Vidyamandir Classes
Solutions to Advanced Problem Package | Physics
KINEMATICS [MOTION IN 1 & 2 DIMENSIONS]
1.(D)
v
dv
a bx
dx
v 2 2ax bx 2
Now v = 0 at x = 0 and x = l (l : the distance between the stations)
l=
2a
b
also v max
a
b
2.(A)
The particle is moving with constant acceleration therefore velocity time graph of the particle will be straight line. From
t = 0 s to t = 1 s slope of given displacement-time graph is negative and decreasing. From t = 1 s to t = 2 s slope is
positive and is decreasing. At time t = 0 and t = 2 s slope of displacement time graph is zero therefore velocity at that
moment will also be zero.
3.(C)
The graph, given in the question shows that a s
dv
Ks (where K is some constant)
dt
v
dv
v Ks
vdv K
0
ds
v 2 Ks 2
v Ks
v K 's
K K'
s
0 sds
vs
4.(C)
2u
a
2 (u sin )
T
g cos
T
In both the cases ,
(ii)
u and a is same, so the time
T1 T2
2u sin
T
g cos
v = u + at
APP | Kinematics [Motion in 1-D & 2-D]
1
Solution | Physics
Vidyamandir Classes
u x u cos α g sin θ t
u y u sin α g cos θ t
For maximum height to the incline Uy = 0
u sin α
1
t
;
h ut at 2
g cosθ
2
u sin α g cosθt
u sin α 1
u sin α 2
g cosθ
h = u sin α
g cosθ 2
g cosθ
h
u 2 sin 2 α
2g cosθ
As u y is same for both the cases and ay is same so h1 h 2
(iii)
1
R (u cos )T ( g sin )T 2 ;
2
u2 u cos g sin t
;
u sin α
u 2 u cos α g sin θ
g cosθ
;
u sin α
u1 u cos α gsin θ
g cos θ
u 2 sin 2 α
2g cosθ
As T is same for both
u1 u cos g sin t
Clearly u1 u 2
5.(B)
6.(A)
u
V
The distance travelled during the period A arrives at nearest distance = d cos
The relative velocity V makes an angle with AB, where cos
P1: A(5, 3)
d cos du
2
V
V
P2 : (B) (7, 3)
v1 2iˆ 3jˆ
v2 xiˆ yjˆ
7.(B)
Required time =
(A)
At t = 2 seconds they collide. It means that their S is same.
8.(B)
5 22 7 x2,
3 y2 3 32
x = 1, y = 3
Displacements of B and C in horizontal direction is same.
VC = VB cos60º
vC 1
……(1)
vB 2
Displacement of A and B in vertical direction is same to
v A t vB sin 60º t
vA
3
vB
2
From (2) and (1)
….(2)
APP | Kinematics [Motion in 1-D & 2-D]
2
Solution | Physics
Vidyamandir Classes
9.(B)
vA : vB : vC 3 : 2:1
dv
a v vdv adx
dx
v
v2 f
Area under a –x curve.
2 v
i
10.(B) Let speed of man be ‘v’ m/sec.
Then, v (time of flight of projectile) = 9 m
v
11.(C) (1)
(2)
2u sin θ
9 ;
g
As the bird starts it’s motion at the same time as the ball, therefore, Vx = u.
For all values of ‘x’ except h = ‘Hmax”, the ball will touch the bird twice.
h = Hmax =
u y2
13.(A) Average velocity =
2u sin
g
1101 / 2
10
2u
g
2h
g
2 sec
roots are tOA and tOB
gt 2 2u sin . t h 0
2
t AB
vx v y
displacement S x, AB
ux u cos
t AB
t AB
h
g
u sin t t 2
2
2
Range = 2
2g
12.(A) Time period = time of projectile =
v = 5 m/sec
2
2 u sin gh
g
t AB diff. in roots
2 u 2 sin 2
g
u 2 sin 2
2
2u sin
g
T/ 2.
As acc. is uniform = g av. Acc. in any interval of time is also g .
Direction of ins. Velocities at A & B are different.
3u
a
4u
14.(C) S x t t 2
;
Sy
t 5t 2
5
2
5
3u
a
4u
at
7u
14u
t t2
t 5 t2
5t
at 10t
5
2
5
2
5
5
3u
4u
3u
4u
Vx
at ;
V y
10t ;
at
10t
5
5
5
5
y
u
10t at
5 add.
;
20t 3u
14u
10t at
5
x
26
20 t
4 26
10 t a t
a 10
m/ s
a
m / s2
3 3
35
3
15.(D) Let V : velocity of buggy
Velocity highest point of rear wheel = 2V
APP | Kinematics [Motion in 1-D & 2-D]
3
wr. t buggey
S y 2b 2a
g
2
t2
Solution | Physics
Vidyamandir Classes
2
S x C 2 b a Vt
16.(ABCD)
4 b a
g C b a C b a
V2
V
g
C b a C b a
4 b a
Change in velocity = final velocity – initial velocity
u cos θiˆ u cos θiˆ u sin θjˆ u sin θjˆ
(A) is correct
= u cos ˆi
(B)
is correct.
Change in velocity = final velocity – initial velocity
= u cos ˆi usin ˆj u cos iˆ u sin ˆj = 2u sin ˆj
(C)
is also correct.
Rate of change of momentum = force
Constant gravitational force is acting on the projectile.
(D)
is also correct.
Average velocity = (total displacement)/(time taken) = Riˆ / Time of flight
17.(AC) At t = 2 sec, projectile reverses its motion.
1
2
Maximum displacement in initial velocity = 10 2 5 2 = 10 m.
2
Distance travelled = Displacement from t = 0 to t = 2 added to magnitude of displacement from t = 2 to t = 3 = 12.5m
dv
18.(BC) a = V
ds
v 2f v i2 1
Or,
vdv
ads area under a curve Or,
10 2 10 4 For S 10m
2
2
vi 0
v f 10m / s
B is correct.
2
Vmax
Vi2 1
30 6
2
2
Max. velocity is attained at S = 30.
vmax 13.4
river width
time
10
1
=
m/s
120 12
19.(BD) For A : VA,y
And, VA,x, river 0
VA, river
1
m/s
12
VA,x,earth Vr VA,x,river
= Vr
30
30
120
VA,x, earth Vr
1
Vr m / s
4
B is correct.
10
1
m/s
120 12
25
5
VB,x,earth VB,x,river Vr
m/s
120 24
For B : VB,y,earth
APP | Kinematics [Motion in 1-D & 2-D]
4
Solution | Physics
Vidyamandir Classes
20.(BD) For first one Minute :
1
2
h1 0 1060 18,000m = 18 km
2
V1 = 0 1060 600 m / s
After first one minute : Rocket motion is under gravity.
2
v12 600
18000m (V = 0)= 18 km
2g 210
Max. ht. reached , Hmax = h1 + h2 = 18+18= 36 km
1
1
Using h u t gt 2
18000 600 t 2 10 t 22
2
2
V2 V12 2gh2
h2
t 22 120t 2 3600 0
t 2 60 60 2 .
Total time required, T t1 t 2 = 60 60 60 2 = 120 60 2 r
21.(ACD) v A,Board v A,earth v Board,earth = 2v – v = v
v B,Board 2v v = –3v
L
L
L
T
vA,B v 3v 4v
d B,Board VB,Board T 3v
L
3L
4v
4
d A,Board VA,Board T v
and
22.(ABC) Δv v f v i 60cos 60º ˆi 60sin 60º ˆj 60iˆ
2
Δv 30 30 3
a ar at
2
L L
4v 4
30iˆ 30 3jˆ
= 60 m/s
dv
a t 0
dt
a a 2r a 2t a r
60
v2
12m / s 2
R
0.31000
v vi
60
a arg f
11.5 m / s 2
π
Δt
300 / 60
3
2
=
23.(ABCD)
vA 4iˆ 4kˆ
vB 3jˆ 4kˆ
v A,B 4iˆ 4kˆ 3ˆj 4kˆ 4iˆ 3jˆ
v A,B = 5 m/s (A is correct)
As the initial velocity and acceleration of both the particles in vertical direction are equal, so they would hit the ground
at same time, B is correct.
2 4 4
4 16
Time of projectile =
sec
Distance covered by A = 4 m
10
5
5 5
4
12
Distance covered by B = 3 m m.
5
5
C is correct.
16 2 12 2
Separation = =
5 5
162 122 20
=
4m
5
5
rA,B rA rB v A t v B t v A,B t 4iˆ 3jˆ t
APP | Kinematics [Motion in 1-D & 2-D]
5
Solution | Physics
Vidyamandir Classes
24.(ACD)
HA
2
3
RB
sin2 1
u 2 sin 2
2g
2
R A max
1
max
u sin 90
g
u
2
2 u 2 sin 2 p
sin1
3 sin2 u sin
g
3
3 1 cos 2 8 sin2
3
4
2
g
25.(ABCD)
T
R 2 R cos /2
2v0
v0
dT
R R sin /2
0
0
d
2v0
v0
Distance = R
26.(ABC)
x2
sin
1
R
60 T 3
2 2
v0 6
3
2 R×
3
2
y2
1 ; So the path is an ellipse
a 2 b2
Vx ap sin pt , V y bp cos pt
a x ap 2 cos pt , a y bp 2 sin pt
So,
V a 0 as V a
So,
a 2 p3 sin pt .cos pt b 2 p 3 sin pt.cos pt
a 2 p3 sin pt .cos pt b 2 p 3 sin pt.cos pt
As,
ab
So,
sin pt.cos pt = 0
2p
The motion is similar to motion of each around sun. So force always towards focus and hence acceleration.
At t = 0 particle is at (a, o)
At
t
particle is at (o, b)
2p
sin p 2t 0
p 2t , 2
t
So distance travelled along X-axis is a not the actual distance, which is the length of the
part of the ellipse between (a, o) to (o, b) you can try out for distance by following
method ds dx 2 dy 2
27.(ABD)
ds
dy
1
dx
dx
2
s
0
0
ds
a
2
dy
1 . dx
dx
If u is the initial speed and the angle of projection. Then v y u sin gt i.e., v y t graph is a straight line
with negative slope and positive intercept. x (v cos )t i.e., x – t graph is a straight line passing through the origin.
1
y (u sin )t gt 2 i.e., y – t graph is parabola i.e., vx t graph is a straight line parallel to t-axis.
2
1
1 2
28.(ABC) u1t a1t
…(1)
2
2
1
1
and u1t ( a2 )t 2
2
2
APP | Kinematics [Motion in 1-D & 2-D]
6
Solution | Physics
Vidyamandir Classes
1
1
u2t a2t 2
2
2
Subtracting (1) and (2), we get
u u
t 2 2 1
a1 a2
…(2)
…(3)
Substituting (3) in (1) or (2) and rearranging, we get
4(u2 u1 )
1
(a1u2 a2u1 )
…(4)
(a1 a2 ) 2
Since the particle P and Q reach the other ends of A and B with equal velocities say v.
For particle P
v 2 u12 2a21
2
…(5)
u22
For particle Q
v
2a11 …(6)
Subtracting and then substituting value of 1 and rearranging, we get (u2 u1 )(a1 a2 ) 8(a1u2 a2u1 )
l
l
2l
urel u u 3u
2
2l 2l
Distance vdt u
3u 3
29.(ABCD)
t
vavg
total disp l / 3
3u
total time 2l /3u
2
;
v AB u u cos 60
3u
2
30.(ABC)The velocity of motor boat is given as vm vmw vw
5 3
sin sin120
1
sin
2
(A), (B) and (C)
30
31.(BC) In the first case BC vt1 and w ut1. In the second caste u sin v and w u cos t2
Solving these four equations with proper substitution, we get w = 200 m, u = 20 m/min, v = 12 m/min and 37
32.(AC) Distance travelled by motor bike at t = 18 s
1
S bike S1 (18)(60) 540 m
2
Distance travelled by car at t = 18 s
Scar S2 (18)(40) 720 m
Therefore, separation between them at t = 18s is 180 m. Let separation between them decreases to zero at time t beyond
18s. Hence, Sbike 540 60t and Scar 720 40t
Scar Sbike 0
720 40t 540 60t
t = (18 + 9) s = 27s from start and distance travelled by both is Sbike Scar 1080 m
APP | Kinematics [Motion in 1-D & 2-D]
7
Solution | Physics
Vidyamandir Classes
33.(ACD)
v0 x0 x0
or
v
dv
Acceleration
0
dx
0
0
v0 dv
;
v (or )
dt
v v0 e t (or ) v 0 for t
34.(ABC)
x
0
dv
v. v
dx
v
t
dv
dt
v
v
0
0
v
v0
1
when t
e
dv
v 0 kx
dt
dx
v v a kv k (v0 kx )
x
dt
v
dv
kv
dt
Further, a
dv
kv
dt
dv
kdt
v
1
v
Since, v v0 kx. Hence slope of velocity displacement curve is
0
t
dv
k dt
v
0
t
v
1
log e 1
k
v0
dv
k.
dx
35.(AC) At highest point angle between a and v is zero. Hence, total acceleration is only normal or radial acceleration.
a an
But
a=g
v2
R
a an
but
v2
R
a=g
2
(u cos )
(u cos )2
g
R
R
2
2
2
u cos
u cos 2
or
R
or
R
g
g
At point of projection component of acceleration (= g) along velocity vector is g cos(90 ) or g sin .
g
36.(C) Time of flight is the time taken by projectile to come back to horizontal level It depends only on vertical
acceleration. As vertical acceleration is unchanged, T is unchanged T = T`
37.(A) To fall vertically below A, ball undergoes a
Projectile from ‘A’ to ‘B’ as shown.
Net displacement (A B) about x-direction = 0
1
0 Sx u x t a t 2 ,
2
a = 26.6 m/sec2
t
time of flight
2
38.(B) As the ball undergoes a parabolic motion from A to B,
VB y u y v sin θjˆ 12ˆj
,
39.(B) x = 5 sin 10t
y = 5 cos 10t
40.(C)
; Vx
;
dx
50 cos10t
dt
dy
Vy 50 sin10t
dt
VB,x VA,y 16iˆ
VB v x 2 v y 2 20 .
vnet v x 2 v y 2 50m / sec
VRx 10 m / s
APP | Kinematics [Motion in 1-D & 2-D]
8
Solution | Physics
Vidyamandir Classes
VRy 10 m / s
Vmy 0
Drops appear vertical to man
41.(D)
V Rx Vmx
Vmx 10 m / s
10 3 m / s
VRY
)2 (VRy
)2
VR (VRx
(10 3)2 102 20 m / s
42.
[A – q s ; B – r ; C – p ; D – q s]
(A) constt. Speed
dx
So,
constant
dt
Position time graph will be straight line
(B)
(D)
43.
dx
2
dt 2
d2x
dt
2
0 (C) is correct match
0 (
(C)
dx
= constant)
dt
[A – q s ; B – p ; C – p ; D – qr]
dv
(A)
constant.
dx
vdv
is increasing uniformly
dx
B, D will be the match
d 2x
dt 2
've'
(A) is correct match.
(B, D) are correct match.
acceleration is increasing (B)
ax
(D)
dv2
dv
constant
2v constant
dx
dx
vdv
So;
constant
dx
Acceleration of particle is constant
(A)
dv
(C)
constant
dt
(A)
a constant α t
dv
dv dt
(D)
constant
or
.
= constant
dt dt 2
dt 2
1 dv
= constant
2t dt
[A – qs ; B – p ; C – r ; D – r]
(B)
44.
APP | Kinematics [Motion in 1-D & 2-D]
9
Solution | Physics
Vidyamandir Classes
2
2
(A)
R=
u sin2θ
R 10 sin60 10 3
=
=
4
g
2
2g
Displacement =
45.
R2
Hmax 2
2
usinθ
g
time
2
1
10
u 2 sin 2 θ
10
2
Hmax =
=
m
2g
210
8
usin θ 1
sec.
g
2
Avg. velocity =
(B)
The time is given by
(C)
R=
(D)
Change in linear momentum = initial momentum – final momentum
displacement
total time
Solve to get answer.
u 2sin2θ
solve to get answer.
g
= 3 10cos30º iˆ 10sin 30 ˆj - 3 10cos30iˆ
[A – q ; B – r ; C – q ; D – r]
If particle is gaining speed in a uniform manner, then it’s tangential acceleration is non-zero and constant.
