r gS A ekuk bl {ks=k esafo| q r {ks=k E = (a + by) ˆj } kj k fn; k x; k gS A ?ku dsvUnj vkos'k Kkr dhft ; s-

Determine the charge inside the cube -

(A) 20b3

(A) 20b3

(C)

0 b 2

(B)

(D) Zero

A cube of side has one corner at the origin

(C)

q –q

(B) 0b3

3

(D) 40b3

0 b 3 2

(B) 0b3 (D) 40b3



deZ djus esa gh vf/kdkj gS] Qy esa ughaA


Page # 19

statement based question (29 - 30) (A) If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement – I. (B) If both Statement- I and Statement- II are true, and Statement - II is the correct explanation of Statement– I. (C) If Statement - I is true but Statement - II is false. (D) If Statement - I is false but Statement - II is true. 29.

Statement I : Value of

amperian loops is

 B . d

for the

dFku v k/kkfj r i z'u (29 - 30) (A)

dk l ghLi "Vhdj .k gS A (B)

29.

(C)

dFku-I l R; gS ] dFku-II vl R; gS A

(D)

dFku-I vl R; gS ] dFku-II l R; gS A

dFku-I : uhpsfp=k esan' kkZ , , fEi ; j y w i ds  B. d dk eku

5  . 2 0

5  gS A 2 0

2A 1 A 2

Amperian loop

Statement–I : For the situation shown in figure, if we displace the charge q within the conducting shell, then nature of distribution of charge on the outer surface of the shell does not change. Q

Amperian loop

dFku-II : fdl hcUn , sfEi; j y w i i j ] pq Ecdh; {ks=kdkj s[kk

Statement II : Line integral of magnetic field over closed amperian loop is equal to 0 times current enclosed in that loop. 30.

dFku-I l R; gS ] dFku-II l R; gS A dFku-II, dFku-I

dk l ghLi "Vhdj .k ughagS A

2A 1 A 2

dFku-I l R; gS ] dFku-II l R; gS A dFku-II, dFku-I

l ekdy u] bl y w i esai fj c) /kkj kvksadk 0 xq uk gksrk gS A

30.

dFku–I : fp=kesan' kkZ ; hfLFkfr dsfy , ] ; fn ge pky dh; dks'kdsvUnj q vkos'kdksfoLFkkfi r dj r sgS ] r c dks'kds ckg~ ; i `"B i j vkos'k dsfor j .k dhi zd `fr ughcny r hgS A Q q

q

Neutral conductor Neutral conductor

Statement–II : Any conducting shell divides the entire space into two regions (inside and outside the shell), which are independent to each other in terms of electric field.

dFku–II : fdl hHkhpky d dks'kdsi jwsLFkku dksnks{ks=kksa (dks ' k dsvUnj r Fkk ckg~ ; ) esafoHkkft r dj l dr sgS a ] t ks fo| q r {ks=kdsi nksaesa, d nw l j sl sLor a =kgksrk gS A



:[email protected]


Page # 20

PART - II [CHEMISTRY] 1.

Kp for the reaction : CO2(g)+ H2(g)

1.

CO(g) + H2O(g) is found to be 16 at a given temperature. Originally equal number of moles of H2 and CO2 were placed in the flask . At equilibrium, the pressure of H2 is 1.20 atm. What is the partial pressure of CO and H2O ?

2.

CO2(g)+ H2(g)

4.

i zfr ok; q e.My

(A) 4.80 atm each

(B) 9.60 atm each

(A) 4.80

(C) 2.40 atm each

(D) 1.20 atm each

(C) 2.40 i z fr

10 ml of

M H2 SO 4 is mixed with 40 ml of 200

(A) 1

(B) 2

(C) 2.3

(D) none of these

CO(g) + H2O(g) i z kj EHkesa

¶y kLd esaH2 r FkkCO2 dhcj kcj eksy gS Al kE; oLFkki j H2 dknkc 1.20 atm gS ACO r FkkCO2 dkvka f' kd nkc D; k gksxk?

