Ch a p t e r # 2
1. Sol.
P h ys ic s & M a t h e m a t i c s
Two vectors vectors having having equal magnitudes magnitudes A make an angle with each other. Find the magnitude and direction of the resultant. The magnitude magnitude of of the resultant will be B= =
A 2 A 2
2 AA cos 4 A 2 cos2
2 A 2 (1 cos ) =
= 2A cos
2
2 The resultant will make an angle with the first vector where
A sin tan = = A A cos
=
or,
2 A sin cos 2 2 2 A cos
2
= tan
2
2
. 2 Thus, the resultant of two equal vectors bisects the angle between them.
2.
Two vectors of equal magnitude 5 unit have have an angle 60º 60º between between them. Find the magnitude of (a) the sum of the vectors and (b) the difference of the vectors.
B 60º B
B A +
B
A
120º
A + B
A
Sol.
Figure shows the construction of the sum A B and the difference A B .
(a) A B is the sum of A and B . Both have a magnitude of 5 unit and the angle between them is 60º. Thus the magnitude of the sum is
| A B | 5 2
5 2 2 5 5 cos 60º
= 2 × 5 cos 30º = 5 3 unit.
(b) A B is the sum of A and (– B ). As shown in the figure, the anlge between A and (– B ) is 120º.
The magnitudes of both A and (– B ) is 5 unit. So,
| A B | 5 2
52 2 5 5 cos120º
= 2 × 5 cos 60º = 5 unit. 3. Sol.
A force of 10.5 10.5 acts on on a particle particle along along a directi direction on making an an angle angle of 37º with with the vertical. vertical. Find Find the component of the force in the vertical direction. The component of the force in the vertical vertical direction direction will will be F1 = F cos = (10.5 N) cos 37º) = (10.5 N)
4.
4 5
= 8.40 N
The work done done by a force F during a displacement r is given by F r . Suppose Suppose a force of 12 N acts on a particle in vertically upward direction and the particle is displaced displaced through 2.0 m in vertically downward direction. Find the work done by the force during this displacement.
manishkumarphysics.in
Page # 1
Chapter # 2
Physics & Mathematics
Sol.
The angle between the force F and the displacement r is 180º. Thus, the work done is
W = F r = Fr cos = (12 N) (2.0 m) (cos 180º) = –24 N – m = – 24 J.
5.
The vector A has a magnitude of 5 unit, B has a magnitude of 6 unit and the cross product of A and
B has a magnitude of 15 unit. Find the angle between A and B . If the angle between A and B is , the cross product will have a magnitude
Sol.
| A × B | = AB sin 15 = 5 × 6 sin
or,
1 2 = 30º or, 150º, sin =
or, Thus,
From the curve given in figure find
6.
dy at x = 2, 6 and 10. dx
P
6-
D
42C –2
2
1
0
2
4
6
8
10
12
14
The tangent to the curve at x = 2 is AC. It slope is tan1 =
Sol.
dy
AB BC
=
5 4
5 at x = 2 dx 4 The tangent to the curve at x = 6 is parallel to the X-axis. Thus,
=
dy = tan = 0 at x = 6. dx
Thus,
The tangent to the curve at x = 10 is DF. Its slope is tan 2 =
Sol.
5 . 4
dy
if y = e x sin x dx y = ex sin x
Find
So
8.
EF
=–
dy 5 =– at x = 10. dx 4
Thus,
7.
DE
dy d d d = (e x sin x) = e x (sin x) + sin x (e x) dx dx dx dx ex cos x + ex sin x = e x (cos x + sin x).
The height reached in time t by a particle thrown upward with a speed u is given by
1 gt 2 2 where g = 9.8 m/s 2 is a constant. Find the time taken in reaching the maximum height. h = ut –
manishkumarphysics.in
Page # 2
Chapter # 2 Sol.
Physics & Mathematics
The height h is a function of time. Thus, h will be maximum when h = ut = dh dt
or,
=u
dt
=u–
dt
= 0. We have,
1 gt 2 2
d d (ut) – dt dt
dt
dh
1 2 gt 2
d 1 g (t 2) dt 2
–
1 g (2t) = u – gt. 2
For maximum h, dh dt or,
=0
u – gt = 0
or,
t=
u . g
6
9.
Evaluate
(2x
2
3 x 5) dx .
3
(2x 3x 5) dx = 2x dx 3 x dx 5 dx = 2 x dx 3 x dx 5 x dx 2
Sol.
2
2
= 2
= 6
Thus,
3
=
2 3
x3 3 x3
0
3
x2 2
5
x 1
3
x 2 5x 2
6
2 3 3 2 ( 2x 3 x 5) dx = x x 5 x 2 3 3 2
2
2
(36 – 9)+ 5(6 – 3) 3 3 = 126 + 40.5 +15 = 181.5. 10.
Round off the following numbers to three significant digits (a) 15462, (b) 14.745, (c) 14.750 and (d) 14.650 × 10 12. (a) The third significant digit is 4. This digit is to be rounded. The digit next to it is 6 which is greater than 5. The third digit should, therefore, be increasesd by 1. The digits to be dropped should be replaced by zeroes because they appear to the left of the decimal. Thus, 15462 becomes 15500 on rounding to three significant digits. (b) The third significant digit in 14.745 is 7. The number next to it is less than 5. So 14.745 becomes 14.7 on rounding to three significant digits. (c) 14.750 will become 14.8 because the digit to be rounded is odd and the digit next to it is 5. (d) 14.650 × 10 12 will become 14.6 × 10 12 becausethe digit to be rounded is even and the digit next to it is 5.