46.(20) At the time of collision, position of the both particles must be same.
So, diff. in x coordinate = 10 .
y coordinate is equal.
x1 – x2 = 10.
…..(1)
20cos 45º t u cos 60º t 10
and
…..(2)
u sin 60º t 20cos 45º t
Solving, we get the answer.
47.(1)
Let t be the instant at which the ball hits rear face AB of the trolley.
38
38
3.8s
v 0 cos 45º u 0 28.28cos 45º 10
Then
t
At
t = 3.8 s, the y-coordinate of the ball is y v0 sin 45º t gt 2 20t 5t 2
1
2
Or
y = 20(3.8) – 5(3.8)2 = 3.8 m
Since 3.8 m > 2m, therefore, the ball cannot hit the rear face of the trolley.
Now, we assume that the ball hits the top face BC of the trolley, and let t be that instant.
Then, y = 2 = 20 t - 5 t 2
or
t 2 - 4 t +0.4 = 0 ;
t = 3.9s
Let d be the distance from the point B at which the ball hits the trolley. Then,
d v 0cos 45º u 0 t t 20 103.9 9.8 1m
1/2
48.(5)
2( )l
t min
1/2
2l
vmax
(0.25 0.5)8 103 2
310 s 5 min10 s.
0.25 0.5
The situation can be roughly shown in the figure. Let C take time t to overtake A.
t min
49.(1)
drel 1000 m, vrel (10 15) 25 ms 1
drel 1000
40 s
vrel
25
Let acceleration of B be a for overtaking
Here t
drel 1000 m ; vrel 15 10 5 ms 1
APP | Kinematics [Motion in 1D & 2D]
10
Solution | Physics
Vidyamandir Classes
50.(2)
drel a and t = 40 s
1
Using d rel urel t arel t 2
2
1
100 5 40 a (40) 2 a 1 ms 2
2
t1 t2 4min, v a1t1 a2t2
1
S 4v 4 2v v 2
2
1 1
1 1
1 1
t1 t2 v 4 2
2
a1 a2
a1 a2
a1 a2
51.(8)
S1
1 2
1
gt ; S 2 ut gt 2
2
2
S1 S2 h ; 4h
52.(4)
u2
u 8 gh
2g
ut h
8gh t h
t
h
8g
0 v0 cos30 g sin 30t
t
v0 cos30
g sin 30
…(1)
1
H cos30 v0 sin 30t g cos30t 2
…(2)
2
By equation (1) and (2), we get
v 2 cot 2
2 gH
H 0 1
4 m /s ( 30)
v0
g
2
5
53.(3)
54.(3)
Vabsolute in vertically downward VHe after collision vertically upwards since collision is elastic so velocity of hail
stones w.r.t. car before and after collision will make equal angles.
V
VHe /1 VH Vc V V1 ; 90 2 1 90
;
a1 . 2 21 tan 2 tan 21 1
V
The horizontal and vertical components of the velocity are the same, let it be u = vcos 45°.
From A to B : 1
u2
u2 2g
2g
At B : d ut1 t1
d
g
d g d2
; 1 ut1 t12 u
u
2
u 2 u2
1 d
g d2
2 u2
1 d
gd 2
4g
4 4d d 2
d 2 4d 4 0
d 2m ; 3d ut2 1t 2
3d
u
1
3d g 9d 2
9 gd 2
9d 2
9
l ut2 gt22 u.
3d
3d
3 2 4 6 9 3
2
2
4 2 4
4g
4g
4
55.(9)
vx v cos 60 2 15 m /s
v y v sin 60 4 15
H max
v 2y
2g
l 3m
v 4 15 m /s
3
2 45 m /s
2
4 45
9m
20
APP | Kinematics [Motion in 1D & 2D]
11
Solution | Physics
Vidyamandir Classes
56.(6)
2V sin( ) 2 3 3 sin(60 30)
T 0
g cos
10 cos30
23 3
10
3
2
1
2 0.6 0.1x ; x 6
57.(15) As seen (from ground, ball rises vertically, so;
……(1)
vcos60º 10m / s
vy vsin 60º
2
u 2 sin 2 θ vsin 60
2u
2g
Solving, we get the answer
Hmax =
58.(96) For =45º,
H max
2
10 sin 2 45º
2.5 2m
2g
45º
R
2
10 sin
2
10
2 g
2
sin 2
sin 2 100 2sin .cos
=
g
g
2
5
96m
2v02 2v0
gv0
g
dy
If y is the height of balloon at any instant t and
its velocity then
dt
Range v0 time of flight (t)
2v0
g
ln v0
g
2
v0
t
dy 2v 1 2v
y 0 g 0
dt g 2 g
60.(5)
Room Height
R will be max for max. ‘ ’ possible for projection.
H max
59.(2)
……(2)
2
dy
g
y v0 0
dt 2v0
y t c
t = 0, y = 0
At
;
dy
dt
g
v0
y
2v0
c
2v0
ln v0
g
2v02
[1 e gt / 2v0 ]
g
Simplifying, we get
y
1
For rat S t 2
2
…(1)
1
For cat S d ut t 2
…(2)
2
Putting the value of S from equation (i) in equation (2),
( )t 2 2ut 2d 0
;
u2
( )
2d
Substituting a, d and u we get
For t to be real,
2.5
t
2u 4u 2 8d ( )
2( )
u2
2d
52
2.5 2.5 5 ms 2
25
APP | Kinematics [Motion in 1D & 2D]
12
Solution | Physics
Vidyamandir Classes
DYNAMICS OF A PARTICLE
1.(A)
2.(C)
3.(B)
Displacement of M with respect to ground = displacement of block with respect to M
Acceleration of m with respect to ground a G a a cos iˆ a sin ˆj
a G 2a sin
2
For the equilibrium of block 150cos 45º 50cos 45º 150sin 45º 50sin 45º
(A) is correct.
0.5
F.B.D. of man and plank are
For plank be at rest, applying Newton’s second law to plank along
the incline Mgsin f
……..(1)
And applying Newton’s second law to man along the incline.
Mg sin f ma
………(2)
M
a g sin 1 down the incline
m
4.(C)
Till both Blocks remain in equilibrium :
F T fr1 0
T fr1 fr2 0
x2 y2 2
dx
dy
2x 2y. 0
dt
dt
5.(B)
vy cot 60º 3
dy x dx
,
dt
y dt
v y 1m / s
v v x 2 v y 2 2m / s
F.B.D. of A
6.(C)
C
a=3
B
A
2T – 10g = 10a A
APP | Dynamics of a Particle
13
(2)
Solution | Physics
Vidyamandir Classes
m A 10kg. mB 5kg
From (1) and (2)
10a A 10a B
let acc of A,B and C be a A ,a B ,a C
aA aB
F.B.D. of B
a A : a B 1:1
T 5g 5a B
(1)
Velocity of block and wedge along contact will be
same.
9 cos37º a sin 37º
4
a 9 = 12
3
7.(B)
Final motion
Nsin 37º ma
3
N 1210
5
8.(B)
N 200N
Draw force diagram of M and se that net force on M in both the cases is zero.
A
9.(B)
Let B : foot of perpendicular drawn from A on the ground.
C : foot of perpendicular drawn from B to OX
5m
O
B
x
C
30
sin
AB
OA
s ut 1 / 2 at
2
1
4
10.(B)
AC 5 sin 30 2.5m
AB AC sin 30 1.25m
acc. of the block g sin 2.5 m / s 2
1
2.5 t 2 t 2 sec
2
Along horizontal MV m Vr cos V
APP | Dynamics of a Particle
Vr
14
10
3
m / s 5.77 m / s
Solution | Physics
Vidyamandir Classes
11.(BD)
12.(ACD)
Maximum fr force between m2 and m1
m2gμ 10 0.2 2N
F – fr – T = m2 a2
If in equilibrium
F fr T
T 2N
For m1 in this case :
T Frmax
[If F < 4N]
fr and T will be equal
(A) is correct
It is not necessary if F > 4
T = 4N. (B) is not correct.
If F > 4, max Fr = 2N
and system will accelerate
system will not be in equilibrium (C) is correct
If F = 6N : Fr will be at max value. Fr = 2N
F – T – Fr = a2
4-T = a2
Since blocks are connected by string there acceleration will be equal
T – fr = a
2a = 2, a = 1
and,
T–2=a
T = 3N
4–T=a
13.(AC) Due to symmetric Structure wedge will be in equilibrium
and therefore acceleration of wedge is = 0
For B : mg T ma …..(1)
T
m a
mg
A
For A :
m T
APP | Dynamics of a Particle
15
Solution | Physics
Vidyamandir Classes
T = ma …..(2)
a
From (1) and (2) ;
g
2
14.(ABCD)
If Fr is not present system will move towards right
fr will act on P towards left
Max fr = 400.6 24N
System will be in equilibrium
If m q g f r m R g
P
4kg
= 0.6
4kg
Q
40 f 20
f 20N
Since for is within limiting value
Q is in equilibrium
2kg
R
f 20 N system is in equilibrium
R is in equilibrium
TA 20N
TB = 40 N
Contact force = 402 202 = 2000 = 20 5
15.(ABC)
System will be in equilibrium until A is in equilibrium.
Max Fr force on A μ s 0.5g
= 2N
F = Kt = t
(k = 1)
If T = f2
B is in equilibrium
f2 max 0.2 0.5g 1N
For
T 3 : system as a whole is in equilibrium
For t upto 3 sec. system is in equilibrium and is at rest.
Options A, B and C are correct.
For t > 3 sec :
F on A = 1N
Fr force max while motion of A k N 1
T – 1 t T 1
T 1
a
2
a
2
t–2=a=
dv
or
dt
APP | Dynamics of a Particle
10
3
v
t 2 dt 0 dv
16
Solution | Physics
Vidyamandir Classes
t2
10
2t V 0 =31.5 m/s
2
3
16.(BCD)
D is incorrect.
To maintain constant velocity, Fnet 0 P fr always
17.(BC) For 10kg block : kx 10 12 120Nt.
For 20 kg block :
200 kx 20 a
200 120
a
20
= 4 m/s2.
18.(AD) Suppose blocks A and B move together. Applying NLM on C, A + B, and D
60 – T = 6a
T – 18 – T’ = 9a
T – 10 = 1a
Solving a = 2 m/s2
To check slipping between A and B, we have to find friction force in this case. If it is less than limiting static
friction, then there will be no slipping between A and B.
Applying NLM on A.
T – f = 6(2)
As T = 48N
f = 30 N
and fs = 42 N hence A and B move together. And T` = 12 N.
19.(ABC) F.B.D of block B w.r.t. wedge
For block A
N cos 45º = 1.7 a
for block B
0.6g sin 45º + 0.6a cos 45º = 0.6b
N + 0.6a cos 45º = 0.6g cos 45º………(iii)
By solving (i), (ii), and (iii) a
……..(i)
………(ii)
3g
23g
and b
20
20 2
Now vertical component of acceleration of B bcos45º
23g
40
And horizontal component of acceleration of B = bsin 45º a
17g
40
20.(BD)
by string constrain
APP | Dynamics of a Particle
17
Solution | Physics
Vidyamandir Classes
Or v B u v A
vA u vB 0
Differentiating both side
21.(AC) (1)
;
aB 0 a A
Balancing forces perpendicular to incline
N = mg cos37º + ma sin 37º
4
3
N1 mg ma
5
5
and along incline mg sin 37º - ma cos 37º = mb1
3
4
b1 g a .
5
5
4
5
3
5
3
5
4
5
Similarly for this case get N2 mg ma and b2 g a
(2)
4
3
N2 mg ma
5
5
4
5
3
5
3
5
3
5
(3)
Similarly for this case get : N3 mg ma and b3 g a
(4)
Similarly for this case get
4
4
N 4 mg ma
5
5
3
3
and b4 g a
5
5
22.(AD) Force of friction on the block = max = N 0.5 20 10 N
Total force exerted by the block on the cube is F 10 ˆi 24 kˆ
(A)
F
10 2 24 2
26N
(D) is also correct.
23.(AD) For equilibrium
/2
Rg
/ 2
cos d Rg
0
sin d
0
1
At the position of maximum tension in the rope
Rd cos (Rd g sin )
45
At any
dT Rd cos Rd g sin
Tmax
0
/4
dT Rg
(cos sin ) d
0
1
1
Tmax Rg[sin cos ]0 / 4 Rg
1 Rg ( 2 1)
2
2
APP | Dynamics of a Particle
18
Solution | Physics
Vidyamandir Classes
24.(AC) Initially the 4kg block experience increasing friction as it tries to prevent relative motion between the 2kg and 4kg
and the force increases with time then there is a discontinuity in the graph of a4vst because the values of frictional
force decrease from limiting to kinetic friction. The friction causes increasing acceleration on 2 kg block but after it
starts relative motion the kinetic friction it constant causing constant acceleration.
25.(AD) Initially both friction and external forces acts opposite to motion.
mg 5 N
F 15N
20
a m /s 2
10
V = at
2 m /s 2
Velocity change direction
At t
10
5s
2
Later after velocity changes direction friction acts opposite to motion (+ve x-axis) and ext force act along motion.
mg 5 N
F = – 15 N
10
a m /s 2 1 m /s 2
10
Hence,
dv
dv
d 2 x2
2 at t = 5 and
1 from then
2
dt
dt
dt 2
26.(BC)
When the block is seen with respect to wedge a pseudo force ma will act horizontal. By normal constrain
ma cos(90 ) mg cos N
ma sin mg cos N
Magnitude of acceleration
a 2 a 2 2 a 2 cos(180 )
2a 2 (1 cos )
4a 2 sin 2 /2
2a sin /2
27.(ABCD)
Taking wall as reference from a pseudo force acts on the block = ma
Where a is acceleration of reference 20 m/s 2
N = ma
= 10 × 20
= 200 N
APP | Dynamics of a Particle
19
Solution | Physics
Vidyamandir Classes
Limiting friction N
= 0.6 × 200 = 120 N
Friction required to prevent sliding is 100 N
f = 100 N
(200) 2 (100) 2
Total contact force
100 5N
28.(ACD)
For breaking off the plane :
at02 sin mg
Fsin mg
t0
mg
a sin
Speed at time of breaking off.
t0
at 2 cos
at 2 cos a cos mg
mg
mg 3
dt 0
m
3m
3m a sin a sin
9a tan 2 sin
0
v
a
Fcos at02 cos
amg cos
g cot .
m
m
a sin m
t0
at 3
a 4
a m2 g 2 cos
mg 2
cos dt
t0
2 2
3m
12m
12m a sin 12a tan sin
0
s vdt
29.(BD) As on the gravity and normal are the only two forces acting
( M m) g N ( M m)acm
g
N
acm
( M m)
By normal constrain
Block m will apply N1 force of the wedge normally.
And the component of the force in x direction ( N1 sin ) will provide acceleration
Max N1 sin
ax
N1 sin
M
30.(BCD) Angular velocity of sleeve =
Radius of rotation = l1
Centrifugal force m2l1
N x be normal in plank of motion N x m2l1
Let N y be normal in direction planking out gravity
N y mg
N 2 N x2 N y2
N (m2l1 )2 (mg ) 2
APP | Dynamics of a Particle
20
Solution | Physics
Vidyamandir Classes
m2 (4l12 g 2 )
m 4l12 g 2
As it starts slipping
F = f N
31.(D) a = 0, since mB mA gsin45º g AmA BmB cos45º
2
mg cos 45º
3
2
And 2mg sin 45º > mg cos 45º
3
Therefore block B has tendency to move downward.
2mg
We have
T FrB 0
2
32.(B) Since mg sin 45º >
FrB
2 mg
3 2
mg
FrA 0
2
34-36. 34.(A) 35.(A) 36.(B)
33.(B) Again T
FrA
T
T
T
m1
mg
downward.