2.

10 ml,

ok; q e.My

(B) 9.60 i z fr

ok; q e.My

(D) 1.20 i z fr

ok; q e.My

M M H SO dk40 ml, H SO dsl 200 2 4 200 2 4

kFk

fey k; k t kr k gS , cuusoky sfoy ; u dhpH D; kgks xh\

M H2 SO 4 . The pH of the resulting solution is 200

3.

j kl k; fud vfHkfØ; kdsfy, Kp dkeku fn; kx; kr ki i j 16 gS A:

(A)1 (C)2.3

To 500 cm3 of water 3.0 × 10–3 kg of acetic acid is added. If 23% of acetic acid is dissociated what will be the depression in freezing point ? Kf and density of water are 1.86 K kg–1 and 0.997 gm cm–3 respectively (A) 0.186 K (B) 0.228 K (C) 0.372 K (D) 0.556 K

3.

Which of the following statement(s) is (are) correct for CaF2 : (a)Ca2+ ions are present only at the corners of a cube (b) c.c.p. type structure (c) F– ions are present in all the octahedral voids (d) The structure has 8 : 4 co-ordination (A) a, b (B) b, c (C) b, d (D) c, d

4.

(B) 2 (D) none of these

500 cm3 t

y esa3.0 × 10–3 kg , fl fVd vEy feyk; k t kr k gS; fn 23% , fl fVd vEy fo; ksft r gksrk gSr ks fgeka d esavoueu D; kgksxk? Kf r Fkkt y dk?kuRo Øe' k% 1.86 K kg–1 r Fkk0.997 gm cm–3 gS (A) 0.186 K (C) 0.372 K

(B) 0.228 K (D) 0.556 K

fuEu esal sdkS ul k dFku CaF2 dsfy , l ghgS: 2+ (a) Ca vk; u ds oy ?ku dsdksuksai j mi fLFkr gksrsgS a i zd kj dhl a j puk (c) F vk; u l H khv"VQy dh; fj fDr ; ksaesami fLFkr gksrs gS a (d)l a j puk 8 : 4 mi l gl a ; kst u l a [ ; k; q Dr gS (b) c.c.p. –

(A) a, b (C) b, d


(B) b, c (D) c, d


Page # 21

5.

An excited electron of H-atoms emits of photon of wavelength  and returns in the ground state, the principal quantum number of excited state is given by (A)

R ( R – 1)

5.

, d H-i j ek.kqdsmÙksft r by sDVªkW u r j a xnS /; ZdsQksVksu mRl ft Z r dj r kgS ] , oavk| voLFkkesai q u%y kS V vkr kgS ] r ks mÙksft r voLFkk esaeq [ ; Doka Ve l a [ ; k gksxh-

(B)

R (R – 1)

(A)

(C)

(D)

(R – 1) R

R ( R – 1)

(B)

R (R – 1)

(D)

(R – 1) R

1

1 (C)

6.

7.

8.

R (R – 1)

Standard enthalpies of formation of O3, CO2, NH3 and HI are 142.2, – 393.2, – 46.2 and +25.9 kJ mol–1 respectively. The order of their increasing stabilities will be : (A) O3, CO2, NH3, HI (B) CO2, NH3, HI, O3 (C) O3, HI, NH3, CO2 (D) NH3, HI, CO2, O3

6.

22.4 litres of H2S and 22.4 litres of SO2 both at STP are mixed together. The amount of sulphur precipitated as a result of chemical calculation is : (2H2S + SO2  3S + 2H2O) (A) 16 g (B) 23 g (C) 48 g (D) 96 g

7.