Sol.
11.
(216 – 27) +
Evaluate
25.2 1374 . All the digits in this expression are significant. 33.3 manishkumarphysics.in
Page # 3
Chapter # 2 Sol.
Physics & Mathematics
We have
25.2 1374
= 1039.7838... 33.3 Out of the three numbers given in the expression 25.2 and 33.3 have 3 significant digits and 1374 has four. The answer should have three significant digits, it becomes 1040. Thus, we write 25.2 1374 = 1040. 33.3
Ex. Sol.
Evaluate 24.36 + 0.0623 + 256.2 24.36 0. 0623 256.2 –––––––––– 280.7 The sum is 280.7.
13.
The focal length of a concave mirror obtained by a student in repeated experiments are given below. Find the average focal lengthwith uncertainty in ± limit. No. of observation 1 2 3 4 5 6 7 8 9 10
Sol.
The average focal length f
focal length in cm 25.4 25.2 26.6 25.1 25.2 25.5 25.4 25.4 25.3 25.7
1
10
f i = 25.37 = 25.4 10 i 1 The calculation of is shown in the table below :
2
i
f i cm
f i – cm
(f 1 – ) cm
1 2 3 4 5 6 7 8 9 10
25.4 25.2 25.6 25.1 25.3 25.2 25.5 25.4 25.3 25.7
0.0 -0.2 0.2 0.3 -0.1 -0.2 0.1 0.0 -0.1 0.3
0.00 0.04 0.04 0.09 0.01 0.04 0.01 0.00 0.01 0.09
1 10
2
( f 1 f )2 cm2
0.33
( f 1 f ) 2 0.033 cm2 = 0.18 cm.
= 0.2 cm. Thus, the focal length is likely to be within (25.4 ± 0.2 cm) and we write f = (25.4 ± 0.2) cm.
manishkumarphysics.in
Page # 4
Chapter # 2
Physics & Mathematics
1.
Is a vector necessarily changed if it is rotated through an angle?
2.
Is it possible to add two vectors of unequal magnitudes and get zero? It is possible to add three vectors of equal magnitudes and get zero?
3.
Does the phrase "direction of zero vector" have physical significance? Discuss in terms of velocity, force etc.
4.
Can you add three unit vectors to get a unit vector? Does your answer change if two unit vectors are along the co-ordinate axes?
5.
Can we have physical quantities having magnitude and direction which are not vectors?
6.
Which of the following two statements is more appropriate? (a) Two forces are added using triangle rule because force is a vector quantity. (b) Force is a vector quantity because two forces are added using triangle rule.
7.
Can you add two vectors representing physical quantities having different dimensions? Can you multiply two vectors representing physical quantities having different dimensions?
8.
Can a vector have zero component along a line and still have nonzero magnitude ?
9.
Let 1 and 2 be the angles made by A and
A with the positive X-axis. Show that tan1 = tan2. Thus, giving
tan does not uniquely determine the direction of A .
10.
Is the vector sum of the unit vectors i and j a unit vector? If no, can you multiply this sum by a scalar number to get a unit vector?
11.
Let A
12.
Can you have A B A B with A 0 and B 0? What if one of the two vectors is zero?
13.
If A B 0 , can you say that (a) A
ˆ
ˆ
3 i 4 j . Write four vectors B such that A B but A = B.
B , (b) A
B?
5 i 4 j and B 7.5 i 6 j . Do we have B k A ? Can we say
14.
Let A
1.
A vector is not changed if (A) it is rotated through an arbitary angle (C) it is cross multiplied by a unit vector
B
A
k?
(B) it is multiplied by an arbitary scalar (D*) it is slid parallel to itself
dks bZ z lfn'k ifjofrZ r ugha gks rk gS] ;fn & tk;sA (A) bldks fdlh LoS fPNd dks .k ls ?kq ek;k (B) bldks fdlh LoS fPNd vfn'k jkf'k ls xq .kk fd;k tk;sA (C) bldk fdlh ,dka d lfn'k ls otz&xq .ku fd;k tk;sA (D) bldks Lo;a ds lekukUrj foLFkkfir fd;k tk;sA 2.
Which of the sets given below may represent the magnitudes of three vectors adding to zero ? (A) 2, 4, 8 (B) 4, 8, 16 (C*) 1, 2, 1 (D) 0.5, 1, 2
fuEu es a ls dkS ulk leq P; O;Dr djrk gS fd blds rhu lfn'kks a ds ;ks x dk ifj.kkeh 'kw U; iz kIr gks xk & (A) 2, 4, 8
(B) 4, 8, 16
(C) 1, 2, 1
manishkumarphysics.in
(D) 0.5, 1, 2 Page # 5
Chapter # 2
Physics & Mathematics
3.
A rFkk B (A) <
(D) > ;fn A = B
dk ifj.kkeh]
A ls dks .k rFkk B ls (B) < ;fn A < B
The component of a vector is (A) always less than its magnitude (C) always equal to its magnitude
dks .k cukrk gS & (C) < ;fn A > B
(B) always greater than its magnitude (D*) none of these
fdlh lfn'k dk ?kVd & (A) lnS o blds ifjek.k ls de gks rk gS (C) lnS o blds ifjek.k ds cjkcj gks rk gS 5.
(D) < if A = B
4.