3 2
T
x1
R
a1
4 mg
3 2
T=
a1
m2
N
a3
a2
N
m2g
x3
x2
m3g
m1g
T N m1a1
. . . . (i)
N m2 a1
. . . . (ii)
m2 g T m2 a2
. . . . (iii)
m3 g T m3 a3
. . . . (iv)
x1 x2 x3 const.
a1 a2 a3 0
. . . . (v)
Solve the equation for T , a1 and a3 to get : T
37.
120
N ;
7
a1
40
m / s2 ;
7
a3
30
m / s2
7
[A – p r] [B – p s] [C – p r] [D – q]
(A)
f1 0.3 20 6 N , f1K 0.2 20 4 N
f 2 f 2 K 0.1 50 5N
For combined block
15 5 10a a 1 m s 2
f1 2 1 2 N
Hence all blocks will have same acceleration. Also f1 f hence [A-p, r] similarly solve others
APP | Dynamics of a Particle
21
Solution | Physics
Vidyamandir Classes
38.
[A – p r] [B – r ] [C – q r s] [D – q r]
mA m B 3kg;mC m D m E 2kg
If spring 2 is cut then block D is momentarily at rest. it will accelerate up.
Block B will only with be at momentarily rest and has zero acceleration because just after
cutting other force remains same.
If spring 1 is cut block A will accelerate down with acc. g and will be at moment’s rest
Similarly for D it will be at moment’s rest and it will accelerate down but not with g due
to stretched spring between D and E.
39.
[A – q s] [B – p r] [C – q r] [D – q s]
Acceleration will be same till N exists. This is possible only if μ1 μ 2
When acceleration will be different, then N = 0 and μ 2 μ1
Now match the option.
40.(10)
30°
N A sin 60º mg
From (1) and (2) NB = 0 incase of amax.
mg
g
ma
a
3
3
41.(5)
NA cos60º N B ma
…..(1) and
….(2)
n 10
v A 1m / s, v B 3m / s
Power delivered by internal force = 0.
2TVA cos180º 2TVB cos0 TVC cos180º 0
2T T 6 VC T 0 VC = 4
Since C is in contact with B.
C will have velocity = 3 m/s along horizontal
Net velocity of C = 32 42 = 5m/s
42.(5)
50m
F2=8N
F1
B
ON
A
For A :
F1 2F2 mAa A
A 4m / s 2
For B :
APP | Dynamics of a Particle
22
Solution | Physics
Vidyamandir Classes
F2 m B a B a B
8
8m / s2
1
Distance covered by B
dist covered by A + 50
1
1
8t 2 4t 2 50
2
2
2t 2 50
t 2 25
t 5sec.
43.(1)
Case (i)
String length is constant and B is in always with A.
a1 a 2
For B : N1 m B .a1 20a1
For A : T N1 m A a1 40a1
44.(1)
…..(1) ;
…..(3)
20g T 20a 2 20a1
….(2)
1
2
1
(1) + (2) + (3) a1 a 2 m / s 2
a B a12 a 22
m / s2 .
4
4
2 2
Case (ii) :
For B : a1 = a2
20g – T = 20 a2
..…(iv)
For A : T = 40 a1 = 40 a2
…..(v)
From (iv) + (v)
1
20g 60a 2
a 2 m / s2
3
1
2
3
Required ratio = 2
n = 1.
1
2 2
3
Max acceleration of system is possible if for value is max
Fr max = 500 0.2 = 100 N
max acceleration of man = 2m/s2 and of plank = 10 m/s2
Their acceleration will always be in opposite directions .
Now, let man acce at 2m/s2 and plank at 10 m/s2 for time t, and let them
decelerate at 2 m/s2 and 10 m/s2 for time t2 .
1 2 1
1
1
2t1 10t12 2t 2 10t 22 300
;
t12 t 22 50
2
2
2
2
Now t1 + t2 will be min if time for acceleration and deceleration are same
t1 t 2
2t12 50
t12 25sec
t1 = 5 sec.
APP | Dynamics of a Particle
total time t 2t1 10 sec.
23
Solution | Physics
Vidyamandir Classes
45.(9)
Particle in gravity free space
m 2.5kg
F = 67.5N
This is a case of circular motion where F is to motion. Since it is uniform circular motion
mv2
R
2.5 9
67.5
R
F
67.5 = m
46.(4)
v2
;
R
R
2.5 9 1
67.5 3
v
9rad / s ;
R
T
2 2
w
9
amonkey/rope g /4
Let acceleration of M a0
So, acceleration of rope a0
a monkey a monkey/rope a rope (a0 g /4)
Now for mass M, T Mg Ma0
…(i)
For monkey mg T m(a0 g /4)
…(ii)
From equation (i) and (ii)
mg Mg a0 ( M m)
47.(1)
mg
4
(5m 4M ) g
5m
a0
M g /(M m)
4( M m)
4
T
M (5m 4M ) g
Mg
4(M m)
In order to achieve a minimum of F, it should be directed as shown in the figure, The value of can be found using
the FBD of the block. Normal to the incline.
N mg cos F sin
…(i)
Along the incline
N F cos mg sin
…(ii)
From (i) and (ii), we get
mg (sin cos )
F
cos sin
For F to be minimum
d
(cos sin )
d
tan
Fmin
mg (sin cos ) 1 2
1
APP | Dynamics of a Particle
24
Solution | Physics
Vidyamandir Classes
48.(6.25)
mg T ma1
2mg 2T 2ma2
So, a1 a2
Relative acceleration of bead with respect to end = 3a
1
l
displacement of block x at 2 6.25m
2
3
49.(2)
At Vmax ,
dv
0 so there won’t be any tangential force, only force will be radial provided by friction
dt
mv 2
r
0 1 mg
r
R
r2
v 2 0 r g
R
;
d ( v) 2
0
;
dr
Therefore at r = 1, v is maximum
50.(2)
r
R
1
2
…(i)
Drawing F.B.D. diagrams
10 g 2T 10aA
5 g T 5aB
T from equations (1) & (2) we get
10a A 10aB 0
a A aB
l1 l2 l3 constant
yE yA y A yB constant
Differentiation twice w.r.t. time,
a 2aA aB 0
2a A aB 3
or
3a0 3
aB 1 m/s2 , aA 1 m/s2
aA aB 1 m/s2 upwards
51.(2)
Block A and B will not move unless tension in the string T A g (48 N )
Let blocks A and B do not move and maximum elongation in the spring be x.
Applying conservation of mechanical energy for the block C and the spring. We get
x
2mc g
2 102 m
k
APP | Dynamics of a Particle
25
Solution | Physics
Vidyamandir Classes
In this situation F.B.D. of the block B is
For blocks to be in equilibrium T kx0 mg 40 N
Therefore maximum distance moved by the block C is x0
52.(2)
Tmax 40 N ( 48 N ) Hence blocks A
2mc g
2 102 m = 2 cm
k
For small values of friction will be directed radially inwards as the tension in the string is zero. The string will
develop tension only if the centrifugal force FC exceeds the limiting friction f e i.e. when
m2 r mg
f e mg
or
g
r
in this case direction of friction will be as shown in the figure.
For equilibrium
F0 T cos 45 fe cos
and T sin 45 fe sin
Eliminating T, we get
Fc f e sin cos
53.(2)
54.(2)
i.e., m2 r mg 2 sin 45 or
2
Maximum value of sin 45 is 1.
a1
F f1 F m1 g
10 m /s 2
m1
m1
a2
F m2 g
1 m / s 2
m2
s
2g
sin 45
r
Maximum value of
1
arel t 2
2
2g
2.
r
1
22 [10 ( 1)]t 2 t 2sec
2
Using conservation of energy principle, if v be the speed of either ball when its radius vector makes angle with
vertically upward direction.
mgR 1 cos
1 2
mv
2
mv 2
2mg 1 cos
R
From F.B.D. (i)
APP | Dynamics of a Particle
26
Solution | Physics
Vidyamandir Classes
mv 2
mg cos 2mg 1 cos
R
From F.B.D. (ii)
N 2 N cos Mg
N mg cos
At the instant tube breaks its contact with ground
N 0
Mg mg cos 2 mg 1 cos 2cos 0
For 60, we get m/M 2.
55.(3)
mg sin f m2l
mg sin mg cos m2l
1
3
g g
2l
2
2
5 2 5 3
3
5
3
2
25
k 3
l
56.(9)
W 2rxdxP rPl 2 (0.3)(3 102 )
0
57.(3)
105
102 9 J
Acceleration of bead ( g sin 37 g cos37)
1
15.3 ut at 2
2
1
15.3 ( g sin 37 g cos37) t 2
2
t = 3 sec
58.(3)
The free-body diagrams are shown in figure. Let’s first determine how F1 is related to N1 and then invoke the
F1 w N1 condition. Balancing torques on the disk around the centre gives F1R F2 R F1 F2 . Balancing
horizontal forces on the disk gives N1 F2 . Combining these4 relations gives N1 F1. So the F1 w N1 condition
becomes.
F1 w F1 w 1
Note that this result has nothing to do with the exact nature of the stick. It could have a different length (as long as it
is at least the radius of the disk), be nonuniform, etcs.
APP | Dynamics of a Particle
27
Solution | Physics
Vidyamandir Classes
Now lets’ determine how F2 is related to N 2 and then invoke the F2 s N 2 condition. Balancing torques on the
stick around the pivot gives N 2 mg. Balancing vertical forces on the disk gives F1 N 2 mg F1 2mg.
(Basically, F1 is the vertical force supporting the whole system. There is no vertical force at the pivot even through
we drew one in the figure to be general, because otherwise there would be a non-zero torque on the stick around its
centre). But F1 F2 from above, so we have F2 2mg. The F2 s N 2 condition therefore becomes.
2mg s (mg ) s 2
We see that we need a larger coefficient of friction with the stick than with the wall. The entire set of forces in this
problem is N1 F1 F2 2mg and N 2 mg. But the actual values weren’t necessary for the w 1 result.
59.(8)
2T sin
d
v2
dN Rd
2
R
( is linear mass density of belt)
Td dN v 2d
so total normal
N 8 newton.
/2
/ 2
dN cos
/ 2
60.(1)
/2
T cos d v 2
/2
cos d
/2
From Collision
mv0 3mvy
v0
3
From COE
1 2
1
1
mv0 3 mv 2y 2 mvx2
2
2
2
vy
1 2
v2 2
mvx 1 2 2 mvx
2
9
v0
3
From frame of ball B,
vx
3T
mvx2
l
T
mvx2 (3)(v02 )
1
l
3 3 1
APP | Dynamics of a Particle
28
Solution | Physics
Vidyamandir Classes
ENERGY AND MOMENTUM
1.(D)
2.(B)
3.(C)
Since the external force is always equal and opposite to the tangential component of mg, there would be no
acceleration of block. i.e. its kinetic energy will remain constant. Now, use work energy theorem. Since there is no
change in K.E., net work done on the block would be zero. Therefore, work done by external force will be negative
of the work done by gravity, i.e. –(mgH) = mgH.
Use conservation of energy to find the speed of block at 600 position. Note that there is no change in spring energy.
Now write the force equation at this position, putting normal reaction as zero.
N – mgsin =
mv 2
R
1 2
mv = mgH
2
mv 2
2mgH
mg
5mg
=
= 2mg N =
+ 2mg =
R
R
2
2
2
25 3
28
Contact force = mg
mg
=
4 2
2
4.(B)
In both COM and ground frame, Kmax is when x is zero in spring, which occurs simultaneously.
Vcm =
m (V ) 0 Vo
=
5m
5
Kmax ground =
Kmax m =
1 2
mvo
2
2
2
2 2
4Vo 1
Vo
5 2 (4m) 5 5 mvo
V2
1
2
Kmin cm = 0, Kmin ground = (m 4m )Vcm
o
2
10
Kmax cm =
;
1
m
2
1
mVo2 (ground frame)
2
Kmin m = 0 (ground frame when energy is shared by spring & 4m only) Hence (B)
5.(D)
vf
The component of velocity along the inclined plane must
remain unchanged.
3 m/s
V f sin 30 3 cos 30 V f 3m / s
30
30
before
after
6.(A)
Parallel to inclined plane, u cos v1 sin
Along horizontal mu 4mv2
v1
u
v2
4
u
3
u
m
4m
Before collision
Along common normal v2 sin v1 cos eu sin
7.(C)
u
3
u 1
3
.
eu
4 2
2
3 2
7
e
12
u1
u2
After collision
1
3g
2 g K 2000 N / m and to lift 3 kg, elongation in spring should be
15 cm .
100
K
1
1
2
2
Let 2 kg is compressed by x K 0.01 x 2 g 0.01 x 0.015 K 0.015
2
2
2
1000 x 0.0001 0.02 x 20 x 0.025 0.225
x 2 625 10 6 x 2.5 cm
In equilibrium, K .
APP | Energy and Momentum
29
Solution | Physics
Vidyamandir Classes
8.(D)
m a m2 a2
acm g 1 1
m1 m2
9.(A)
Fore =
or
( m m2 ) g m1a1
a2 1
m2
m 6iˆ 4 ˆj 5kˆ iˆ 2 ˆj
15 5iˆ 6 ˆj 5kˆ
m (v f Vi )
=
=
= 150 5iˆ 6 ˆj 5kˆ
t
.1
.1
10.(D) Since the cloth is sliding under the dishes, frictional force acting on dishes is kinetic friction. Hence magnitude of
this force is fixed (i.e. it is independent of velocity with which cloth is pulled). Hence the momentum imparted by
cloth to dishes is proportional to time alone. The faster you pull the cloth, the lesser momentum you impart to the
dishes.
11.(BC) Resolve the initial and final velocities parallel and perpendicular to the ground. Since the ground is frictionless,
the parallel component will be conserved. Also, perpendicular component becomes e times in magnitude,
after collision.
12.(ABC)
v1 =
m1 m2
2m2u1
u1 ; v2 =
m1 m2
m1 m2
13.(AC) Maximum possible velocity of ball occurs if e = 1. In this case, if v denotes the velocity of ball after collision,
e 1
v u1
;
u0 u1
This will give v = u0 + 2u1
Also, work done by racket = change in K.E. of the ball. This gives (C).
14.(ACD)
Initially
After collision
MV1 = MV/2
V1 = V/2
MV/2 + MV/2 = 3MV'/2
V' = 2V/3
2V V MV
=
2
6
3
Impulse = M
V D
D
D 3D
4D
2 V
Time =
=
=
=
V
V (V / 2)
V
V
V
D
15.(C) Since the ball is moving along the inclined plane after collision, we can say that the normal component of their
relative velocity has become zero after collision. Hence (A) is correct.
During collision, wedge applies a force on the ball along its normal .Hence linear momentum of ball can be
conserved only along the inclined plane.
Momentum of (ball + wedge) can be conserved only along horizontal direction as an impulsive normal reaction acts
on the wedge from ground.
APP | Energy and Momentum
30
Solution | Physics
Vidyamandir Classes
16.(AC)8 sin = v cos
2g×3.2=8 m/s
8cos
v sin
2
1
2
2 tan = cot tan =
8
v=
= 4 2 m/s
2
2
1
k = 1 4 2 82 = –16 J
2
90–
V
Projectile never travels vertically downward.
17.(ABC)
vA
A
30°
10m/s
B
30°
B
vB
vB = 10 cos 30° 5 3
18.(ACD)
and
vA = 10 sin 30° = 5
Since the pulley is frictionless, string will not be able to exert any tangential force on the pulley. Hence
pulley will not rotate. Rest can be solved by energy conservation.
19.(AD) As particles stick after collision, so their initial momentum are the impulses imparted to 10 m.