M FeSO4 was titrated A solution of 10 ml 10

Which solution will show highest resistance during the passage of current ? (A) 1 NNaCl (B) 0.1 NNaCl (C) 2 NNaCl (D) 0.05 NNaCl

O3, CO2, NH3 o HI dsfuekZ . kdhekud

, UFkS fYi ; kaØe' k% 142.2, – 393.2, – 46.2 o +25.9 kJ mol–1 gS a A budsc<+ r sgq , LFkkf; Ro dkØe gksxk(A) O3, CO2, NH3, HI (C) O3, HI, NH3, CO2

(B) CO2, NH3, HI, O3 (D) NH3, HI, CO2, O3

ds22.4 y hVj , oaSO2 ds22.4 y hVj nksuksadks STP i j , d l kFk fey k; k t k; s A l YQj vo{ksi dhek=kk j kl k; fud x.kukdsi fj .kkeLo: i gS: H2S

(2H2S + SO2  3S + 2H2O)

8.

(A) 16 g

(B) 23 g

(C) 48 g M 10 ml FeSO4 10

(D) 96 g

dsfoy ; u dksKMnO4 foy ; u ds

l kFk vEy h; ek/; e esami pkfj r dj usi j ¼ fey kusi j ½ i z;q Dr KMnO4 dhek=kkgksxh-

with KMnO4 solution in acidic medium. The amount of KMnO4 used will be (A) 5 ml of 0.1 M (B) 10 ml of 0.1 M (C) 10 m of 0.5 M (D) 10 ml of 0.02 M

9.

R (R – 1)

9.

(A) 5 ml of 0.1 M

(B) 10 ml of 0.1 M

(C) 10 m of 0.5 M

(D) 10 ml of 0.02 M

fuEu esal sdkS ul k foy ; u fo| q r /kkj k i zokfgr dj usi j l okZ f/kd i zfr j ks/kn' kkZ , xk? (A) 1 NNaCl (C) 2 NNaCl


(B) 0.1 NNaCl (D) 0.05 NNaCl


11.

Page # 22

50 % of a zero order reaction completes in 10 minutes. 100 % of the same reaction shall complete in (A) 5 min (B) 10 min (C) 20 min (D)  time

10.

Predict the product of the following reaction

11.

(A)

(C)

, d ' kw U; dksfV vfHkfØ; kdk50 % 10 feuV esai w . kZgksrk gS A vfHkfØ; kdk100 % i w . kZgksxk(A) 5 min (D) 20 min

(B) 10 min (D)  time

fuEu vfHkfØ; kesamRikn cr kb; s

(A)

(B)

(C)

(D)

(B)

(D)



Page # 23

12.

The product Z of the following sequence of reaction.

12.

fuEu vfHkfØ; kØe esamRikn Z gS A p-Dy ks j ksVkW yq bZ u

Z

p-chlorotoluene Z

(I)

13.

(II)

(III)

(IV)

(II)

(III)

(A) I

(I)

(IV)

(B) II

(C) III

(A) I

(B) II

(C) III

(D) IV

In the reaction below, the product would be

13.

uhpsnhxbZvfHkfØ; kesamRikn gksxk + HCN

+ HCN

(A) A recemate (B) Optically active (C) A meso compound (D) A mixture of the diastereomers

(A) j s l sehd feJ.k (B) i z d kf' kd l fØ; (C) , d ehl ks; kS fxd (D) foofj e l eko; oh; ks adkfeJ.k


(D) IV


Page # 24

What would be the principal product of nitration of 2-bromoanisole ?

14.

16.

The principle product of the reaction between methyl butanoate and 2 moles of CH3MgBr after hydrolysis is : (A) C3H7COCH3 (B) C3H7C(OH)(CH3)2 (C) C3H7CHOHCH3 (D) C3H7COCH(CH3)2

15.

What is product of Baeyer Villiger oxidation below ?

16.

(A)

(B)

(C)

(D)

ukbVªhdj .k dj kusi j eq [ ; mRikn

(A) 2-ukbVªks , uhl ksy (B) 2-cz kseks-3-ukbVªks,uhl ksy (C) 2-cz kseks-4-ukbVªks,uhl ksy (D) 2-cz kseks-5- ukbVªks,uhl ksy

(A) 2-nitroanisole (B) 2-bromo-3-nitroanisole (C) 2-bromo-4-nitroanisole (D) 2-bromo-5- nitroanisole 15.