The result of A and B makes an angle with A and with B , (A) < (B) < if A < B (C*) < if A > B
(B) lnS o
blds ifjek.k ls vf/kd gks rk gS (D) bues a ls dks bZ ugha
A vector A points vertically upward and B points towards north. The vector product A x B is (A*) along west (B) along east (C) zero (D) vertically downward
lfn'k A m/okZ/ kj Åij dh vks j ba fxr gS rFkk B mÙkj dh vks jA lfn'k xq .ku A x B gS & (A) if'pe dh vks j (B) iw oZ dh vks j (C) 'kw U; (D) m/okz Z/ kj Åij dh vks j
6.
The radius of a circle is stated as 2.12 cm. Its area should be written as (A) 14 cm2 (B*) 14.1 cm2 (C) 14.11 cm2 fdlh o` r dh f=kT;k 2.12 ls eh crkbZ tkrh gSA bldk {ks =kQy n'kkZ ;k tk;s xk &
(D) 14.1124 cm2
(A) 14 ls eh 2
(D) 14.1124 ls eh 2
(B) 14.1 ls eh 2
(C) 14.11 ls eh 2
Objective - II 1.
A situation may be described by using different sets of co-ordinate axes having different orientations. Which of the following do not depend on the orientation of the axes ? (A*) the value of a scalar (B) component of a vector (C*) a vector (D*) the magnitude of a vector
ifjfLFkfr funs 'kka Z d v{kks a ds fofHkUu >q dko okys dbZ funs 'k Z v{kks a ds fofHkUu leq P;ks a }kjk O;Dr dh tkrh gSA fuEu es a ls dkS ulh v{kks a ds >q dko ij fuHkZ j ugha djrh gS & (A) fdlh vfn'k dk eku (B) fdlh lfn'k dk ?kVd (C) dks bZ lfn'k (D) fdlh lfn'k dk ifjek.k
2.
Let C = A + B
(A) | C | is always greater than | A | (C) C is always equal to A + B
ekukfd
C
lnS o | A | ls cM+ k gks xk lnS o A + B ds rq Y; gks xk
(B) ;g
lEHko gS fd | C | | A | rFkk | C | | B | gksA (D) C dHkh Hkh A + B ds cjkcj ugha gks xk
3.
A B
(A) | C | (C) C
(B*) It is possible to have | C | < | A | and | C |< | B | (D) C is never equal to A + B
Let the angle between two nonzero vectors A and B be 120o and its resultant be C . (A) C must be equal to | A - B| (C*) C must be greater than | A - B |
(B) C must be less than |A - B| (D) C may be equal to | A - B |
ekukfd nks v'kw U; lfn'kks a A rFkk B ds e/; dks .k 120° gS rFkk bldk ifj.kkeh C gS & (A) C fuf'pr :i ls | A – B | ds cjkcj gks xk (B) C fuf'pr :i ls | A – B | ls de gks xk (C) C fuf'pr :i ls |A – B| ls vf/kd gks xk (D) C, |A – B| ds cjkcj gks ldrk gSA
4.
The x-component of the resultant of several vectors (A*) is equal to the sum of the x-component of the vectors (B*) may be smaller than the sum of the magnitudes of the vectors (C) may be greater than the sum of the magnitudes of the vectors (D*) may be equal to the sum of the magnitudes of the vectors. dq N lfn'kks a ds ifj.kkeh dk x- ?kVd & (A) lfn'kks a
ds x- funs 'kka Z dks a ds ;ks x ds rq Y; gks xk (C) lfn'kks a ds ;ks x ds ifjek.k ds cjkcj gks ldrk gS
(B) lfn'kks a
ds ;ks x ds ifjek.k ds ;ks x ls de gks ldrk gS (D) lfn'kks a ds ;ks x ds ifjek.k ds cjkcj gks ldrk gSA
manishkumarphysics.in
Page # 6
Chapter # 2
Physics & Mathematics
5.
1. Sol.
The magnitudes of the vectors product of two vectors A and B may be (A) greater than A B (B*) equal to AB (C*) less than AB
nks lfn'kks a
A rFkk B
ds lfn'k xq .ku dk ifjek.k gks ldrk gS &
(A)AB ls
vf/kd
(B)AB ds
cjkcj
(C) AB ls
de
(D*) equal to zero.
(D) 'kw U;
ds cjkcj
A vector has component along the X-axis equal to 25unit and along the Y-axis equal to 60 unit. Find the magnitude and direction of the vector. The given vector is the resultant of two perpendicular vectors, one along the X-axis of magnitude 25 unit and the other along the Y-axis of magnitude 60 unit. The resultant has a magnitude A given by
A
(25)2 (60)2 2 25 60 cos 90º =
(25)2
(60)2 = 65.
The angle between this vector and the X-axis is given by tan = 2.
60 . 25
Find the resulatant of the three vectors shown in figure Y
2.0m
0 m 5.
3.0m
X
Sol.
Take the axes as shown in the figure, The x-component of the 5.0 vector = 5.0 m cos 37º = 4.0 m, the x-component of the 3.0 m vector = 3.0 m and the x-component of the 2.0 m vector = 2.0 m cos 90º = 0. Hence, the x-component of the resultant = 4.0 m + 3.0 m + 0 = 7.0 m. The y-component of the 5.0 m vector = 5.0 m sin 37º = 3.0 m, the y-component of the 3.0 m + 0 = 7.0 m. The y-component of the 3.0 vector = 0 and the y-component of the resultant = 3.0 m + 0 + 2.0 m = 5.0 m. The magnitude of the resultant vector =
(7.0 m)
(5.0 m)2
= 8.6 m. If the angle made by the resultant with the X-axis is , then tan =
y - component 5.0 = or, q = 35.5º, x component 7.0
manishkumarphysics.in
Page # 7
Chapter # 2 3. Sol.