IT
Along vertical net impulse to 10m is zero
mu
3mu
IT mu cos 60 3mu cos 60 = 2mu
60
60
Along horizontal, 3mu sin 60 mu sin 60 10 1 1 mv
(v : velocity of combined mass just after the collision) v
3u
12
10m
2
3u
Impulse diagram of 10m
39mu 2
m 3u mu 12 m
2
2
2
8
12
u
20.(BCD) At the time of impact, angle between the line following the centre of A , B and A , C is 90
Loss in energy =
1
2
1
2
1
By symmetry (C) is also correct.
v
Let u : velocity of B and C after collision.
mV 2mu cos 45
Initial KE
1
2
mV 2
APP | Energy and Momentum
B
A
Net impulse on A = Change in momentum = mV.
Along the initial line of motion of A
u V / 2
C
u
1
Final KE 2 mu 2
2
1 V 2 1
2
2 m
2 mV initial KE
2
2
31
Solution | Physics
Vidyamandir Classes
21.(B)
x displacement = distance of centre of mass from the centre of initial position of the
sphere
M 0 M R / 2
M M
x
R
2
22.(ABC)From momentum conservation
mu = mv cos30 + mv cos 30 ; v
u
3
So, choice (C) is correct
For an oblique collision, we have to take components along normal i.e., along AB for sphere A and B.
vB v A e(u A uB ) ; v – 0 = e [u cos30 – 0]
;
v eu
3
3
2
; v e.v 3.
; e
2
2
3
So, choice (A) is correct. Also, loss of kinetic energy
2
1
1 u 1 2
1
1
K mu 2 2 mv 2 mu 2 2 m
mu
2
2 3 6
2
2
23.(ABCD)
The velocity of bob just before the impact is v 2 gl along the horizontal direction
From momentum conservation
mv mv1 5mv2
v1 v2
v1 v2 v
v
2v
v
Solving above equations, we get ; v1 , v2
3
3
From coefficient of restitution equation, 1
For tension in string ; T mg
T mg
2 gl
mv 2
, v2
l
3
mv12
17
T mg
l
9
(T = 3 mg)
(i) Let the maximum height attained by the bob be h, then
24.(ABC)
(A)
mv ( M v )V cos
(B)
Mv (m M )V sin
(C)
Initial kinetic is ; kl
;
mv12
4l
mgh h
2
9
(m M )V (mv ) 2 (MV ) 2
1 2 1
mv MV 2
2
2
1
Final K.E. is k f (m M )v2
2
mM
Decrease in K .E. Kl k f ; k
(v 2 V 2 )
2(m M )
Fraction of initial kinetic transformed into heat is
APP | Energy and Momentum
k
mM
kl m M
32
v2 V 2
2
2
mv MV
Solution | Physics
Vidyamandir Classes
25.(AC) Area of (a – t) curve 32 ms 1 V f Vi
;
V f 32 Vi 32 6 38 ms 1
Work done by all forces KE
;
1
1
m(V f2 Vi 2 ) (382 6 2 ) 704 J
2
2
Work done by conservation forces
U i U f 320 J
Work done by external forces= 704 – 320 = 384 J
26.(AC) The spring is compressed by x
Block will not return if mg Kx
mg (0.3)(1)(10)
0.30 m
K
10
Work done against friction Ei E f
So,
xmax
1
1
mg ( x 2) mv02 Kx 2
2
2
1
1
(0.3)(1)(10)(0.3 2) (1)v02 (10)(0.3)2
2
2
;
On solving, v0 3.8 m /s
27.(AC) In case of both A and C they are path independent.
Now consider B
3
2
y dx xy dx it dependent on now ‘y’ is related to ‘x’. So case B fails same with D.
28.(BCD)
Work is said to be done in a frame any when the point of application of force undergo displacement.
29.(ABC)Between A and B
1
mgl cos mvB2 ; VB 2 gL cos
2
ar 2 g cos ; at g sin
Now, at B ; TB mg cos
mvB2
L
Put VB TB 3mg cos
tan(90 )
30.(ACD)
at 1
tan
ar 2
tan 2
cos
1
3
Applying conservation of total energy
1 2
1
mu mga (1 cos ) mv 2
2
2
For particle to lose contact N = 0
;
mg cos N
mv 2
a
v 2 ag cos ; u 2 ga (2 3cos ) 0
31-33. 31.(B) 32.(B) 33.(D)
Let V1 be speed of combined mass just after collision.
From COM in horizontal direction.
2m V cos 45 3m v1 v1 5 g
Vmin 3
2
3m
At 60 Let velocity = V2.
APP | Energy and Momentum
5g
33
V1
Solution | Physics
Vidyamandir Classes
3 mg 1 cos
1
2
1
3mV12 3mV22
2
V2 2 g
Hence velocity at highest point = V2 cos = 2 g
Maximum height = 1 cos
V22
1
2
= g
sin
2g
V2
V22
2
= 2
34-35. 34.(D) 35.(B)
Let velocity of 2m and m be V and V1 respectively then V cos V1 (using constraint relation)
V1 = 0.6 V
Using conservation of energy 2m 10 3 2 m101
36.
37.
1
2
2m V 2 2
5
5
V 10
m / s and V1 0.6V 6
m/ s
17
17
5
36
V12
17 1.53 m
H max 1
1
2g
2 10
1
m 0.6V
2
4
5
v1 m
1m
2
4
3
m
2m
m v1
1m
v
[A – s ; B – r ; C – q ; D – p]
A corresponds to the case where velocities are exchanged. This matches with S.
B corresponds to a perfectly inelastic collision. This matches with R as the putty is expected to be perfectly
inelastic.
[A – q ; B – r s ; C – q p ; D – r p]
Final common velocity= 4 m/s
(from cons. of momentum)
As KE of 1 kg block decreases, work done by friction on it is –ve.
Similarly, work done by friction on 2 is +ve
Total work done by friction = change in KE (2 kg + 1 kg)
=
1
1
1
× 1 × 62 +
× 2 × 32 –
(1 + 2) (4)2 = 27 – 24 = 3J
2
2
2
38.(1)
Form linear momentum conservative (for collision B & C)
mv0 = 2mv v =
v0
2
FBD
APP | Energy and Momentum
34
Solution | Physics
Vidyamandir Classes
aA = g , aBC =
g
2
3g
aA/ B =
v 2A/ B = U A2 / B + 2aA/B SA/B
2
2
v
3g
0 = 0 x
L
2
2
v2
L= 0
= 1m
12g
39.(2)
his happens of after collision both ball and inclined
plane have same horizontal velocities say V2, say V1
be initial velocity of ball Vy be vertical velocity of
ball after collision, m1 – mass of ball, m2 – mass of
inclined plane.
Component of velocity of ball along the inclined
plane remain same
V1 cos V2 cos Vy sin
Vy sin (V1 – V2 ) cos
using (2)& (3)
Vy cos V1sin
V1 V2
2
Tan =
V1
Conservation of linear momentum in horizontal
direction
m1V1 ( m1 m2 )V2
... (1)
V2 = V1(1– tan2)
(m1 + m2) V2 = m1V1
Coefficient of restitution = 1
1=–
[V2 sin (V2 sin Vy cos )]
0 V1 sin
V y cos V1 sin
... (3)
Divide M1 + m2 =
... (2)
... (4)
... (1)
m1
1 tan 2
m1
= cot2 –1; cot2 30° –1 = 2
m2
Ans.
40.(1)At the moment of collision
After collision
0.25 v0 = –0.25 v1 + 0.5 v2
As collision is elastic,
e=1=
41.(2)
v 2 v1
v0
2
v2 = v0
3
or
v2 + v1 = v0
v1 =
2v2 – v1 = v0
... (1)
... (2)
v0
=1
3
Velocity A is 1 m/s backward
v2
g
(d1 + d2) = v1
v2
g
d1 = (ev1)
APP | Energy and Momentum
35
Solution | Physics
Vidyamandir Classes
2ev2
g
v v
d2 = 2e2 1 2
g
d2 = ev1
d2 = 2ed1
Solving e =
42. (3) mg cos =
1
2
d1 + d2 =
;
Therefore, 1/e = 2
1
mv2 – 0
2
... (1)
mv 2
... (2)
T – mg cos=
Tsin µN
Tcos+ Mg = N
On solving µ
T
ar
... (3)
... (4)
sin 2
M
2
cos 2
3m
RHS is maximum when = 45°
43.(3)
d1
d
1
d1 + 2ed1 = 1 1 + 2e =
e
e
e
;
T
N
mg v
;
µ
1
~ 3m 3×10–3
2M
1 2M
3m
µN
mg
P = kl1
1
P(l1 + l2) = kl22 0 –
2
1 2
kl1 0
2
k 2 2
l2 l1
2
2l12 + 2l1l2 – l22 + l12 = 0
kl12 + kl1l2 =
44.(6)
3 l12 + 2l1l2 – l22 = 0
;
l2 = 3l1
(1 + 3) v = (1) (8) + (3) (4) = 20 ; v = 5 m/sec
1
39
For block A,
W f (1) (52 82 ) J
2
2
1
27
For block B,
W f (3) (52 4 2 ) J
2
2
Net work done by friction = – 6J
45.(10) Loss in gravitational = gain in KE + gain in elastic potential energy + work done against friction
1
1
mgx sin 53 mv 2 Kx 2 (mg cos53) x …(1)
2
2
kx mg cos mg sin
…(2)
Solving (1) and (2)
46.(5)
T mg sin
mv 2
R
APP | Energy and Momentum
;
4mg mg sin 30
36
m(v02 2 gl sin30)
l
;
v0
5g
2
Solution | Physics
Vidyamandir Classes
47.(24) FBD of the block,
;
Displacement of the block w.r.t observes
Work done by friction w.r.t observes
f 4N
2 5 3 m /s 2
1
3 4 6 m
2
= – 24 Joule
Acceleration of the block with respect to observes
48.(2)
f L 6 N , Fpseudo 4 N
Free body diagram is:
N cos mg
N sin m (2 g )
tan 1 (2)
tan 2
Maximum possible angular displacement
2 2 tan 1 (2)
49.(5)
3 (5cos37) 5 0
1.5 m /s
8
3 (5sin 37) 5 5
(Vcm ) y
2 m /s
8
Vcm (1.5 i 2 j ) m /s
Collision at origin hence initial position of C.M. is ri 0
(Vcm ) x
50.(2)
(rcm ) f (rcm )i V cm t 3i 4 j
(rcm ) f 9 16 5 m
Force F on plate = Force exerted by dust particles
= Force on dust particles by the plate
= Rate of change of momentum of dust particles
= Mass of dust particles striking the plate per
Unit time × change in velocity of dust particles A(v u) (v u ) A(v u ) 2
51.(4)
When simple pendulum released from position A strikes the wall with velocity v then by conservation of mechanical
energy.
L
mgL 0 mv 2 0, i.e. v 2 gL
2
Now as coefficient of restitution is e so, speed of pendulum after first collision will be
v1 ev e 2 gL
Now after completing oscillation in accordance with conservation of mechanical energy it will strike the wall with
same velocity and so its velocity after second collision will be
v2 ev1 e(e 2 gL ) e 2 2 gL
So the velocity of the pendulum after n collision will be vn en v e n 2 gL
Now if it rises to a height h, by conservation of mechanical energy
1
m(vn )2 mgh, i.e., e 2n 2 gL gh
2
2
APP | Energy and Momentum
37
Solution | Physics
Vidyamandir Classes
or,
e2n
h L(1 cos )
L
L
2
5
or,
2n
2
1 cos as e
5
n
or
4
1 cos
5
1
Now for to be lesser than 60, cos
2
1
i.e.
1 cos
2
n
n
1
4
5
or,
2 or n(log10 – 3log2) > log2
2
5
4
0.301
or
n
[as log10 = 1 and log2 = 0.3010]
or
n > 3.1
0.097
As n (number of collisions) must be integer so for 60, n 4
So,
52.(2)
Let Vrel be the final velocity of the ball w.r.t. wedge and V be the final velocity of the wedge w.r.t. ground.
Now, velocity of ball w.r.t. ground
Horizontal component Vx Vrel .cos V
Vertical component V y Vrel sin
COM in horizontal direction gives
mu m(Vrel cos V ) MV
…(1)
Since velocity of ball along wedge remains constant
u cos Vrel V cos
…(2)
Solving (1) and (2) we get
53.(4)
;
V
mu sin 2
M m sin 2
2 m /s
Let velocity of I ball and II ball after collision be v1 and v2 ; v2 v1 0.5 10
mv2 mv1 m 10
v2 v1 10
Solving equation (1) and (2) v1 2.5 m /s, v2 7.5 m /s
54.(4)
Ball II after moving 10 m collides with ball III elastically and stops. But ball I moves towards ball II. Time taken for
10
second collisions between ball 1 and 2 ;
4 sec
2.5
After elastic collision
m 2m
VA1
9 3 m / s
m 2m
By conservation of linear momentum after all collisions
m(9) m(3) 3m(Ve )
or
Ve 4 m /s
55.(5)
Velocity of first block before collision
v12 12 2(2) 0.16 1 0.64 ; v1 0.6 m /s
By conservation of momentum ; 2 0.6 2v1 4v2
Also v2 v1 v1 for elastic collision. It gives v2 0.4 m /s ; v1 0.2 m /s
Now distance moved after collision s2
APP | Energy and Momentum
(0.4)2
(0.2)2
and s1
; s s1 s2 0.05 m 5 cm.
2 2
2 2
38
Solution | Physics
Vidyamandir Classes
56.(2)
Consider first collision between M and 4 M on the right, the velocities after collision are
3
2
VM u ; V4 M u
5
5
Particle of mass M will move to the left and collide with 4 M. The velocities after collision are
2
2 3
3
VM u; V4 M u
5 5
5
So in all there are two collisions.
57.(7)
h0 3
u B2 sin 2
2 g (h hB )sin 2 30
3
2g
2g
4(given) 3 h sin 2 30 hB sin 2 30 3
58.(4)
h 3) 7 h
;
4 4
4 4
h 7m
Decrease in mechanical energy = work done
1
1
Against friction
m2 kx 2 (mg ) x
2
2
2gx k
m
v
Putting m = 0.18 kg, x = 0.06 m, k 2 Nm1
01 we get
0.4 m /s
4
m /s
10
N 4
59.(6)
After colliding with ground, horizontal component of velocity, i.e., 10 sin30° = 5 m/s will remain unchanged while
its vertical component will become zero. Collision with wall is elastic.
Hence, it will only reverse the direction of velocity of ball, magnetic will remain unchanged,
BC CB BA 30
i.e., 5 m/s ; Therefore t
6s
V
5
60.(6)
By Energy conservation v12 = v02 2 g (1 – cos ) = 0
Tangential at = gsin= 6 m / s2
=37°
v1
v0
Radial ar = 0
Resultant acceleration =
APP | Energy and Momentum
ar2 at2 = 6 m/s2
39
Solution | Physics
Vidyamandir Classes
ROTATION AND GRAVITATION
1.(C)
Since all the particles on a helix are equidistant from the axis, we can use I = mR2, where m is the total mass of
wire. Length of helical wire can be found by unrolling the helix into a straight line.
2.(A)
Use the formula for moment of inertia of a triangular plate about its base (I = mh2/6). Note that the two diagonals of
a rectangle will not be in general perpendicular to each other. Hence perpendicular axis theorem cannot be used.
m 2
6F
2
12
m
F
F mac ac
m
3F F
aB
ac
2m / s 2 (right)
2
m m
F.
3.(A)
I 2
40 I 2 1
3 I 2 1
2
m
r
m2 r 2
100 2
2
2
5
2
5
4.(B)
I=
2 2
mr
3
Relative acceleration 2µg
5.(D)
m
µmg
µmg
m
L
cos = v0
2
L
sin = v
v = v0 tan
2
L 1
mg = mL2
2
3
3g
=
and x = g
2L
6.(D)
7.(B)
C
3g
2L
x=g x=
2L
3
v0
v=?