2-cz kseks, uhl kW y ij gksxk?

esfFky C;w Vsuks,V vkS j 2 eksy CH3MgBr dhvfHkfØ; kr r ~ i ' pkr ~t y vi ?kVu dj kusi j eq [ ; mRikn i zkIr gksxk: (A) C3H7COCH3 (B) C3H7C(OH)(CH3)2 (C) C3H7CHOHCH3 (D) C3H7COCH(CH3)2

cs;j fofy xj vkW Dl hdj .kmRikn gS?

(A)

(B)

(C)

(D)



Page # 25

17.

Select the best statement for the situation below :

17.

(A) This mixture gives an optical rotation of zero. (B) A will rotate plane polarized light in clockwise direction. (C) B will rotate plane polarized light in clockwise direction. (D) None of the above statement is true. 18.

Arrange the following in the increasing order of stability

(A) D < B < C < A (C) A < C < B < D

19.

(A) ; g feJ.ki z d kf' kd ?kw . kZ u ' kw U; nsrkgS A (B) A l er y /kq zfor i zd k' kdksnf{k.kkor Z?kw ekr kgS A (C) B l er y /kq zohr i zd k' kdksnf{k.kkor Z?kw er kgS A (D) mi j ks Dr esal sdksbZHkhdFku l R; ughagS A

18.

(B) D < C < B < A (D) D < B = C < A

POCl3   (a) + H O ; HCONHR Pyridine 2

(a) in the above reaction is (A) RCH=NOH (B) R–N=C=O (C) R–CN (D) R–NC

fuEu dsfy , l okZ f/kd l ghdFku cr kb; s:

fuEu dksmudsLFkkf; Ro dsc<+ r sØe esaO ; ofLFkr dhft ,

(A) D < B < C < A (C) A < C < B < D

19.

(B) D < C < B < A (D) D < B = C < A

POCl3   (a) + H O ; HCONHR Pyridine 2

mi j ksDr vfHkfØ; kesa(a) gS (A) RCH=NOH (C) R–CN


(B) R–N=C=O (D) R–NC


Page # 26

Match the List I with List II and pick the correct matching from the codes given below: List I List II (Reactants) (Products)

20.

fupsfn; sx; sLr EHkI dkLr EHkII dsfn; sx; sdw Vksadsl kFk l ghfey ku dhft , Lr EHkI Lr EHkII (vfH kdeZ d) (mR i kn) ethanol (A) RX + KCN    1.

ethanol (A) RX + KCN    1.

(B) RX + CH3COOAg  2. RCN

(B) RX + CH3COOAg  2. RCN LiAlH4 (C) CH3CN  

(D)

(E) R–NH2

LiAlH4 (C) CH3CN  

3. CH3COOR

(E) R–NH2

22.

5. RNC

5. RNC (A) A—3, B—1, C—2, D—5, E—4 (B) A—2, B—3, C—4, D—1, E—5 (C) A—3, B—2, C—1, D—5, E—4 (D) A—5, B—3, C—2, D—4, E—1

(A) A—3, B—1, C—2, D—5, E—4 (B) A—2, B—3, C—4, D—1, E—5 (C) A—3, B—2, C—1, D—5, E—4 (D) A—5, B—3, C—2, D—4, E—1 21.

Hofmann ' s  4. CH3CH2NH2 degradation

(D)

Hofmann ' s  4. CH3CH2NH2 degradation

3. CH3COOR

Correct order of second ionization potential of carbon, nitrogen, oxygen and fluorine is : (A) O > F > N > C (B) F > O > N > C (C) C > N > O > F (D) O > N > F > C

21.