Physics & Mathematics
The sum of the three vectors shown in figure is zero. Find the magnitudes of the vectors OB and OC . Take the axes as shown in the figure. Y
C
45º X 0
B 5m
A
The x-component of OA = (OA) cos 90º = 0. The x-component of OB = (OB) cos 0º = 0.
The y-component of OC = (OC) cos 135º = –
1 2
OC.
Hence, the x-component of the resultant = OB –
1
OC.
....(i)
2 It is given that the resultant is zero and hence is x-component is also zero. From (i), OB =
1
OC.
....(ii)
2 The y-component of OA = OA cos 180º = – OA. The y-component of OB = OB cos 90º = 0.
1
The y-component of OC = OC cos 45º =
2
OC.
Hence, the y-component of the resultant
1
=
OC – OA
2
.....(iii)
As the resultant is zero, so is its y-component. From (iiii),
1 2 From (ii), OB =
4.
1 2
OC = OA, or OC =
2 OA = 5 2 m.
OC = 5 m.
The magnitude of vectors OA , OB and OC in figure are equal. Find the direction of OA + OB – OC , Y
C
A 45º
30º 0
X 60º 5m
B
Sol.
Let OA = OB = OC = F.
manishkumarphysics.in
Page # 8
Chapter # 2
Physics & Mathematics
3
x-component of OA = F cos 30º = F
.
2
F 2
x-component of OB = F cos 60º =
F
x-component of OC = F cos 135º = –
2
x-component of OA + OB – OC =
F 3 F F 2 2 2
=
F ( 3 +1+ 2
2 ).
y-component of OA = F cos 30º =
F . 2
y-component of OB = F cos 150º =
F 3
y-component of OC = F cos 45º = –
2 F 2
y-component of OA + OB – OC
F 3 F F = 2 2 2 =
F (1 – 2
3 –
2 ).
Angle of OA + OB – OC with the X-axis
F (1 3 1 2 = tan F (1 3 2
5.
2) = tan –1
2)
(1 3
2) (1 3 2 )
Find the resultant of the free vectors OA , OB and OC shown in figure. Radius of the circle is R. C B
45º 45º
A
0
Sol.
OA = OC.
OA + OC is along OB (bisector) and its magnitude is 2R cos 45º = R/2. ( OA + OC ) + OB is along OB and its magnitude is R 2 + R = R(1 +
2 ).
manishkumarphysics.in
Page # 9
Chapter # 2 6.
Physics & Mathematics
The resultant of vectors OA and OB is perpendicular to OA (figure). Find the angle AOB. Y B
6m
0
Sol.
X
4m
A
Take the dotted lines as X, Y axes, x-component of OA = 4 m, x-component of
OB = 6m cos x-component of the resultant = (4 + 6 cos ) m But it is given that the resultant is along Y-axis. Thus, the x-component of the resultant = 0 4 + 6 cos = 0
cos = –
or ,
2 3
7.
Write the unit vector in the direction of A = 5 i + j – 2 k .
Sol.
| A | =
52
12 ( 2)2 = 30 .
The required unit vector is
5
1
+ 30 i
| A | 2
A
30
j –
. 30 k
8.
If | a + b | = | a – b | show that a
Sol.
We have | a + b |2 = ( a + b ).( a – b )
b.
= a .a + a .b +b .a + b .b
= a2 + b2 = 2 a . b .
Similarly,
| a – b |2 = ( a – b ).( a – b )
= a2 + b2 – 2 a . b
If
| a + b | = | a – b | ,
a2 + b2 + 2 a . b = a2 + b2 – 2 a . b
or ,
a.b = 0
or ,
a
b.
9.
If a = 2 i + 3 j + 4 k and b = 4 i + 3 j + 2 k , find the angle beween a and b .
Sol. We have
a . b = ab cos
or ,
a.b cos = ab
where is the angle between a and b .
Now
Also
a . b = a xbx + ayby + azbz = 2 × 4 + 3 × 3 + 4 × 2 = 25.
a= =
ax
2
ay 2 az 2
4 9 16 =
29
manishkumarphysics.in
Page # 10
Chapter # 2
Physics & Mathematics
and
bx
b=
b y 2 b z 2 = 16 9 4 =
29 .
25 29
cos =
Thus ,
2
25 = cos –1 29 .
or ,
10.
If A = 2 i – j + 7 k , B = i + 2 k and C = j – k , find A .( B × C ).
Sol.
B × C = ( i + 2 k ) × ( j – k )
= i × ( j – k ) + 2 k × ( j – k )
= i ×
j – i × k + 2 k × j – 2 k × k
= k + j – 2 i – 0 = – 2 i + j + k
A .( B × C ) = (2 i – 3 j + 7 k ) . (– 2 i + j + k ) = (2) (–2) + (–3)(1) + (7)(1) = 0.
11.
Thevolume of a sphere is given by 4 R3 3 where R is radius of the sphere. (a) Find the rate of change of volume with respect to R. (b) Find the change in volume of the sphere as the radius is increased from 20.0 cm to 20.1 cm . Assume that the rate does not appreciably change between R = 20.0 cm to R = 20.1 cm. V=
Sol.
(a)
V= dV
4 3
R3 4
=
d
4
. 3R 2 = 4 R2. dR 3 dr 3 (b) At R = 20 cm , the rate of change of volume with the radius is or ,
(R)3 =
dV = 4 R2 = 4 (400 cm 2) dR = 1600 cm 2. The change in volume as the radius changes from 20.0 cm to 20.1 cm is dV R dR = (1600 cm 2) (0.1 cm) = 160 cm 3 .