Distance from B =
L
3
Impulse = mV mV0 m V V0 and about centre of mass angular impulse
8.(A)
= m V V0
2
cos
mV 2
12
. (= change in angular momentum)
T
mg T ma
9.(A)
Tr
mg
a
N
2
mr 2
2
6 V V0 cos
ma
a 2g / 3
a r
T
f
1
mg
3
T mg
2mg
3
mg
3
T f
N mg
mg
APP | Rotation and Gravitation
f N
mg
3
mg
40
1
3
Solution | Physics
Vidyamandir Classes
10.(C) About centre of earth, m
5
ΔL 0
ve R mvr
6
Where v is the velocity at maximum distance r
5 GM
11.(C)
6 R
Vs2 Vs1
r2 r2
R
5 GM
.
6 R
R2
r
Vs1 / 2 Vs1
4 Rs1 Rs1
Rs2 4 Rs1 as V0
12.(A)
GM
2
Vs1
6 Rs1
GM
Let
r
2
r
x
R
2
T R
6TS1
Vs2
R
and
GM
2
5
GM m
1
GM m
m
ve
m v2
2 6
R
2
r
1
2 4
10
1
Rs
8
2
3
Vs1
2
x2 6 x 5 0
2
6Ts1
x 5 or 1
3
Rs2 40,000 km
/ 3 rad / hr
P
W ΔV V V p V p
2
(4 R) x
To find V p we considering of radius x and thickness dx.
GdM
dV p
x 2 16 R 2
VP
M 2 xdx
, dM
2
4 R 3R
4R
GdM
x 2 16 R 2
7R2
3R
2
4R
2 Mxdx
7R
2 MGxdx
2
x 16 R
2
2
2
2GM
7R
4
2 5
x
dx
13.(C) Work done for rotation is minimum when, moment of inertia is minimum and MI is minimum when axis passes
through the centre of mass. Let it be at a distance x from 0.3 kg.
x
0.3 0 0.7 1.4 0.98
0.3 0.7
GM
14.(A)
x
2
G 81M
60 R x 2
x 6R
15.(B)
16.(A) If another hemisphere(identical) is added so that it becomes a complete sphere then total intensity at both point P
and Q becomes same = I P IQ where I p and I Q are intensities at P and Q respectively due to only given
hemisphere
17.(B)
1
2
I P IQ = intensity due to complete sphere =
m u2
GM m
R
8 64
GM m
Rh
64 106 6.4
6
40002
2
G 2m
2R
Gm
2R
2
IQ
Gm
2R 2
6.4 10
10
6
10 6.4 106
41
IP
2
6.4 106 h
6.4 106 7 7h 8 6.4 106
6.4 10 h
APP | Rotation and Gravitation
2
h
64
7
105 m 914.3 km
Solution | Physics
Vidyamandir Classes
18.(A)
19.(A)
GM
r
V
2
v r mv1r1 (r1 min distance)
3
1
2
GMm 1
GMm
Also m v 2
MV12
2
3
r
2
r1
m
Solving r1
. . . . (i)
. . . . (ii)
r
2
20.(B) For earth,
Ve
2GM
R
v
For Sun + Earth,
GM
3 105 GM
Potential energy =
m
R
2.5 104 R
GMm
13GMm
1 12
R
R
2GM
. 13 13 ve
R
v
1 2 13GMm
mv
2
R
ve 40.4 km / s 42 km / s.
21.(BC) N F sin
GMmx
R
R
const.
2x
F cos GMx
3
m
R
a
3
GM
R
22.(AC) V0
3
x2 R2 / 4
x
x2 R2 / 4
1
GMm
GMm
mVe2
K
2
r
r
23.(ABCD)
24.(BC) I =
1
GMm
mVe2
0
2
r
GM
1
GM
GMm
, K0 m
, E
;
r
2
r
2R
Ve
2GM
r
2V0 1.41 V0
speed increases nearly 41%
Work done by kinetic friction on ONE body may be positive/negative/zero. Direction of frictional force
in B,C,D is correct for providing the necessary torque.
1 2
Ml
3
1 2
I = Mg
2
=
25.(AC) NA + fB = 2mg ; NB = fA
APP | Rotation and Gravitation
3g
l
;
mgR + –fA R – fBR = 0
42
;
mg = fA + fB
Solution | Physics
Vidyamandir Classes
26.(BC) The angular momentum of the system is conserved. Kinetic energy will not be conserved because friction is there.
Parallel Axis theorem, check the distance carefully. ID = IB (symmetric)
27.(ABC)
28.(BCD)Since normal is impulsive, friction will also be impulsive and it will reduce and give some horizontal velocity to
C.M. v r friction cannot act when there is no tendency of relative motion.
29.(AC) mgr = fR
N1 sin + f = mg
... (i)
... (ii)
N1 cos = N2
f µN2
f
g
R
N2
r
... (iii)
... (iv)
N1
mg
Due to torque of friction about CM eventually decreases to zero, initially there is no translation. Friction
is sufficient for pure rolling therefore after sometime pure rolling begins. There is no external force in ×
direction therefore momentum is conserved along × direction.
30.(AD)
31.(AD) Linear impulse = mv0
Angular impulse = (2R/3) Linear impulse. This will give the angular speed of sphere just after collision.
Impulse of friction DURING collision is negligible.
32.(CD)
2
3
4
T1 r1 1
T1 : T2 1: 8
T2 r2 4
33.(ABC)
GM
1 1
V1 : V2
2 :1
r
R 4R
L mvr L1 m 2v r , L2 mv 4r L1 : L2 1: 2
4
4
Field at P G PC1 G PC2 ; : density of massive sphere.
3
3
V0
34.(CD)
35.(AC) AB is a
AP is R
At point P, velocity is out of the plane
Angular momentum L0 L mv ( a R ) mv
L0
(R mv )
Constant in direction
magnitude
(a mv )
Constant in
magnitude only
36.(AC) I 0 0
APP | Rotation and Gravitation
43
Solution | Physics
Vidyamandir Classes
ml 2
mx 2 mgx
12
12 gx
2
l 12 x 2
For maximum
d
0
dt
…(i)
12 gx
d 2
l 12 x 2 0
dx
;
12 g (12 x 2 l 2 )
2
2 2
(12 x l )
0
Now put x in (i)
12 g
2
12 x l
;
12 x 2 l 2 0
;
2
;
288 gx 2
(12 x 2 l 2 )2
x
0
l
2 3
g 3
l
37.(AB) Velocity of connected mass will be v0 due to conservation of linear momentum
Drawn FBD with respect to P,
(i)
For 2m
TR I
T = 2 ma
(ii)
2mv02
v2
and a 0
L
5l
5l
Also after certain value of F, torque on cylinder will be constant hence D is corerct.
38.(ABD) 0 R vcom
L mv R I
L mv R (k ) I (k )
(ii)
C
T m(a dR)
com
mv02
For m
T
com 0
LD mvcom R (k ) Icom 0 ( k )
LC LD
L mv R (k ) I (k )
0
com
com 0
APP | Rotation and Gravitation
44
Solution | Physics
Vidyamandir Classes
R
( k ) I com (k )
2
LB Icom (k )
LA mvcom
So LA is minimum
L0 LC
39.(ABCD)
R
v2
;
ac
v
J
M
JL
6J
L J 3J 4J
22
vA v
ML
2
M M M
MI
12
v2
J 2 M 2L 8
L 2 J
R A A 16 2
L ; vB v
ac
2
M
M 18 J 2 9
;
acA
A
2 L 18 J 2
2
2
M L
RB
4J 2 M 2L 2
L
M 2 18 J 2 9
Basic knowledge to write angular momentum and kinetic energy of the system.
2a
2 dy
x
2
41.(BC) x
y
;
2x
a
;
tan 3
a
3
3 dx
g sin
g
mg sin mg
a
;
f
60 ;
I cm
mR 2 2 3
3
1
1
mR 2
I cm
40.(ABCD)
42.(AD) r r0 ct
…(i)
( distance covered by thread in time t)
By conservation of angular momentum about O,
I 00 I
v
v
(mr02 ) 0 mr 2
r
r
0
v0r0 vr
v
v0 r0
r
…(ii)
T at any time mv 2 velocity and radius, r at that instant
2
v r
m 0 0
mv
mv 2 r 2
r
T
03 0
r
r
r
2 2
mv0 r0
T
…(ii)
(r0 ct )3
2
T and time
at time
v v0 r0
2
r
r
v0 r0
(r0 ct ) 2
[From (i) and (ii)]
43.(ABCD)
2
vcom
(R )2 2vcom R cos
In frame of R com each point has velocity and hence same KE.
APP | Rotation and Gravitation
45
Solution | Physics
Vidyamandir Classes
( no slipping)
…(i)
44.(BCD) a R
For block, ma = mg – T
For disc, TR fR I
mR 2 a 1
2
R R
ma
Tf
2
f ma
Tf
And
…(ii)
…(iii)
2
Put (i) and (iii) in (ii)
;
a g
5
2
For block, acceleration
gj
5
2
For disc, acceleration
gi
5
Acceleration of block in frame of disc
aBD aB aD
2
2
g j gi
5
5
From (i),
T = mg – ma
2 mg
3mg
T mg
;
T
5
5
45.(CD) Frictional force on disc = mg
a g , gR
Let the velocity and angular velocity during radian be v and
and
v u at and 0 t
v at and 0 t
v gt
…(i)
0 gRt
…(ii)
Put (i) and (ii),
0 vR
vR 0
Now, vR
( rolling)
Time taken
t
t
v
g
2 0
;
0
2
v R
0 R
2
;
S
[From (i)]
0 R
2g
Displacement till rolling begins
1
S ut at 2 ;
2
1
2 R 2
S 0 mg 02 2
2
4 g
20 R 2
8g
2 R 2 2 R 2
mg 0 0
8g
8
Velocity v and work done by friction do not depend on value of coefficient of friction
Work done by friction
= f is
APP | Rotation and Gravitation
;
46
Solution | Physics
Vidyamandir Classes
To the right of B, there is no friction therefore, no torque acts on the body
The angular velocity remains constant (no angular acceleration)
Rotation KE will be constant to the right of B
To the right of B, F is still being applied; therefore the object will still undergo constant linear acceleration.
46.(ACD)
47.(ABCD)
(A)
(B)
Is correct since v R at P any horizontal acceleration at P will cause slipping a R
If R 2 acom
aQ will be downward
(C)
aR can not be horizontal.
(D)
Correct
48.(BC)
Conservation of linear momentum
MV 2MVCOM
V
2
Conservation of angular momentum
MVl / 4 2MVCOM (0) I COM
VCM
Ml 2 Ml 2
MVl / 4 2
8
12
49.(ACD)
;
3V
5L
For M
M h p C q G r
M M 1L2T 1
APP | Rotation and Gravitation
p
q
L1T 1 M 1 L3T 2
r
47
Solution | Physics
Vidyamandir Classes
M M p r L 2 p q 3r T p q 2r
pr 1
…(i)
2 p q 3r 0
…(ii)
p q 2r 0
…(iii)
On solving (i), (ii) & (iii) we get
1
1
1
p , r
and q
2
2
2
M
h
M
C
M
1
G
Similarly For [ L ]
pr 0
…(iv)
2p + q + 3r = 1
p q 2r 0
…(v)
…(vi)
On solving (iv), (v) & (vi)
1
3
1
p ,q
,r
2
2
2
50.(AD) m1 m, m2 2m
Gm(2 m)
r2
r1
L
;
h
L
1
C 3/2
;
L
G.
2mr
2r
m 2m
3
mV12
2Gmr1
V12
r1
r2
2r1
4 2 r 2 r1 4 2 r 3
r2
T12 4 2 r12
;
T12 r 3 , T12 m1
V1
2Gmr1
2Gm
3 Gm
51.(D) 52.(B) 53.(B)
Let us first understand some general concepts of the problem.
Speed of the particle remains constant. Since the only force acting on the
particle is tension and this force is always perpendicular to the instantaneous
velocity of the particle. Hence tension does no work on the particle and by
work energy theorem; speed of the particle remains constant.
Let us denote the point at which the thread touches the cylinder by P. As we
can see, the speed as well as acceleration of this point is zero. Hence, at an
instant, in the reference frame of this point, the particle can be taken to be
performing circular motion.
(A)
Torque on the particle is due to T and obviously NOT zero about B, C, or midpoint of BC. Hence answer to
‘a’ is none of these.
T1
(B)
mv02
(r length AP).
r
r continuously decreases whereas m, v0 remain constant, T continuously increases. Torque on cylinder
T
due to T is TR. So, this torque also continuously increases. Hence, the external torque required to keep the
cylinder stationary (by balancing the torque TR) should also be increased continuously.
APP | Rotation and Gravitation
48
Solution | Physics
Vidyamandir Classes
(C)
At any instant, angular speed of segment AP is: ω
dθ
v
0
dt l Rθ
So,
θ goes from zero to θ
v0
where r Rθ l r (l Rθ)
r
… (i)
l
.
R
l/R
T
(l Rθ)dθ v0 dt
0
We can get T
0
l2
2 Rv0
PA is always perpendicular to PC, angular speed of PC and PA are same. That is why θ on
Note:
both sides of equation (i) are taken to be same.
54.(B) 55.(C)
Just after release.
For block, mg T ma
Also
a
…(i) ;
For rod, T mg
2
m 2
3
ac
F
T
mg
So let force exerted by hinge = F (up) then F T mg mac F
2G
mv
m
r
r
…(ii)
…(iii)
T 5mg / 8
3g
3g
,a
8
8
3g
For rod ac
(up)
2
16
56.(C)
5mg
8
mg
3 mg
16
F
9 mg
16
2
Where mass per unit length A R 2
3R
57.(B)
R
2G R 2 v 2
r
r
v R 2G
2G m
1
dr mv 2 v 2 R g ln 3
r
2
58.(D)
T
2d
G
d RT
v
2
59.
[A- qrs, B-prs, C-qrs, D-prs]
Conserve angular momentum about the hinge and use the equation for e to get angular speed of rod and speed of
particle just after collision. Thereafter, you may calculate the linear momentum of rod using P = M.Vcm.
60.
[A-p, r] [B-p, q, r, s] [C-p, q, r, s] [D-p, r, s]
61.(5)
Conservation of angular momentum
mV0 sin 5R mVR
. . . . . (i)
Conservation of energy
GMm 1
GMm 1
m V02
mV 2
5R
2
R
2
. . . . . (ii)
1
8GM
Solving sin 1 1 2
5
5V0 R
APP | Rotation and Gravitation
49
Solution | Physics
Vidyamandir Classes
62.(2)
For a particle at a distance r from the center of Earth, force is given by,
Gm1m2
F
r2
Force becomes one fourth, when r = 2R (R = radius of Earth)
2GM
R
Using conservation of energy for the given particle,
Escape velocity, Ve
1
GmM
GmM
mV 2
2
R
2R
This gives,
GM
R
Ve V 2
V
63.(5)
P is neutral point
GM
r
2
G (16M )
(10a r ) 2
r 2a
GMm G16Mm 1
GMm G16Mm
mV 2
8a
2a
2
2a
8a
64.(2)
Hence, n = 2
;
V
3 5GM
2
a
Mgr 1 Mr 2 2
=
2
2 3
K 5
;
=
3g
r
5 gr m r
;
Mr 2
3g
3
r
m = 2kg
65.(6)
By linear momentum conservation impulse (J) = mV.
By angular momentum conservation, angular impulse = J
6v
mv
mv
=ISo mv = I or =
=
=
= 6rad/s
2I
m 2
2
2
2
12
66.(1)
mass of Cavity M
Fnet F full Fcavity
M
3
4 R
M
3 2
8
4 3
R
3
GMm
GM m
23 GMm
2
2
100 R 2
(2 R )
(2.5 R )
GMm 1
GMm
mV 2
3R
2
R
4GM
3R
K 23,
n4
K
1
23
67.(4)
68.(5)
Time in which C.M. reaches its highest point = 1 sec. (from v = u + at, putting v = 0, u = +10, a = 10 m/sec2) after
projection angular velocity will not change as the torque of external forces is zero.
V
In 1 sec., the rod will rotate by an angle = t =
× 1 rad.
2
The rod will be vertical with point A at the lowest point.