The major role of fluorspar (CaF2) which is added in small quantities in the electrolytic reduction of alumina dissolved in fused cryolite (Na3AlF6) is : 1. as a catalyst 2. to make the fused mixture very conducting 3. to lower the temperature of melt 4. to decrease the rate of xidation of carbon at the anode (A) 2,3 (B) 1,2 (C) 2,3,4 (D) 3,4

22.

dkcZ u] ukbVªkst u] vkd Wl ht u vkS j Dy ksjhu dsf} r h; vk; uu Åt k dk l ghØe gS: (A) O > F > N > C (B) F > O > N > C (C) C > N > O > F (D) O > N > F > C

dks(Na3AlF6) esafey kusdk eq [ ; dkj .k D; k gS ] t ksdh Al fo| q r vi ?kVu mRiknu esai z;q Dr gksrkgS : 1. mR i zsjd dhr j gA 2. foy ; u dksvf/kd pky d cukusdsfy , A 3. /kkr qdsr ki dksde dj usdsfy , A 4. dkcZ u dsvkW Dl hdj .kdks, uksMi j de dj usdsfy , A CaF2

(A) 2,3


(B) 1,2

(C) 2,3,4

(D) 3,4


Page # 27

23.

Which alkali metal ion has lowest mobility in aqueous solution ? (A)

Rb(H2O)n

(C) Li(H2O)n 24.

(B)

23.

Cs(H2O)n

t yh; foy ; u esadkS ul h{kkj h; /kkr qde xfr ' khy gksrhgS ? (A) Rb(H2O)n

(B) Cs(H2O)n

(C) Li(H2O)n

(D) Na(H2O)n

(D) Na(H2O)n KO2 dks' okl

y susdsmi dj .k esaO2 mRiUu dj usr Fkk

Potassium superoxide finds use in breathing equipment and safeguards the user to breathe in oxgen generated internally in the apparatus and exposed to the toxic fumes outside. The supply of oxygen is due to 1. Slow decomposition of KO2 2. Reaction of superoxide with moisture in the exhaled air 3. Reaction of KO2 absorb CO2 (A) 1, 2 and 3 are correct (B) 2 and 3 are correct (C) 3 only is correct (D) 1 and 2 only are correct

24.

25.

Ammonia can be dried over : (A) conc. H2SO4 (B) P4O10 (C) CaO (D) anhydrous CaCl2

25.

26.

The number of lone pair and the number of S–S bonds in S8 molecule are respectively : (A) 8,8 (B) 16,8 (C) 8,16 (D) 8,4

26.

27.

The shape of O2F2 is similar to that of : (A) C2F2 (B) H2O2 (C) H2F2 (D) C2H2

27.

O2F2 dhl a j puk l eku gksrhgS: (A) C2F2 (B) H2O2 (C) H2F2 (D) C2H2

28.

In perchloric acid, Cl–O bond contains : (A) d-d bonding (B) p–d bonding (C) p–p bonding (D) None of the above

28.

i j Dy ksjhd vEy esa , Cl–O cU /kj [ kr kgS (A) d-d ca /k (B) p–d ca /k (C) p–p ca /k (D) bues al sdksbZugha

vuq i;q Dr xS l ksadksckgj fudky usesai z;q Dr fd; kt kr kgS ] D; ksa fd 1. KO2 dk/khehxfr 2. KO2 uehl

l sVw VukA

svfHkfØ; kdj dsO2 mRiUu dj r kgS A

3. KO2 ] CO2 dksvo' kks f"kr (A) 1, 2 (B) 2 (C)

dj r kgS A

r Fkk 3 l ghgS

r Fkk3 l ghgS

dsoy 3 l ghgS

(D) ds oy 1 r Fkk2 l

ghgS

veksfu; kdks' kq "d fd; kt kl dr kgS: (A) conc. H2SO4 (C) CaO

(B) P4O10 (D) fut Z y h; CaCl2

S8 ; kS fxd

esavcU/khr by sDVªkW u;q Xe dhl a [ ; kr FkkS–S cU/k dhl a [ ; kØe' k%gS: (A) 8,8


(B) 16,8 (C) 8,16 (D) 8,4


30.

Page # 28

In whi c h of the fol l owi ng oct ahed ral complexes of Co(at. no. 27), wi ll the magnitude of 0 be highest ? (A) [Co(NH3)6]3+ (B) [Co(CN)6]3– (C) [Co(C2O4)3]3– (D) [Co(H2O)6]3+

29.