V =
12.
Find the derivative of the following functions with respect to x (a) y = x 2 sinx , (b) y =
Sol.
(a) y = x2 sinx
sin x and (c) y = sin (x)2. x
dy d d = x2 (sinx) + (sinx) (x2) ] dx dx dx = x2 cosx + (sinx) (2x) = x (2 sinx + x cosx). (b) y =
dy dx
sin x x x =
d dx (sin x ) sin x dx dx x2 manishkumarphysics.in
Page # 11
Chapter # 2
Physics & Mathematics =
x cos x sin x x2
.
d dy d( x 2 ) 2 = (sinx ). dx dx 2 dx
(b)
= cosx2 (2x) = 2x cosx2.
13.
Find the maximum or minimum values of the function y = x +
Sol.
y=x+ dy dx
=
d
1 for x > 0. x
1 x (x) +
d
dx dx –2 = 1 + (– x ) =1–
(x –1)
1
. x2 For y to be maximum or minnimum , dy =0 dx or , Thus
1–
1
=0 x2 x = 1 or – 1.
For x > 0 the only possible maximum or minimum is at x = 1 , At x = 1 , y = x +
1 = 2. x
1 is very large because of the term x. Thus , x = 1 must correspond to a minimum. Thus, x y has only a minimum for x > 0. This minimum ocurs at x = 1 and the minimum value of y is y = 2. Near x = 0 , y = x +
Figure shows the curve y = x2. Find the area of the shaded part between x = 0 and x = 6.
14.
x
x
Sol.
The area can be divided into strips by drawing ordinates between x = 0 and x = 6 at a regular interval of dx. Consider the strip between the ordinates at x and x + dx. The height of this strip is y = x 2. The area of this strip is dA = ydx = x 2 dx. The total area of the shaded part is obtained by summing up these strips -areas x varying from 0 to 6. Thus, 6
A =
x dx 2
0
6
x3 216 0 = = = 72 . 3 3 0 t
15.
Evaluate A sin t dt where A and are constants. 0
t
Sol.
A sin t dt 0
manishkumarphysics.in
Page # 12
Chapter # 2
Physics & Mathematics t
cos t A =A = (1 – cost). 0 16.
The velocity v and displacement x of a particle executing simple harmonic motion are related as
dv = – 2 x. dx At x = 0 , v = v0 . Find the velocity v when the displacemet becomes x. Sol. We have v
or ,
dv
= – 2 x dx v dv = – 2 x dx
v
v
or ,
x
v dv =
v0
2
x dx
.....(i)
0
When summation is made on – w 2 x dx the quantity to be varied is x. When summation is made on vdv the quantityto be varied is v. As x varies from 0 to x the velocityvaries from v0 to v. Therefore , on the left the limits of integration are from v 0 to v and on the right they are from 0 to x. Simplifying (i) , v
1 2 2 v = – 2 v0 or , or ,
v
1 2 2 v v0
1 2 x2 2 2 (v –v0 ) = – 2 2 2 2 v = v 0 – 2 x2
or ,
v=
v0
2
2 x 2 .
17.
The charge flown through a circuit in the time interval between t and t + dt is given by dq = e – t / dt where is a constant. Find the total charge flown through the circuit between t = 0 to t = .
Sol.
The total charge flown is the sum of all the dq's for t varying from t = 0 to t = t0 . Thus , the total charge flown is
Q=
e
t /
dt
0
et / 1 = 1 . = e 1/ 0 18. Sol.
Evaluate (21.6002 + 234 + 2732.10) × 13.
21.6002
22 234 2732
234 2732.10
2988
The three numbers are arranged with their decimal points aligned (shown on the left part above). The column just left to the decimal has 4 as the doubtful digit. Thus , all the numbers are rounded to this column. The rounded numbers are shown on the rigid part above. The required expression is 2988 × 13 = 38844. As 13 has only two significant digits the product should be rounded off after two signigicant digits. Thus the result is 39000.
manishkumarphysics.in
Page # 13
Chapter # 2
Physics & Mathematics
EXERCISE
1.
A vector A makes an angle of 20º and B makes an angle of 110º with the X-axis. The magnitudes of these vectors are 3 m and 4m respectively. Find the resultant. x-v{k
ls ,d lfn'k A , 20° dks .k cukrk gS rFkk nw ljk lfn'k B , 110° dks .k cukrk gSA bu lfn'kks a ds ifjek.k Øe'k% o 4 gSA ifj.kkeh Kkr dhft;sA
Let A and B be the two vectors of magnitude of 100 unit and inclined to the X-axis at angles 30º and 60º respectively, find the resultant.
ekukfd nks lfn'kks a Kkr dhft;sA 3.
3
2.
A o B ds
iz R;s d ifjek.k 10 bdkbZ gSA ;fn budk x-v{k ls >q dko Øe'k%
30° rFkk 60° gS]
ifj.kkeh
Add vectors A , B and C each, having magnitude of 100 unit and inclined to the X-aixs at angles 45°, 135°
and 315° respectively.
A , B rFkk C iz R;s d
dk ifjek.k 100 bdkbZ rFkk buds
x-v{k
ls >q dko Øe'k% 45°, 135° rFkk 315° gS] budk ;ks x dhft;sA
4.