APP | Rotation and Gravitation
50
2 4
aA = g – 2L/2 = 10 –
= 5 m/sec2
4 2
Solution | Physics
Vidyamandir Classes
69.(8)
3mg
mg N
2
For point O
N
5mg
2
;
R acom
;
5
f mg
2
f
acom
m
3
I mg 2 R fR
2
mR 2
3
5
6 g 5g
mg 2 R mgR
;
2
2
2
R
f
5
R 6 g 5g
;
6 mg 5g mg
m
2
12 4
;
10 8
15 5
70.(7)
R ( R r )2
;
r
Also,
VC r
VC ( R r )
Where is the angular velocity of cylinder about C
( R r )
r ( R r )
r
ax of point P = 0 (pure rolling) acceleration of P along Y w.r.t. C is 2 r vertical
upward.
However C itself has acceleration in Y 2 ( R r ) vertically downward
(aP /c )Y aC aP
2 r 2 ( R r ) aP
2
( R r )
2
r ( R r ) aP
r
71.(1)
R ( R r ) 2
r
aP
;
7 R( R r )2
kr
Before cutting, apply equations of translational and rotational equilibrium about point P.
N1 T mg and T (2l )sin mgl sin
N1 mg /2
…(i)
After cutting, the acceleration of centre of mass will be vertically downwards.
Z about POC
T
ml 2
3
and
acom
…(ii)
l
2
Now, along vertical
macom mg N 2
…(iii)
…(iv)
Solve equation (i), (ii), (iii) and (iv)
APP | Rotation and Gravitation
51
Solution | Physics
Vidyamandir Classes
72.(3)
If the COM reaches P, the triangular frame will complete circle. By conservation of energy.
(Assuming initial P.E. = 0)
Initial K.E. = find P.E.
5
1
2mg ( L cos30 cos37 L cos30) 2 mL22
4
2
10 10 3 4
3
2
2.5
4 2 5 2
;
4
2 3 3 2
5
4 3 5 3 2
;
(9 3)1/2 3(3)1/ 4
x3
73.(8)
For critical case
Initial KE = P.E. at max height
l 1
mg I Hinge2
4 2
Angular momentum conservation about hinge
l
mv I
2
l ml 2 ml 2
m 2 gh
(v 2 gh v 2 u 2 2 gh )
2 12
4
6
2 gh
4l
74.(3)
;
2mg
l 1 ml 2 ml 2 36
2 gh
4 2 12
4 16l 2
;
2l
2
h 12 8
3
3
;
h=8
Length = 2l
Apply pseudo force ma to left at centre of mass of rod by translational equilibrium,
N1 ma
…(i)
By rotational equilibrium about point P,
mal sin 30 mgl cos30 N1 (2l )sin 30
mal mgl 3
N1l
2
2
Put (i) in (ii),
ag 3
APP | Rotation and Gravitation
…(ii)
R3
52
Solution | Physics
Vidyamandir Classes
75.(4)
Moment of inertia of each rod
ml 2
l
m
12
2
2
2mr 2
3
For entire object,
( l 2r )
2mr 2 8mr 2
I 4
3
3
Now,
4ma 4mg sin f
and
fR I
f
Ia
R2
8ma
f
3
…(ii)
Putting (ii) in (i),
20ma
4mg sin
3
Putting in (2),
f
a
f mg cos 4
M(2d cos – d) – m
2md cos –
78.(2)
79.(6)
80.(3)
8mg mg
4
5 2
2
4
10
Angular momentum of the particle is conserved about the vertical centre line
mv0 r0 mvr cos
where conservation of mechanical energy gives,
1 2
1
r 2 r02 h 2
mv0 mgh mv 2
;
2
2
77.(7)
12 g
20 2
8m 12 g
8mg
3 20 2 5 2
f N
Now,
76.(3)
…(i)
;
1
L2
d
3md
7 md
– 2md + 2md cos = 0 ; 4md cos =
2
2
m1v1r1 m2 v2 r2
R
1
m2 v2 r2
M
3
d
M
R
Ve2
2
11
2
3
2 gh
h2
1 2
2
v0
r0
d
– 2m (d – d cos ) = 0
2
2 total energy = G.PE TE 2MJ At , total energy = 0
L
1
cos
2
6
121
.
1
R
m1v1 r1
m2 v2 r2
Ve2 9
APP | Rotation and Gravitation
;
cos =
7
8
additional energy provided = 2MJ.
11
1 1
6
2 .2
g
d g 2
ve
ve2 d g 2
Ve 3
53
Solution | Physics
Vidyamandir Classes
LIQUIDS
1.(C)
Height of water which exerts same pressure as oil is : h w g 5 oil g h 5 0.8 4m
Total height = 10 + 4 = 14 m
2.(B)
v 2 10 14 280 m / s .
Let mercury level drops in the left side by x mercury on right
x
side would rise by 2
n
Equating pressure at A & B
x
x
h
h g x 2 sg
2
2
n
n
n 1 s
3.(D)
The whole system must have a horizontally rightward acceleration
for heights of both side to be same
1gl 2la 3 gl
4.(A)
= density of material ;
0 = density of water when the sphere has just started sinking, the weight of the sphere = weight of water displaced
5.(C)
V3
R
4
4
R 3 r 3 g R3 0 g
3
3
3
r3
7
r
R
2
0
1/ 3
R3
2 gh ; using continuity equation at section '2' and section '3'
A
A
1
hg
V2 V3 V2 V3
2
4
2
2
Using Bernoullie's theorem at section '2' and at the opening end of pipe '3'
1
1
1
1
gh
3 gh
P0 V32 P2 V22 P0 V32 V22 P2 P0 2 gh ; P2 P0
2
2
2
2
2
4
6.(B)
Initially, w 2kx 0
a a3
Finally, w 2k x
. 2g 0
2 2
. . . . .(i)
w w w0 w0 a k a 2 g
7.(A)
4
Mg 2T1 cos 45 Fb R3 w g
3
8.(C)
Loss in GPE =
m
h
gh1 h1
APP | Liquids
. . . . .(ii)
4 3
3 R w g Mg
T1
2
m
h
mg h h1
gh1 h1
P
9.(A)
L
dP . A
Adx
2
x P
0
0
F PA
AL 2
M 2L
L
2
2
54
1
2 L2
2
Solution | Physics
Vidyamandir Classes
2 3
r 1 g
3
2 3
( P0 1 gh)r 2 F
r 1 g
3
2
F P0 r 2 h r r 21 g
3
11.(B) Applying Newton’s law in vertical direction
mg FB
dSH g w Sh g
10.(A)
PA F FB
h
dH
Now when force F is applied, for minimum work a 0 ( For a 0, F is minimum)
F mg Sxg 0
W
F dx
(mg Sxg ) dx mg
dx Sg
x dx
2
Sgd 2 H 2
Sgh2
Sgh 2 Sgh 2 Sg dH
(Sh) gh
2
2
2
2
2
12.(A) Taking cylinder and the ball as system
W mgh
4 3
4
r 2 g Ah 1 g r 3 w g Ah1 w g
3
3
1/3
3 A(h1w h1 )
r
4 ( 2 )
A 11cm 2 ; h1 4 cm; w 1gm / cm3
1 0.5 gm / cm3 ; 2 8 gm / cm3
1/3
1/3
3 11(4 1 6 0.5)
3
cm
r
4 22 (8 1)
8
7
4T
13.(B) Inside pressure must be
greater than outside pressure in bubble. This excess pressure is provided by charge on
r
bubble.
4T 2
r
2 0
;
4T
Q2
r
162 r 4 2 0
;
Q
...
4r 2
Q 8r 2rT 0
14.(C)
r R sin
Required force [2r ]T sin 2R sin 2 T
15.(D) Let the density of water be , then the force by escaping liquid on container S ( 2 gh )2
Acceleration of container a
Now
APP | Liquids
Sh
v
a
2Sgh vg 2Sh
g
v
u
Sh
g
v
55
Solution | Physics
Vidyamandir Classes
16.(ABD)
PD = PB (Pascal's Law)
Let be the side of the square PB PD PC g
Similarly, PB PD PA g
17.(ACD)
V12
V22
P1
P
2
2
2
But, P1 P2 gh
18.(AC) tan
2
V22 V12
2
. . . . .(i)
2
. . . . .(ii)
P1 P2
V22 V12
2
PB PD
PA PC
2
V22 V12 2 gh
h a
hg
a
c
g
c
Maximum volume that can be retained =
1
hcb
2
hcb hg
And F M
2 c
WF WFB Wmg 0
19.(ABD)
WF WFB Wmg
WB ugravitational
WB ugravitational
20.(CD) T B m1 g
N B m2 g
T
N
N m1 m g T
B
B
m2g
m1g
21.(BC) Since body is floating, Buoyant force is same in both liquids and is equal to the weight of the body.
Pb
same
PL
22.(AD) (A)
T 2
l
geff
(B)
Fraction
(C)
T 2
m
k
(D)
P P0 hPgeff
23.(BD) In both cases VA VB
PA v 2 PB v 2
h
Pg
y Pg
y
PA PB in both cases.
;
24.(AB) Total displacement of mercury against atmospheric pressure
l 5l
Process l l
2 2
PEinitial S
l l
2 0 g
2 4
{Assuming zero at ground level}
l
PEmax S 2 Sl l g
2
APP | Liquids
25.(AB)
56
Solution | Physics
Vidyamandir Classes
26.(ABD)
FBuoyant (mA mB ) g ; 2vd Fg v (d A d B ) g
d A d B 2d F . Therefore A, B, D
27.(AB) By equation of continuity,
A1V1 A2V2
S1 u S2 u cos
Now range R
u 2 sin 2 u 2 2 sin cos 2u 2
g
g
g
H max
S
cos 1 1
S2
sin 1
S12
S 22
S1
S2
1 1
S2
S 22
u 2 sin 2 u 2 S12
1
2g
2 g S22
Rate of flow volume is ( S1 u ) or ( S2 u cos )etc.
1
1
28.(AD) P1 gh1 V12 P2 gh2 V22
2
2
h1 h2 0
(Horizontal pipe)
and
A1V1 A2V2
4(4) V2
16 V2
1
1
2.80 105 (4)2 P2 (16) 2
2
2
5 1
2.80 10 (16 256) P2
2
1
2.80 105 900(240) P2
2
172 103 N/m 2 P2
F1 P1 A1 56 103
d Poil
F1 F2 Fpipe
dt
ˆ
F P1 A1i
F2 P2 A2 (cos 37iˆ sin 37 ˆj )
F2 P2 A2 8.6 103
d Poil
Fpipe
( F1 F2 )
dt
;
;
d Poil dm
V (A1V1 ) (V2 V1 )
dt
dt
;
V2 16[cos 37(iˆ) sin 37( ˆj )]
V1 4iˆ
Solving this for Fpipe we get, | F | 76 103 N
A
29.(BD) t
A0
APP | Liquids
2
( H H )
g
;
t1
t2
H
H
2
H
0
2
57
2 1
2
Solution | Physics
Vidyamandir Classes
30.(AB) dA 2 dy
1
cos
and
dF
2 R 2 h 2 V
dy
h
t
2 R 2 h2
Rt
h
; dy dx
R
R x
tan
h y
;
d
2
R 2 h2 x3 dy
ht
R
x3 dy, P
0
h
g = 4r 2
.g
3
3
Force due to pressure (P1) created by liquid of height h1 above the wooden block is
2
31.(C) Weight of cylinder W 2r h
2
2
= P1 2r P0 h1 g 2 r P0 h1 g 4r 2
Force acting in upward direction due to pressure P2 exerted from below the wooden
2
2
2
block and atm pressure is = P2 2r r P0 r
= P0 h1 h g 3r 2 r 2 P0
At the verge of rising P0 h1 h g 3r 2 r 2 P0 4r 2 h g P0 h1 g 4r 2
3
32.(B) Balancing forces h 2
5
h1 h
3
4h
9
33.(A) When the height h2 of water level is further decreased then the upward force acting on the wooden block decreases.
The total force downward remains the same. This difference will be compensated by the normal reaction by the tank
wall on the wooden block. The block does not move up and remains at its original position.
34.(D)
35.(C) mg Fb
2
a 0.4a
5
a3 0.4 g a 2 hg
;
h
m a3 0.4
;
Fnet a 2 xg
;
T 2
T 2
a3 0.4
2
a g
2a
5g
36.(B) Displacement must be less than submergence depth of cube.
37.(A) v1w g (v v ) i g v (m ) g
v2w g (v v ) i g v (m ) g
v
1000
1 i m
1 0.9 4.9
v1 (v v ) i v m v
49
20
v2 (v v ) i vm v
46
1000
1 i m
1 0.9 1.9
v
20
38.(C) Mass of cavity is more in cube A than in cube B.
APP | Liquids
58
Solution | Physics
Vidyamandir Classes
39.(D) So long as cubes are floating, respective water levels do not change.
Let at t t0 , cube A sinks.
vw g (v v )i g vm g
v is volume of cube which is changing linearly with time at t t0 .
vw g N A (v v )i g vm g
After sinking water level decreases due to melting of ice,
dh
1 dv
A 1 ; A -cross-section area of vessel
10 dt
dt
Let at t t0 , cube B sinks
;
v w g (v v )i g vm g
dh
1 dv
A 2
10 dt
dt
Final heights are same in both reach.
40.(B) p p0 pgeff
dv dv
dt
dt
dh1 dh2
dt
dt
2
v2
geff g 2 3g
R
g2
41.(D)
v2
R2
8g 2
;
100 4
R2
8g 2
;
R
1004
25 500
100
250 2
8 100
2
2
P P0 P h g eff (0.30) (103 ) (3 10) 9 103 Pa
42.(D) Only in case D, geff g
43.(A) P 2 ; Q 1 ; R 1 ; S 3
Fx W Fy
W Fy Fx which is less than 2h g A 2
44.(C) (P)
(Q)
(S)
F Pc A
h g
A
2
FB V g hA g
FR
(R)
P V t u
u2 A
t
t
1 2
u P2 0
2
1
P2 P1 u 2
2
P1
F P2 P1 A
APP | Liquids
59
1
u2 A
2
Solution | Physics
Vidyamandir Classes
45.
46.(5)
(A-r); (B-p); (C-q); (D-s)
F
( R )
(2Rl )R
;
F
A
h
h
;
;
FR
Power
The force on small element of the slit dF dA v 2
dF bdx 2 g h x F 2 bg
h x dx 2bg h
0
2
2
On putting values we get F 5 N
47.(2)
FB y A w g
mg 1 A
Balancing torque yA w g
48.(8)
49.(6)
w
g
2
y
1
sin 1 A w g cos
2
2
2
4
4
R3 n r 3 R n1 / 3 r or R 2n1 / 3 mm
3
3
v = speed of efflux
Bernoulli’s equation between A & B,
1
0 0 gx v 2 P0 0
2
2P
v 2 gx 0
At equilibrium height V 0
2 gx f
cos 2
1
2
V01 R 2 4n 2 / 3
4
V0 r 2
4
1
cos 2
2
n8
2 P0
P
105
x f 0 4 m 10 n 4
n6
g 10
1
5
50.(1.50)
1 Vd w 1 Vdl
3
9
2 9
Therefore, dl d w 1.5 1000 1500 kg/m3
3 4
51.(2) Applying Bernoulli’s equation at cross-section 1 and 2.
Patm gh 0 P2 0 0
P2 Patm gh
…(i)
Again applying Bernoulli’s equation at section 2 and 3
h
1
P2 0 2 g Patm 2V 2
…(ii)
2
2
V 2 gh
…(iii)
This is required velocity of efflux
Applying continuity equation between 3 and 4 cross-section,
aV a1V1
Again applying Bernoulli’s equation between (iii) and (iv)
APP | Liquids
60
Solution | Physics
Vidyamandir Classes
1
h
1
Patm (2)V 2 2g Patm (2)V12 0
2
2
2
6 2 gh
aV
a1
2 cm 2
V1
3 gh
V12 3gh
52.(300) Let the cube dips further by y cm and water level rises by 2 mm.