The coordination number and the oxidation state of the element 'E' in the complex [E(en)2 (C2O4)] NO2 (where (en) is ethylene diamine) are respectively : (A) 6 and 3 (B) 6 and 2 (C) 4 and 2 (D) 4 and 3

30.

fuEu esal sfdl v"VQy dh; l a dq y Co esa0 dk eku l okZ f/kd gS(Co dki j ek.kqØeka d 27) ? (A) [Co(NH3)6]3+ (B) [Co(CN)6]3– (C) [Co(C2O4)3]3– (D) [Co(H2O)6]3+

la dq y ; kS fxd esar Ro 'E' dhl eUo; l a [ ; kvkjSvkD W l hdj .k voLFkkØe' k%gS [E(en)2 (C2O4)] NO2 (t gk¡(en) , fFky hu Mkb, ehu) Øe' k%gS: (A) 6 r Fkk3 (B) 6 r Fkk2 (C) 4 r Fkk2 (D) 4 r Fkk3


2016 SAMPLE PAPER

SAMPLE TEST(4)


JEE Main 2016 Mathematics Sample Paper by MC Sir Important Instruction. 1.

Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited.

2.

The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully.

3.

The test is of 1 hours duration.

4.

The Test Booklet consists of 31 questions. The maximum marks are 124.

5.

There are Only One part in the question paper consisting of Mathematics having 31 questions in each part of equal weightage. Each question is alloted 4 (four) marks for correct response.

6.

Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each question. 1 4 (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction form the total score will be made if no response is indicated for an item in the answer sheet.

7.

There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.

8.

Use Blue/Black Ball Point Pen only for wrinting particulars/making response on Side-1 and Side-2 of the Answer Sheet. Use of pencil is strictly prohibited.

9.

No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronicdivice etc. except the Admit Card inside the examination room/hall.

10.

Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each page and in one page (i.e., Page 6) at the end of the booklet.

11.

On completion of the test, the candidate must have over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.

12.

Do not fold or make any stray mark on the Answer Sheet.

Name of the Candidate (in Capital letters) : ____________________________________________ Role Number

: in figures : in words ______________________________________________________

Examination Center Number : Name of Examination Centre (in Capital letters) : _______________________________________ Candidate's Signature : ________________

1. Invigilator's Signature : __________________ 2. Invigilator's Signature : __________________

ETOOS Education Pvt. Ltd. 3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005, Tel. 09214233303 | 09214233343 | website : www.etoosindia.com

Read the following instruction carefully : 1.

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2.

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3.

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4.

Out of the four options given for each question, only one option is the correct answer.

5.

For each incorrect response, one-fourth 14 of the total marks allotted to the question would be deducted from the total score. No deduction from the total score, however, will be made if no response in indicated for an item in the Answer Sheet.

6.

Handle the Test Booklet and Answer Sheet with care, as under no circumstance (except for discrepancy in Test Booklet code and Answer Sheet Code), another set will be provided.

7.

The candidates are not allowed to do any rough work or writting work on the Answer Sheet. All calculations/writing work are to be done in the space provided for this purpose in the Test Booklet itself, marked ‘Space for Rough Work’. This space is given at the bottom of each page and in one page (i.e., Page 6) at the end of the booklet.

8.

On completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.

9.

Each candidate must show on demand his/her Admit Card to the Invigilator.

10.

No candidates should not leave the Examination Hall without handing over their Answer Sheet to the Invigilator on duty and sign the Attendance Sheet again. Cases where a candidates has not signed the Attendance Sheet second time will be deemed not to have handed over the Answer Sheet and dealt with as an unfair menas case. The candidates are also required to put their left hand THUMB impression in the space provided in the Attendance Sheet.

12.

Use of Electronic/Manual Calculator and any Electronic device like mobile phone, pager etc. it prohibited.

13.

The candidates are governed by all Rules and Regulations of the JAB/Board with regard to their conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the JAB/Board.

14.

No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.

15.