Let a
4 i 3 j and b 3 i 4 j (a) Find the magnitudes of
ekukfd
ˆ
ˆ
a 4i
ˆ
ˆ
3 j rFkk b 3 i 4 j gSA Kkr dhft;s &
ˆ
ˆ
ˆ
(A) a
5.
ˆ
(C) a b
(B) b
(D) a b
Refer to figure. Find (a) the magnitude, be (b) x and y components and (c) the angle with the resultant of OA ,
BC and a DE .
fp=k ds vk/kkj ij Kkr dhft;s & (a) ifjek.k (b) x rFkk y ?kVd
(c) OA , BC ,oa DE ds 6.
ifj.kkeh dk x-v{k ls dks .k
Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them if the magnitude of the resultant is (a)1 unit, (b) 5 unit and (c) 7 unit
nks lfn'kks a ds ifjek.k Øe'k% 3 bdkbZ o 4 bdkbZ gSA buds e/; dks .k fdruk gks xk ;fn budk ifj.kkeh gS & (a) 1 bdkbZ (b) 5 bdkbZ (c) 7 bdkbZ 7.
A spy report about a suspected car reads as follows. "The car moved 2.00 km towards east, made a perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped". Find the displacement of the car.
,d tklw l lw puk ns rk gS fd la ns g okyh dkj dh xfrfof/k bl iz dkj gS] dkj iw oZ dh vks j 2.00 fdeh- pyh] ck;ha vks j ,d yEcor~ ?kq eko fy;k] 500 eh- nkS M+ h] nk;ha vks j ledks .k ij ?kq eh] 4.00 fdeh pyh rFkk :d x;hA dkj dk foLFkkiu Kkr dhft;sA 8.
A carrom board (4 ft × 4ft square) has the queen at the centre. The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen (a) from the centre to the front edge, (b) from the front edge to the hole and (c) from the centre to the hole, ,d ds jecks MZ (4 Qq V x 4 Qq V oxkZ dkj ) es a ds Unz ij Dohu gSA LVª kbdj ls pks V yxus ij] Dohu lkeus okys fdukjs dh vks j
xfr'khy gks rh gS] iyVrh gS rFkk LVª kbd js [kk ds ihNs fLFkr Ns n es a pyh tkrh gSA Dohu ds foLFkkiu dk ifjek.k Kkr dhft;sA (a) ds Unz fcUnq ls lkeus okys fdukjs rd (b) lkeus okys fdukjs ls Ns n rd rFkk manishkumarphysics.in
Page # 14
Chapter # 2
Physics & Mathematics
(c) ds Unz
ls Ns n rd
2 4 x 2x 2 – x = 2x 3x = 2
Sol.
x=
S1 >
2
2
S2 4
4
+ 2 j 3 i
ˆ
ˆ
4
|S1| =
9
10 Ans.
) – 4 j 3 i
ˆ
ˆ
4
– 4 j 3 i
ˆ
ˆ
3
2
4
2
4 =
| S 2 | = (2– =
>
>
S3
3
S1 =
2-x
x
10
3
| S 3 | = S1 S 2 =
2 4 i 2 j i – 4 j 3 3 ˆ
ˆ
ˆ
ˆ
| S 3 | = 2 i – 2 j
ˆ
ˆ
| S3 | = 2 2 9.
A mosquito net over a 7 ft × 4 ft bed is 3 ft high. The net has a hole at one corner of the bed through which a mosquito enters the net. It flies and sits at the diagonally opposite upper corner of the net.(a) Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length of the bed as the X-axis, its width as the Y-axis, and vertically up as the Z-axis, write the components of the displacement vector. 7 Qq V x 4 Qq V vkdkj okys iya x ij 3 Qq V Å¡ ph ePNjnkuh yxh gq bZ gSA iya x ds ,d dks us ij ePNj nkuh es a ,d Ns n
gS] ftlls ,d ePNj] tkyh es a iz fo"V gks tkrk gS a ;g mM+ rk gS rFkk tkyh ds fod.kZ r% foijhr dks us ij cS B tkrk gSA (a) ePNj ds foLFkkiu dk ifjek.k Kkr dhft;sA (b) Ns n dks ew y fcUnq] iya x dh yEckbZ dks x-v{k rFkk pkS M+ kbZ dks y-v{k ,oa m/oZ/ kj Åij dh vks j z-v{k ekurs gq ,] foLFkkiu lfn'k ds ?kVd fyf[k;sA 10.
Suppose a is a vector of magnitude 4.5 unit due north. What is the vector (a) 3 a , (b) –4 a ?
ekukfd lfn'k a dk ifjek.k 4.5 bdkbZ] mÙkj dh vks j gSA lfn'k D;k gks xs a \
(A) 3a
11.
(B)
4a
Two vectors have magnitudes 2m and 3m.The angle between them is 60º. Find (a) the scalar product of the two vectors, (b) the magnitude of their vector product.
nks lfn'kks a ds ifjek.k 2 eh- o 3 eh- gSA buds e/; dks .k 60° gSA Kkr dhft;sA (a) nks uks a lfn'kks a dk vfn'k xq .ku (B) buds lfn'k xq .ku dk ifjek.k 12.