Then equating the volume (/// volume = \\\ volume in figure)
Volume of water raised = volume of extra depth of wood
100 y (1500 100)
2
2
1400 280
10
10
y 2.8 cm
Extra upthrust
water (2.8 0.2) 100 g mg
m 300 gm
53.(40) As block goes down by distance x, water comes up by distance y. As both are measured from initial level of water,
compression in the spring is x but the block is in depth ( x y ) in water.
Applying conservation of volume
0.2 x (1m 2 0.2 m2 ) y
x 4y
y
x
4
x
5x
x
4
4
Thus total depth of block in water
Free body diagram in equilibrium :
5x
Fb (0.2) (1000) (10)
4
For equilibrium :
mg kx f B
5x
1800 2000 x (0.2) (1000) (10)
4
18
m = 40 cm
45
54.(0.45)Pressure of gas inside the balloon is same as the pressure of surrounding. Also gas inside the balloon obeys isothermal
process, then :
18 20 x 25 x
( P0 gh)V1 P0V2
x
103 10 40
V2 1
0.09 0.45 m3
5
10
55.(0.29)Force of buoyancy
APP | Liquids
61
Solution | Physics
Vidyamandir Classes
FB A[ x0 ( L x )0 ] g
Weight W LAs g
For equilibrium FB W
LAs g A[ x0 ( L x )0 ] g
s
56.(0.5)
57.(10.50)
x
( 0 w ) w
L
Ls x0 ( L x ) w
x w s 1000 800 2
L w 0 1000 300 7
76 L 42(76 57) L 10.5
10.5 cm 1cm 2 10.5
4
4
4
58.(11.11) 1000 r 3 g 500 r 3 g 6rv 500 r 3 a
3
3
3
0.2
6 1104 v
3
59.(10) R 2 cm
;
200
v
18
4S 4 0.05
10 Pa
r
2 102
Let v be the velocity of the movable plate and F is equal to viscous force
P
60.(3)
v
v
F 1 2
A
h h1
h1
APP | Liquids
dF
0
dh1
62
h1
h
3
Solution | Physics
Vidyamandir Classes
PROPERTIES OF MATTER
1.(A)
Tension at point x mw2 r
m
x
T x w2
2
mw2 2
x2
2
Let x be latent heat of fusion
w be water equivalent of apparatus
T
2.(A)
200 70 40 w 70 40 50 x 40 .........
250 40 10 w 40 10 80 x 10 ..........
Solving i & ii x 90 cal / g 3.8 105 J
2
q 1 aT 3 dT
3.(B)
dq msdT
4.(B)
d
1%
dA
2 1% 2%
A
5.(C)
d dx T
1
Integrating 1.208 cm
6.(C)
R1
2
& R2
kg
15a
4
d 3x 2 106 20 dx
length = 2.0120 m
K1 r
3K 2 r 2
Both rods are in parallel
1
1
1
Req R1 R2
7.(B)
i
ii
4 r 2 K eq
r 2 K1 3 r 2 K 2
Keq
K1 3K 2
4
Kx
dT
H K x A
dx
Kx dT
H 2K
A
dx
Integrate and get temperature as a function of x
Alternative approach
K x 2K
Initially K is higher
Later K decrease
dT
should be smaller in magnitude
dx
dT
increases
dx
8.(B)
R 2 l n r 2 l............... i and n 2 rl 2 2 Rl .............ii
9.(B)
10
K 55 To ............ i
4
10
K 35 To ............ ii
8
APP | Properties of Matter
Solve for n
n4
Newtons low of cooling
average method
Solving
i & ii
63
To 15C
Solution | Physics
Vidyamandir Classes
10.(B) Let T : tension at mid point, for the outer half of the rod
m
3
T
3m 2
T 2
and
Y . strain
strain
2
4
A
8 AY
11.(B)
12.(D)
2.4 108 3 10 4
1200 10 1200 a a 10 m / s 2
3
d
ms
A T 4 T04 m1 3 m2 r1 31 / 3 r2
dt
1
13.(A) Error . 86400 4.32 sec
2
14.(A) According to Newton’slaw of cooling
15.(A)
dT
T
dt
tan 1 35 20 15 3
tan 2 25 20 5 1
tan 2 1 9 tan 2 2
sec 2 1 1 tan 2 1
sec 2 1 1 9 tan 2 2
Y
Stress
for same strain (stress) A (stress) B
Strain
YA YB
Breaking point of B Breaking point of A
So, A is less ductile.
16.(ABC) In the case of a perfectly rigid body and incompressible liquid, volume strain is zero.
W ms
or,
m
W
4.5
50 g
s 0.09
The thermal capacity and the water equivalent of a body have the same numerical value. Also,
17.(BCD)
Q 4.5 8 36 cal
Since, the temperature remains constant, during the process of melting, no heat is exchanged with the calorimeter and
hence, Q 15 80 1200 cal . Hence, the correct choices are (B), (C) and (D).
18.(AC) (Stress)s =
F
F
, (Stress)Cu =
2A
A
strain s
stress s / Ys
strain Cu stress cu / Ycu
Given that
YS
2
YCu 1
F / 2 A 1 1
F / A 2 4
So, options (A), (B) and (C) are correct.
19.(BCD)
Thermal force = YA d Y r 2 d
r1 = r, r2 = r 2 , r3 r 3 , r4= 2r
F1 : F2 : F3 : F3 = 1: 2 : 3 : 4
Thermal stress = Y d
As Y and are same for all the rods, hence stress developed in each rod will be the same. As strain = d , so strain will
1
2
also be the same. E = Energy stored Y strain A L
2
E1 : E2 : E3 : E4 1: 2 : 3 : 4
So, option (B) and (C) are correct.
APP | Properties of Matter
64
Solution | Physics
Vidyamandir Classes
20.(BC) We can see the slope of second step (liquid heating) is less than that of first step (solid heating) hence specific heat of
liquid is greater than that of solid. We can also see that the horizontal portion of first part is smaller than that of second
part which implies that solid melting is taking less time than liquid vaporization hence (C) is also correct.
A
21.(AD) H T1 T2
(A)
Temp. diff. quadrupled, c / s area halved, H doubles (correct)
(B)
Temp. diff. doubled, length quadrupled, H is halved (incorrect)
(C)
c / s area halved, length doubled, H becomes 1/4th (incorrect)
(D)
Temp. diff. doubled, Area doubled, length doubled, H is doubled (correct)
2
22.(AC) mA 4mB
S A 4 3 SB
(mass)
(surface area)
H A SA
dT
S
also
H B SB
dt
m
23.(AD) Hg Fe density of Hg decreases more on heating. Hence cube will be submerged more.
If fraction submerged is f
HgVfg FeVg f
Fe
Hg
Fe Fe
Hg Hg 1 Hg 25
Change of fraction
1
Fe
1 Fe 25
Hg
1 180 106 25
(180 35) 106 25
1
3.6 103
6
1
1 35 10 25
dQ
eA(T 4 T04 ) same
;
dt
26.(ACD) Considering heat flow at junction A
Heat in = heat out
x 1 3 5
x 1
x is positive so heat flows in from E to A
25.(AD)
TE TA
TC TA TD
TB TA TE TA
0.36 % 0.4%
24.(BD)
d eA 4
(T T04 ) different
dt
ms
TB TE
27.(ABCD)
When a body is heated all dimension of the material as well as the enclosed lengths, areas and volumes also
increase.
28.(AC) As the two ends of the rod are at constant temperature that means the rod is in steady state of thermal conduction hence
throughout the length the temperature gradient will remain constant and in steady state of thermal conduction the rate of
heat flow (heat current) is given by the product of thermal conductivity, cross sectional area and the temperature
gradient.
APP | Properties of Matter
65
Solution | Physics
Vidyamandir Classes
29.(BC) When temperature increases of a bimetallic strip, the one which is having more value of coefficient of expansion will
expand more so that strip will bend toward the metal with low value of coefficient of expansion and it will bend toward
high value of coefficient of expansion if it cooled.
30.(AD) By wein’s displacement law we use
mT constant
mA TB
mB TA
TA mB 800
2
TB mA 400
By Stefan’s law, we use
E A AATA4
4rA2
EB ABTB4
4rB2
4
TA
4
TB
31-32. 31.(B) 32.(A) 33.(C)
Power of Heater H H 2 H1 =
80 0
100 80
10
dm
dm
H2 L f
rate of melting
dt
dt
80 0
80
dm
dm
10
H1 80 C water H2
6 cal / s
100 C water
0 C ice
H
Heater
0.1 gr / sec
10
dt
dt
Since temperature of middle container is constant, H2 is constant, hence the rate of melting of ice will be constant
40 36
39 x
36 x
k (36 30);
k
30
10
10
2
4
8
x
5x
5x
2 84 168
;
12 72 2 x; 72 12;
84; x
33.6
36 x x
2
2
2
5
5
30
18
2
d
kA
35.(D)
(T T0 ) Magnitude of slope will decrease with time.
dt
ms
40 36
4
1
36.(C)
k (38 30
k
10
10 8 20
When the block is at 38 °C and room temperature is at 30°C the rate of heat loss
d
ms
ms k (38 30)
dt
1
Total heat loss in 10 minute
dQ ms k (38 30) 10 2 8 10 8 J
20
Now heat gained by the object in the said 10 minutes.
Q ms 2 4 8 J
Total heat required = 8 + 8 = 16 J
37.(B) Material which is most ductile is easy to expand is used for making wire.
38.(C) Material which breaks just after proportional point.
39.(D) Material which retains permanent deformation.
34.(D)
40.(B)
V V0 (1 L T ) (10) (100) [1 5 105 20] 1000 (1 0.001) 1001 cm3 1001 cc
41.(B) Cross-sectional area of vessel at 40 °C
A A0 (1 2 g T ) 100 (1 2 10 5 20) 100.04 cm 2
Actual height of liquid
APP | Properties of Matter
Actual volume of liquid
Cross-sectional area of vessel
66
Solution | Physics
Vidyamandir Classes
42.(B)
1
1001
1001 0.04
(1001) (100 0.04)1
1
100.04
100 100
(1001) 0.04
1
1000.6
(1001 0.4)
10.006 cm
1
100 100 100
100
TV SR (1 g T ) (TV ) (1 g T )
SR (TV ) (1 g T ) 1 (10.006) (1 105 20) 10.006 0.002 10.004 cm
43.
[A-r; B-p; C-q; D-s]
F cos F
cos 2
A / cos A
F sin F
cos sin
A / cos A
F
sin 2
2A
;
T will be max if 2 90
will be max if 0
44.
[A – p r s ; B – p r s ; C – p ; D – p q r s]
45.
[A-q; B-r ; C-s; D-p]
10 A
20 A
(a)
(100 T )
(T 0)
;
2
1
10 A(100) 20 A(100)
(b)
;
1
1
(0.2 0.1)
(c)
Rth
40 SI units
4 0.2 0.1
F F
(T) Tension at any point x F2 1 2 x
46.(1)
Integrating
47.(2)
F1 F2
2 AY
Tension in ring = A 2 r 2
A r
48.(3)
Conduction rate
H
50.(8)
3000 100 10 4 30 W
dt 100 0
10 C/m
dx
10
T
d
dx
AY
(d)
;
2
Water = 3 kg
radius
H (max) = K
A
2
length
Let S power be lost to surrounding
16 S 10 m
. . . .(i) and
Solving we get S = 8 watts.
APP | Properties of Matter
2 rad / s
A
r2
Let x kg of ice remaining
2 15 500 2 x 80,000 2.5 251000
T 20
[A = C/s are of ring]
x 1.4 kg
49.(8)
100 T 4T 0 ;
1 10 9 m x 1
2 2
For breaking
45
4r 2
; H min K
32 S 10 3m
67
r2
2
H max
H min
8
. . . .(ii)
Solution | Physics
Vidyamandir Classes
51.(3)
52.(2)
mg
A
dp
3dr
dr
mg
K
K
dp
n3
dv / v
r
r
3 AK
1
1
5 T 15 86400
. . . .(i)
;
10 33 T 86400
. . . .(ii)
2
2
Where T is temperature at which clock shows correct time. Solving T 21C and 2 105 / C
dp
T
53.(4)
54.(3)
t
10
k Tavg T0
k 70 30
. . . .(i)
2
t
Solving (i) and (ii) t = 4 min.
For no heat flow in AB temperature of Junction B must be 20C
Material and cross-section are same
80 20 20 0
BD
BC
16000 20000 9000 (540) ms
58.(4)
. . . .(ii)
Temperature gradient is also same
BD
3
BC
;
45000
ms
540
;
ms
250
83.33
3
Steam left 87.33 83.33 4
Heat used for evaporation = 900 kJ
900 103
J/kg
0.2
L
0.0650
0.0350
Using
; 1
; 2
;
L0 T
30 100
30 100
Solving L01 23 cm and L02 7 cm
Lv
Mass evaporated = 0.2 kg
57.(7)
k 50 30
Qab 200 80 200 1100 (200 1 45)
Qloss ms (540)
55.(4)
56.(9)
10
and
L01 1 T L02 T 0.0580 and L01 L02 30
mg
2sin 37
Change in tension T
2T sin 37 mg
;
T
But
mg L
L
2 sin 37 AY
m A Y 2sin 37 / g
l L
;
l
TL
AY
2 105 10 10 6 5 1011 2 3
12
5 10
From Newton’s law of cooling
ms 1 2 1 2 0
t
2
59.(3)
50 45 50 45
In the first case ms
0
5 2
45
41.5
45
41.5
In second case
0
5
2
From eq. (i) and (ii) 0 33.3C
…(i)
…(ii)
60.(0.18)
APP | Properties of Matter
68
Solution | Physics
Vidyamandir Classes
GASEOUS STATE & THERMODYNAMICS
1.(B)
Apply gas equation PV= nRT to identify decrease or increase in temp and volume for each process
2.(A)
As the system is thermally insulated, Q 0
Further as here the gas is expanding against vacuum (surroundings), the process is called free expansion and for it,
W PdV 0
[As for vacuum P = 0]
So in accordance with first law of thermodynamics, i.e. Q U W , we have
0 U 0,
3.(A)
i.e. U 0 or U = constant
ab and cd are adiabatic
PbVb
Pd Vd
bc and da are isothermal. Hence
PcVc
PaVa
4.(A)
5.(B)
6.(C)
For adiabatic process (from A to B)
W 322 J
U W 322 J
For another process (from A to B)
Q U W
100 J 322 J W W 222 J
RT dV
3 R RT dV
C Cv
;
R
V dT
2
V dT
V
from PV nRT
PV V P nR T
V T
dV 1 dT
0 VT 1/ 2 constant
V
2 T
;
nR 1
P T
At constant pressure Q nC p T ; also Q U w nC p T nCV T w
Since ‘P’ is constant
7.(B)
PcVc
PaVa
Pb
P
a
Pd Vd
Pc
Pd
PbVb
n C p CV T w
PV nR T
nRT 25
6
RT 75 J & Q 25 75 100 J
2
RT dV
R
Q U W C CV
C CV
1 V
V dT
V
U n
8.(B)
10.(A) Sina WA WB and U is same in both process
11.(B)
PV constant
P
12.(B)
U a bPV a nbRT
But
13.(C) Along a to b, Δu 0 and along b to c, Δu w 4 j
14.(A)
5R
3R
U 2
T 4
T 11RT
2
2
QA QB
constant
dU
nbR
dT
9.(D)
7
P
32 5
p
dU
dT
nR
1
nbR
nR
1
b 1
b
a to c, Δu 0 4 4 J (for any path)
RT
3RT
2
. 0.68 1.3872 , f
5.16
M
M
1
APP | Gaseous State & Thermodynamics
69
Solution | Physics
Vidyamandir Classes
2
15.(C)
P0V0 PV1 PV2
and V1 V2 2V0
T0
T1
T0
V1 = x1A, V2 = x2A
x1 = 44 cm and x2 = 40 cm
16.(ABC) For adiabatic process : T i Vi 1 T f V f 1
For isothermal process : Pi Vi Pf V f
17.(ABCD)
18.(ABD)
In case of cyclic process : 1
QR = Heat rejected ;
QR
Qg
Qq Heat given
19.(ABD)from the graph we can conclude that
Wgiven process Wisothermal process
For AB process
k 2
P kV C T
V CV
nR
Which is the equation of parabola
1
20.(CD) For A, D
T1Va 1
T2Vd 1
For B, C
T1Vb 1
T2Vc 1
Va
Vd
Vb
Vc
1
T2
T1
1
T
2
T1
Vb T2 1
Vc T1
Va
Vd
Vb
, i.e., choices C and D are correct. Hence choice A and B are wrong.