Candidates are not allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, electronic device or any other material except the Admit Card inside the examination room/hall.

Mathematics 1.

The value of the expression 2(1 + ) (1 + )2 + 3(2 + 1) (22 + 1) + 4(3 + 1) (32 + 1) + ..... + (n + 1) (n + 1) (n2 + 1) is ( is the cube root of unity) (A)

2

 n(n  1)  (B)   n  2 

n 2 (n  1) 2 4 2

 n(n  1)  (C)   n  2 

(D) None of these

2.

The number of number pairs (x, y) which will satisfy the equation x2 – xy + y2 = 4 (x + y – 4) is (A) 1 (B) 2 (C) 4 (D) None of these

3.

Value of L  lim

1 n  n 4

(A) 1/24

  n   n 1   n 2   1  k   2   k   3   k   .....  n.1 is   k 1   k 1   k 1   (B) 1/12

(C) 1/6

4.

The number of divisors of 22 . 33 . 53 . 75 of the form 4n + 1, n  N is (A) 46 (B) 47 (C) 96

5.

If x 





(D) 94

n

3  1 , then [x] is (where [x] denotes the greatest integer less than or equal to x)

(A) 2k, where k  I (C) 4n 6.

(D) 1/3

(B) 2k + 1, where k  I (D) 8n

Two number x and y are chosen at random from the set {1, 2, 3,......30}. The probability that x2 – y2 is divisible by 3 is (A) 3/29 (B) 4/29 (C) 5/29 (D) None of these Space for Rough Work

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7.

8.

 a 4  5 The set of all values of a for which the function f (x)    1  x  3x  log 5 decreases for all  1 a  real x is

(A) ( , )

 3  21  (B)  4,   (1,  ) 2  

27   (C)  3, 5    (2, ) 2  

(D) [1, )

1 (1  x)1/x  e  ex 2 The value of lim is x 0 x2 (A)

9.

11 e 24

(B) 

11e 24

(C)

e 24

(D) None of these

Tangent are drawn from the origin to the curve y = sinx, then their point of contact lie on the curve (A) x2 + y2 = 1

(B) x2 – y2 = 1

(C)

1 1  2 1 2 x y

(D)

1 1  2 1 2 y x

2[x] 3x  [x] dx , where [.] denotes the greatest integer function is The value of the integral   2[x]  10  3x  [x]    0

10.

(A) 0

(B) –10

(C) 10

(D) None of these

Space for Rough Work


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11.

The equation of a plane passing through the line of intersection of the planes x + 2y = 3, y – 2z + 1 = 0 and perpendicular to the first plane is (A) 2x – y – 10z = 9 (B) 2x – y – 9z = 10 (C) 2x – y + 10z = 11 (D) 2x – y + 7z = 11

12.

If z1, z2, z3  C, iz3 + z2 – z + i = 0, where i  1 then the area of triangle formed by z1, z2 and z3 is equal to (A)

13.

1 2

(B)

The equation

(x  3)

2

(C) 2 2

2

 (y  1) 2 } 

(x  3)

2

(D)

1 2 2

 (y  1) 2 }  6 represents

(A) an ellipse (B) a pair of straight lines (C) a circle (D) a straight line joining the point (–3, 1) to the point (3, 1)

14.

The length of the common chord of the ellipse (x – 1)2 + (y – 2)2 = 1 is (A) zero (B) one

15.

(x  1)2 (y  2) 2   1 and the circle 9 4

(C) three

(D) eight

 ˆ bˆ and cˆ are unit vectors satisfying aˆ  3bˆ  cˆ  0 , then the angle between the vectors aˆ and If a,

cˆ is (A)

 6

(B)

 4

(C)

 3

(D)

 2



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16.

The system of linear equations x + y + z = 2, 2x + y – z = 3, 3x + 2y + kz = 4 has a unique solution, if (A) –1 < k < 1 (B) –2 < k < 2 (C) k  0 (D) k = 0

17.