LetA1, A2, A3, A4, A5, A6, A1 be a regular hexagon. Write the x-components of the vectors represented by the six sides taken in order. Use the fact that the resultant of these six vectors is zero, to prove that cos 0 + cos
3
+ cos
2 3 4 5 + cos + cos + cos = 0 . Use the known cosine values to verify the result. 3 3 3 3
ekukfd A1, A2, A3, A4, A5, A6, A1 ,d fu;fer "kV~ H t kq gSA bldh N% Hkq tkvks a dks Øe es a j[krs gq ,] Hkq tkvks a ls O;Dr lfn'kks a ds x- ?kVd fyf[k;sA ;g rF; iz ;q Dr djrs gq , bu N% lfn'kks a dk ifj.kkeh 'kw U; gS] fl) dhft;s fd & cos 0 + cos
3
+ cos
2 3 4 5 + cos + cos + cos =0 3 3 3 3
manishkumarphysics.in
Page # 15
Chapter # 2
Physics & Mathematics
ifj.kke dks dks T;kvks a ds eku dk mi;ks x djds lq fuf'pr dhft;sA
Let a 2 i 3 j 4k and b
13.
ˆ
ekukfd
ˆ
3 i 4 j 5k . Find the angle between them. ˆ
ˆ
ˆ
a 2 i 3 j 4k rFkk b 3 i 4 j 5k
ˆ
ˆ
14.
ˆ
ˆ
ˆ
ˆ
ˆ
buds e/; dks .k Kkr dhft;sA
Prove that A . ( A x B) 0
fl) dhft;s fd A . ( A x B) 0
15.
If A
2 i 3 j 4k and B 4 i 3 j 2k , find A x B . ˆ
ˆ
ˆ
ˆ
;fn A 2 i 3 j 4k rFkk
ˆ
ˆ
B 4 i 3 j 2k Kkr
ˆ
ˆ
dhft;s
ˆ
ˆ
ˆ
ˆ
A x B
If A, B, C are mutually perpendicular, show that C x ( A x B) 0 . Is the converse true?
16.
;fn A, B, C ijLij yEcor~ gS] O;Dr dhft;s fd C x ( A x B) 0 D;k bldk O;q RØe Hkh lR; gS \
17.
A particle moves on a given straight line with a constant speed u. At a certain time it is at a point P on its
straight line path. O is a fixed point. Show that OP x v is independent of the position?
,d d.k nh xbZ ljy js [kk ds vuq fn'k fu;r pky v ls xfr'khy gSA fdlh fuf'pr le; t ij ;g ljy js [kh; iFk ds fcUnq P ij
gSA O ,d fLFkj fcUnq gSA O;Dr dhft;s fd
OP x v , P dh
fLFkfr ij fuHkZ j ugha djrk gSA
The force on a charged particle due to electric and magnetic fields is given by F qE q v x B . Suppose E
18.
is along the X-axis and B along the Y-axis. In what direction and with what minimum speed u should a positively charged particle be sent so that the net force on it is zero?
fdlh vkos f'kr d.k j fo|q r ,oa pq Ecdh; {ks =kks a ds dkj.k cy]
F qE qv x B
}kjk O;Dr fd;k tkrk gSA ekukfd E ]
x-
v{k ds vuq fn'k gS rFkk B , y-v{k ds vuq fn'k gSA ,d /kukos f'kd d.k dks fdl fn'kk es a rFkk fdl U;w ure pky v ls xfr djokbZ tk;s fd bl j ifj.kkeh cy 'kw U; gks \
19.
A . B C . B
20.
Give an example for which A . B C . B but A
C
fdUrq A C dk ,d mnkgj.k nhft;sA
Draw a graph from the following data. Draw tangents at x = 2, 4, 6 and 8. Find the slopes of these tangents. Verify that the curve drawn is y = 2x 2 and the slope of tangent is tan =
dy = 4x. dx
fuEufyf[kr vka dM+ aks dk mi;ks x djds ys [kkfp=k cukb;sA x = 2 , 4 , 6 rFkk 8 ij Li'kZ js [kk,a [kha ph;s aA bu Li'kZ js [kkvks a ds dy
21.
1 2
2 8
3 18
4 32
5 50
6 72
A curve is represented byy = sin x. If x is changed from ,d
7 98
8 128
9 162
oØ y = s i n x 3
to
3
dx
= 4x gSA
10 200
+
100
, find approximately
the change in y
manishkumarphysics.in
Page # 16
Chapter # 2
Physics & Mathematics
}kjk O;Dr fd;k tkrk gSA ;fn x dk eku ifjofrZ r djds
ls 3
3 100
dj fn;k tkrk gS] y ds eku es a vuq ekfur ifjorZ u
Kkr dhft;sA 22.
The electric current in a charging R-C circuit is given by i = io –t/RC where io, R and C are constant parameters of the circuit and t is time. Find the rate of change of current in at (a) t = 0, (b) t = RC, (c) t = 10 RC. R-C ifjiFk
es a vkos 'ku ds le; /kkjk i = i0e –t/RC }kjk O;Dr dh tkrh gS] tgk¡ i0, R rFkk C ifjiFk dh fu;r jkf'k;k¡ gS rFkk t le; gSA le; (a) t = 0, (b) t = RC, (c) t = 10 RC ij ifjiFk es a /kkjk ds eku es a ifjorZ u dh nj Kkr dhft;sA 23.
The electric current in a discharging R-C circuit is given by R-C i = i 0 e –t/RC where io, R and C are constant parameters and t is time. Let i o = 2.00 A, R = 6.00 × 105 and C = 0.500 F. (a) Find the current at t = 0.3s. (b) Find the rate of change of current at t = 0.3 s. (c) Find approximately the current at t = 0.31 s.
ifjiFk ds fy, foltZ u dky es a fo|q r /kkjk i = i0 e –t/RC }kjk O;Dr dh tkrh gS] tgka gS rFkk t le; gSA ekukfd i0 = 2.00 A, R = 6.00 x 105 rFkk C = 0.500 µF (a) t = 0.3 ls ij /kkjk Kkr dhft;sA (b) t = 0.3 ls ij /kkjk es a ifjorZ u dh nj Kkr dhft;sA (c) t = 0.31 ls- ij /kkjk dk eku ¼yxHkx½ Kkr dhft;s A
i0, R rFkk C ifjiFk
dh fu;r jkf'k;ka
24.