If the sum of the series 2 4 1 25  24  23  23  ..... is maximum, then 5 5 5 (A) lest term of the series is 0

1 5 (D) number of terms of the series are 40

(B) last term of the series is

(C) number of terms of the series are 42 18.

The contrapositive of the conditional “If Gopal works hard, then he gets a first class” is (A) If Gopal get a first class, then he does not works hard (B) If Gopal does not get a first class, then he works hard (C) If Gopal does not get a first class, then he does not work hard (D) If Gopal does not work hard, then he does not get a first class.

19.

If A be a 3 × 3 matrix with real entries such that det. (adj A) = 16, then det. (adj(adj A)) is equal to (A) 16 (B) 64 (C) 256 (D) 512

20.

A common tangent to the conics x2 = 6y and 2x2 – 4y2 = 9 is (A) x + y = 1

21.

(B) x – y =1

(C) x + y =

9 2

(D) x – y =

3 2

A tangent to the hyperbola x2 – 2y2 = 4 meets x-axis at P and y-axis at Q. Line QR are drawn such that OPRQ is a rectangel (where O is origin). The locus of R is (A)

4 2  2 1 2 x y

(B)

4 2  2 1 2 x y

(C)

2 4  2 1 2 x y

(D)

2 4  2 1 2 x y



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22.

The solution of differentiable equation 2ysin x (A) y2 = 1 + cosx

23.

24.

25.

Let A {1, 2, 3, 4}. The number of different ordered pairs (B, C) that can be formed such that B  A, C  A and B  C is empty is (A) 42 (B) 34 (C) 43 (D) 24

 B  If two events A and B are such that P(A ' ) = 0.3, P(B) = 0.4 and P(A  B ' ) = 0.5, then P    AB equals 3 5 1 3 (A) (B) (C) (D) 4 6 4 7 Let Q be the foot of perpendicular from the origin to the plane 4x – 3y + z + 13 = 0 and R be a point (–1, 1, –6) on the plane. The length QR is (A) 3

26.

(B) y = sin2x

dy    sin 2x  y2 cos x , satisfying y    1 is dx 2 2 2 (C) y sinx = 1 + cosx (D) y = sinx

7 2

(B) 14

(C)

19 2

(D)

3 2

x  x  2  ax  b If the function f (x)   x 2  5x  6 2  x  3  px 2  qx  1 3  x  

is differentiabl in ( , ) , then (A) a  1, p  (C) a = 1, b = 2

4 9

(B) b  2, q 

5 3

(D) a = –1, q 

5 3



5

27.

x5 x 4   5 has 20 12 (A) no local extremum, one point of inflection (B) two local maximum, one local minimum, two point of inflection (C) one local maximum, one local minimum, one point of inflection (D) one local maximum, one local minimum, two point of inflection

The graph of function f (x) 

 /4

28.

 sin x d(x  [x])

is equal to (where [.] denote the greatest integer function)

0

(B) 1 

(A) 1/2 29.

1 2

(C) 1

(D) None of these

Consider the integrals 1

I1   e x cos 2 x dx, 0

1

2

I3   e  x dx 0

1

0

1

and

2

I 2   e  x cos2 x dx, 2

I 4   e  x /2 dx 0

Let I be the greatest integral among I1, I2, I3, I4, then (A) I = I1 (B) I = I2 (C) I = I3 30.

If y = f(x) passing through (1, 2) satisfies the differential equation y(1 + xy) dx – x dy = 0, then (A) f (x) 

31.

(D) I = I4

2x 2  x2

(B) f (x) 

x 1 x2 1

(C) f (x) 

x 4 4  x2

(D) f (x) 

4x 1  2x 2

Equation of the circle cutting orthogonally the three circles x2 + y2 – 2x + 3y – 7 = 0, x2 + y2 + 5x – 5y + 9 = 0 and x2 + y2 + 7x – 9y + 29 = 0 is (A) x2 + y2 – 16x – 18y – 4 = 0 (B) x2 + y2 – 7x + 11y + 6 = 0 2 2 (C) x + y + 2x – 8y + 9 = 0 (D) None of these Space for Rough Work


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