Find the area bounded under the curve y = 3x2 + 6x + 7 and the X-axis with the ordinates at x = 5 and x = 10. oØ y = 3 x2 + 6x + 7 rFkk funs Z 'kka dks a x = 5 rFkk x = 10 ds chp x-v{k }kjk ifjc) {ks =kQy Kkr dhft;sA
25.
Find the area enclosed by the curve y = sin x and the X-axis between x = 0 and x = .
oØ y = s i n x rFkk x = 0 ,oa
x = ds
chp x-v{k }kjk ifjc) {ks =kQy Kkr dhft;sA
26.
Find the area bounded by the curve y = e –x , the X-axis and the Y-axis oØ y = e –x, x-v{k rFkk y-v{k }kjk ifjc) {ks =kQy Kkr dhft;sA
27.
A rod of length L is placed along the X-axis between x = 0 and x = L. The linear density (mass/length) p of the rod varies with the distance x from the origin as p = a + bx. (a) Find the SI units of a and b. (b) Find the mass of the rod in terms of a,b,and L. L yEckbZ dh ,d NM+ x-v{k ds vuq fn'k x = 0 rFkk x = L ds e/; j[kh gq bZ gSA NM+ dk jS f[kd ?kuRo ¼nz O;eku@yEckbZ½ P
ew y fcUnq ls nw jh x ds lkFk P = a + bx ds vuq lkj ifjofrZ r gks rk gSA (a) a rFkk b dh SI bdkbZ ;ka Kkr dhft;sA (b) a, b rFkk L ds inks a es a NM+ dk nz O;eku Kkr dhft;sA
28.
The momentum p of a particle changes with time t according to the relation
dp = (10N) + (2N/s)t. If the dt
momentum is zero at t = 0, what will the momentum be at t = 10s ?
fdlh d.k dk la os x] le; ds lkis{k lw =k
dp = (10 U;w Vu ) + ( 2 U;w Vu@ls -)t ds dt
vuq lkj ifjofrZ r gks rk gSA ;fn t ds vuq lkj
ifjofrZ r gks rk gSA ;fn t = 0 ij la os x 'kw U; gks] rks t = 1 0 ls- ij la os x fdruk gks xk \ 29.
The changes in a function y and the independent variable x are related as dy/dx = x 2. Find y as a function of x.
Qyu y es a ifjorZ u rFkk Lora =k pj x es a lEcU/k dy/dx = x2 gSA x ds Qyu ds :i es a y Kkr dhft;sA 30.
Write the number of significant digits in
lkFkZ d la [;k,a fyf[k;s & (a) 1001
(b) 100.1
(c) 100.10
manishkumarphysics.in
(d) 0.001001
Page # 17
Chapter # 2 31.
Physics & Mathematics
A metre scale is graduated at every millimetre. How many significant digits will be there in a length measurement with this scale?
,d ehVj iS ekuk iz R;s d feyhehVj es a foHkkftr gSA bl iS ekus }kjk ekih xbZ yEckbZ es a fdruh lkFkZ d la [;k,a gS \ 32.
Round the following numbers to 2 significant digits.
fuEufyf[kr la [;kvks a dks 2 lkFkZ d va dks a rd lhfer dhft;s & (a) 3472 33.
(b) 84.16
(c) 2.55
(d) 28.5
The length and the radius of a cylinder measured with a slide callipers are found to be 4.54 cm and 1.75 cm respectively. Calculate the volume of the cylinder.
LykbM dS fyijlZ dh lgk;rk ls ,d cs yu dh yEckbZ rFkk f=kT;k ekih tkrh gS] buds eku Øe'k% 4.54 ls eh ,oa 1.75 ls eh iz kIr gks rs gSA cs yu dh vk;ru dh x.kuk dhft;sA 34.
The thickness of a glass plate is measured to be 2.17 mm, 2.17 mm and 2.18 mm at three different places. Find the average thickness of the plate from this data.
,d dka p dh ifV~ Vdk dh rhu fHkUu&fHkUu LFkkuks a ij ekih x;h eks VkbZ ;ka 2.17 feeh- 2.17 feeh rFkk 2.18 feeh gSA bu vka dM+ ks a dh lgk;rk ls ifV~Vdk dh vkS lr eks VkbZ Kkr dhft;sA 35.
The length of the string of a simple pendulum is measured with a metre scale to be 90.0 cm. The radius be 2.13 cm using measurements with a slide callipers. What is the effective length of the pendulum? (The effective length is length of the pendulum between the point of suspension and the centre of the bob.) ,d ehVj
iS ekus dh lgk;rk ls ljy yks yd dh Mks jh dh yEckbZ 90.0 ls eh ekih tkrh gSA LykbM dS fyilZ ls ekius ij ckW c ,oa gq d dh lfEefyr yEckbZ 2.13 ls eh iz kIr gks rh gSA yks yd dh izHkkoh yEckbZ fdruh gS \ ¼fuyEcu fcUnq rFkk ckW c ds ds Unz ds chp dh nw jh izHkkoh yEckbZ dgykrh gS½
manishkumarphysics.in
Page # 18