Vc
m 2 c
21.(CD) P mV c T
T3 T1 T2
V nR V
nR
1 3 expansion process accompanied by heating
So
3 2 expansion process accompanied by cooling
22.(ABD)Degree of freedom of He 3
Degree of freedom of H 2 5
Average degree of freedom
3 2 5 2
4
22
2 3
;
f 2
From first law U w
23.(AD) This process is free expansion so
Q U W 0
Now 1
For adiabatic process Q 0 and PV cost or TV 1 cost
[ Q 0]
and PV
1 1 P2V2
Only initial and final points are defined, process in between is not defined.
APP | Gaseous State & Thermodynamics
70
Solution | Physics
Vidyamandir Classes
24.(AD) As, Wby Q 0, So U constanct so
5
5
2 R T0 3 3R 2T0 2 R 3 3R T
2
2
23RT0 14 RT
By
n1 n2
n
n
1 2
1 mix 1 1 1 2
23T0
14
19
14
T
mix
25.(ABC) Q U W
For process A to B, Q W
For process B to C , Q U
For process C to D, U W
For process D to A, U W
26.(BD)
P2
k
P 2 RT
k
PM
P 2 P 2
P
kM
PT
R
…(i)
P
2
Hence from eq. (i) T T 2
PT constant hence P-T curve is a hyperbola.
27.(AC) PV 1 RT
( V 2 )V RT
;
For maximum value of T ;
T
V V 2
R
R
dT
0
dV
dT 3V 2
dV R
R
;
V
3
28.(AB) Work done in the process A to B nCV T
3
nCV (TA TB ) 2 8.314 150 3741J
2
For the process B to C :
PB PC
TB TC
TC
29.(ACD)
PC
TB 425 K
PB
PC
TB 425 K
PB
TC
;
Q nCV T
3
2 8.314 (425 850) 10600 J
2
(a) Process AB : PT constant
nRT 2
constant
V
B
;
APP | Gaseous State & Thermodynamics
W
B
PdV
A
A
71
Constant
dV
T
Solution | Physics
Vidyamandir Classes
100
Constant 2nRT
dt
T
Constant
dV
2nRT
dT Constant
;
PA TB
PB TA
1 TB
300
TB
100
3 TA
3
W 2nR (100 300) WAB 400 nR
300
(b) Process CA : Isochoric
W
P
constant :
T
TA PA
TC PC
;
TA 1
TC 3
TC 900 R
TA PA
TC PB
TC 3TA
3
3
U nCV T (1) R (TA TC ) R (300 900)
2
2
| U | (900 R)
(c) Process BC : Isobaric Q nCP T
5
5
Q (1) R (TC TB )
;
Q R (900 100)
2
2
30.(AD) Point A and C are on the same line passing through origin
PA PC
V A VC
Also TA 200 K
From eq. (i) and (ii)
31.(C)
Vrms
32.(D)
P
Q
5
R 800 Q 2000 R
2
…(i)
PAV A
PV
and also TC 1800 K C C
nR
nR
PAV A 1
PCVC 9
…(ii)
VA 1
VC 3
3RT
M
n
50
RT 6
8.314 20
V
10 6.03 10 23
1.4 1014 N/m 2
33.(B) Heat gain by left part = heat lost by right part
3T
3
3
nR T T0 nR 2T0 T T 0
2
2
2
3P
p P0
Let final pressure = p
P 0
T T0
2
3 3T
3 V
3
3
34.(C) Heat flow = nR 0 T0 nRT0 P0 0 P0V0
2 2
4
2 8
4
3P
35.(C) P (final in left) = 0 P (final in Right part). So when pin is removed, piston will not move.
2
APP | Gaseous State & Thermodynamics
72
36.(C)
Solution | Physics
Vidyamandir Classes
37.(C)
39.(D)
11R
Q1 nC T n
( 2T0 T0 )
2
7R
Q2 nC p T n
(T0 2T0 )
2
V f nRT0 ln 2
Q3 nRT ln
2
Vi
Total work done by the system
Positive energy supplied to the system
For cyclic process, total work done (W ) total energy ( Q
nRT0
[4( 2 1) ln 2]
2
1ln RT0
( 2 1)
2
4( 2 1) ln 2 4(0.40) 0.7 1.6 0.7
9
100% 20.5%
11(0.40)
4.4
44
11( 2 1)
38.(A)
kx
p
(V Ax)
p1V1 1 A 1
T1
T2
;
kV
pV T
kx 2 p1 A 1 x p1V1 1 2 2 0
A
T1
2000 x 2 4600 x 480 0
;
x 0.1 m
Wgas Watmosphere Wspring 0
;
1
Wgas Pe Ax kx 2 0
2
1
Q U pe Ax kx 2
2
1
Wgas pe Ax kx 2 310 J
;
2
f p1V1
5 105 Pa 0.024 m3
U
T
60 K = 1200.0 J
2 T1
2
300 K
40.(D) 41.(C)
42.(D) (1 to 3)
;
U
f
Nk T
2
Since U aV
U Cv T (a ) V (1) V
…(i)
Where V is the change in volume during the process
This gives
U V
U
V
aV
U
U
V
V
V
V
…(ii)
V U PV U U PV
i.e., W P
U U
U
( 1) [nCvT ] ( 1) U
Work done during the process is W P V
But PV nRT nCv ( 1)T
or
;
;
1
Hence W
U
1
Therefore Q U W U 1
APP | Gaseous State & Thermodynamics
…(iii)
73
Solution | Physics
Vidyamandir Classes
Let C be the molar specific het in the process.
1
1
Q C T U 1
Cv T 1
43.
[A – q, r, s ; B – r ; C – r ; D – s ]
44.
[A – q ; B – p s ; C – s ; D – q r]
Temperature at :
30 10
10 10
20 10
,
K
,
L
,
nR
nR
nR
For JK : W 0 ; U 0 Q 0
J
45.
For KL : W 10 10 ; U 0
For LM : W 0 ; U 0
Q 0
For MJ : W 0 ; U 0
Q 0
M
C Cv
Cv ( 1) R R
1
20 20
nR
Q 0
[A – r ; B – p ; C – s ; D – q]
1
W Area of triangle 2 P0 V0 P0V0
2
3 P0V 2 P0V0
3
3
5
For CA : W P0V0 U R
P0V0
P0V0 & Q P0V0 P0V0
2 R
R
2
2
2
PV
1
3 2 P V 3P V
3
3P0 P0 V0 2 P0V0 U R 0 0 0 P0V0 & Q 0 0
2 R
R
2
2
2
Maximum temperature will be for process BC.
2 P0
2 P0 2 5P0
For BC : P
V 5P0
Using gas equation : T
V
V
V0
V0 R
R
For BC : W
By using maxima/minima : Tmax
25
P0V0
8R
RT m A
RT m A
46. (2) For gas in A, P1
and P2
M V1
M V2
Putting V1 V and V2 2V
47.(7)
48.(1)
49.(5)
We get P
1 1
RT
P P1 P2
mA
M
V1 V2
RT m A
M 2V
RT mB
Similarly for Gas in B, 1.5 P
From eq. (I) and (II) we get 2mB = 3mA
M 2V
For cylinder A.
For cylinder B
dQ = nCP dT
dQ = nCV dT
nCP dT = nCV dT
C 30
dT P
= 30 1.4 = 42 K
CV
Volume of the gas is constant V = constant
PT
i.e. pressure will be doubled if temperature is doubled
P = 2P0
Now let F be the tension in the wire. Then equilibrium of any one piston gives
F = (P – P0)A = (2P0 – P0)A = P0A
Let T be the temperature of the mixture.
APP | Gaseous State & Thermodynamics
74
Solution | Physics
Vidyamandir Classes
f
f
f
(n1 n2 ) RT (n1 ) RT0 (n2 )( R)(2T0 )
2
2
2
5
or
(2 + 4) T = 2T0 + 8T0 (n1 = 2, n2 = 4)
or
T T0
3
Process A to C
Q = 210 J
10 4
Work done WAC area under AC 10 4
60 J
2
From Ist law of thermodynamics.
U Q WAc
U C U A 210 60
U C U A 150 30 150 180J
Path A to B
U B 60 J
U Q WAB
Then U =U1 + U2 or
50.(2)
51.(3)
UB U A Q 0
52.(6)
;
60 30 Q
5
1
nRT n Mu 2
2
2
53.(100) Process is polytropic C Cv
T
Q 30 J
60 10
Mu 2 30 10 3 10 4
6 x 6
5R
5 R
R
R
R
m 1
R 3
R
R
m2
2 2
m 1
PV 2 C
40 V02 P(2V0 ) 2
P 10 kPa
10 2V0 40 V0
PV P0V0
T 100 K
T
T0
T
200
54.(2)
Given,
V
dV
0
dt 400
t
V V0 1
400
From first law of thermodynamics
dQ dU
dV
P
dt
dt
dt
R 3 dT
R
4 2 dt
V
RT
0
t 400
V0 1
400
dT 2 T 1
dt 3 t 400 6
At t 0, T 40 K t thus
55.(2)
dT 2 40 1
dt 3 400 6
dT 1
1
3
1
dt 6 15 30 10
n1Cv1 dT n2Cv2 dT PdV 0
1
7R
dV
2 RdT 4
dT 4 RT
0
2
4
V
APP | Gaseous State & Thermodynamics
;
2
75
dT
T
dV
0
V
Solution | Physics
Vidyamandir Classes
T
1
2ln
ln 0
300
4
T 600 K
U
56.(75)
;
ln
1
7R
2R (600 300) 4
(600 300)
2
4
P2 V2
P1 V1
T
ln 2
300
8R 300 8
25
300 2 104 J
3
P2V1 PV
1 2
W P2 (V2 V1 ) P1 (V2 V1 )
nRT2
nRT1
, P1
V2
V1
W = nRT2 nRT1 P2V1 PV
1 2
and
P2
P2V1 P1 V2 nR T1T2
W nR ( T1 T2 )2 1
25
(20 17) 2 75 J
3
57.(205) Wgas Wspring Watm 0
1
Wgas 25 103 (0.02)2 105 0.05 0.04 0 ;
Wgas 205 J
2
58.(22) Initial pressure of the gas P0 ( Patm 40) cm of Hg = 36 cm Hg
Final pressure of the gas 1.5 P0 54 cm ofHg (sicne process is isochoric)
Difference in height (76 54) 22 cm of Hg
59.(0.25) Pressure of air ( P ) Hg g (76 x )
P Hg g
constant
V
A
Molar heat capacity (C)
R
R
1 n 1
R
R
3R
1.4 1 1 1
25
Q nC T (103 ) 3 (10) 0.25J
3
1
60.(3)
VP k constant
PV k constant : n
;
dU dW
1
k
50% heat as work
dQ dW dU
dQ
f
nR T
2
2
dQ 2dU fnRT nC T
:
C fR 3R
Now :
C
R
R
;
1 n 1
3R
R
5
1
3
APP | Gaseous State & Thermodynamics
R
n 1
76
n
1
3
k
1
3
n
Solution | Physics
Vidyamandir Classes
SIMPLE HARMONIC MOTION
1.(B)
When collision occurs then velocity of both the body get interchanged and hence
T
Tspring
2
Tspend
2
1
m 1
l
2
2
k 2
g
2
2.(B)
m a
mv 2
T = mg +
= mg +
L
L
3.(B)
T = 2
4.(D)
For upper half of oscillations, the block oscillates only with the upper spring and for the lower half of oscillation both
springs are in parallel.
5.(A)
L
= 2 sec
g
Period
x 2a sin t
a2
ma2 g
. = mg 1 2
L L
L
L
= 8 sec
g
T = 2
1
m 1
2
2
2
k
2
m
2k
T
m
k
=
T
=4
T
m
2k
v 2a cos t
At t 0, x a, v a 3
/6
6.(A)
= mg +
sin
1
2
and 2 cos 3a
1 rad / s
x 2a sin t / 6 2a sint cos cot sin a
6
6
2 sint cos t
y sin r 3 cos t 2 sin t At highest point if acceleration is greater than g, it breaks off
3
A 2 g
g
and it occurs when sin t 1 t t
2
3
6
6
2
3
1
1
2
KA2 9 5 K 1 K 8 N / m
2
2
7.(B)
Total energy U 0
8.(C)
Use x = A sin t find min. time when x =
9.(A)
X = (A + x)
A
2
Time period 2
2
g
m
2
2
sec
K
8
and x = A/2 after that compare them
x A cos t
X A A cos t
10.(D)
T 2
T1
T2
l
g
g2
GM
and g
; x is the height from earth surface.
2
g1
R x
11.(CD)In case of S.H.M net force is always opposite to displacement from mean position.
12.(ABCD)
yˆ aˆ as Y increases V decrease and U increase
APP | Simple Harmonic Motion
77
Solution | Physics
Vidyamandir Classes
13.(AC) Net force on the ball will be zero at = 0
i.e. the mean position is at a depth h0 =
or,
h0 = 0
or,
h0 =
0
0
Net force at a depth h0 + x will be
F = ( 0)Vg
or,
F is proportional to x
u=0
F = xVg
h0
Thus motion of the ball is simple harmonic hmax = 2h0 =
20
h0
v=0
E
3E
3
U
x
A
4
4
2
15.(BD) The motion of the particle is somewhat like.
x=0
x=1
x=4
x=7
The minimum value of x can be 4 3 = 1 cm and maximum value of x can be
3 cm
3 cm
4 + 3 = 7 cm
i.e., the particle oscillates simple harmonically about point x = 4 cm with amplitude 3 cm.
14.(ABD)
K
x(m)
7
4
3
t(s)
The x-t graph will be as show
16.(ABC)Free body diagram of the truck from non-inertial frame of reference will be as follows:
This is similar to a situation when a block is suspended from a vertical spring.
m
.
k
Therefore, the block will execute simple harmonically with time period T = 2
Amplitude will be given by
ma0
k
A=x=
(ma0 = kx)
mg
2
Energy of oscillation will be
E=
1 2 1 ma0
kA k
=
2
2 k
Pseudo
force = ma0
m2 a02
2
T = (tPA + tAP) + (tPB) + tBP) = 0.5 s + 1.5 s = 2s =
= s1
Let tOP = t then tOAP = t +
1
2
x = a sin t
t=0
O
a =
3
3 2 m/s = vmax
cos 4
APP | Simple Harmonic Motion
;
A
a
. . . (1)
x = a sin t = a sin
78
P
B
and
1
Then 3 = a cos t = a cos t = a cos t = a sin t
2
2
or
kx
2k
17.(AC) Let the displacement equation of particle is x = a sin t
Time period of particle would be
kx
a
4
2
x
v = (a) cos t
ax
. . . (2)
t = /4
AP a x a a
BP a x a a
2
2
2 1
2 1
Solution | Physics
Vidyamandir Classes
18.(BD) x1 = A cos t and x2 = A sin t
Equating x1 = x2 , we get : A cos t = A sin t
or,
tan t = 1
or,
t =
3
4
or,
2 3
t
4
T
or,
t=
3
A
3T
x2 = A sin
8
4
2
19.(A) E changes equilibrium position only.
20.(BCD)
F
du
10 x 20 0
dx
x2
kE is max at x = 2m. F 10 x 2
d 2u
;
10 0
dx 2
2
10
0.1
T
x 2 is a point of minima
2
/5
21.(ABD)
Description of motion is completely specified if we know the variation of x as a function of time. For simple
harmonic motion, the general equation of motion is x A(t ) . As is given, to describe the motion completely,
we need the values of A and .
From option (b) and (d), we can have the values of A and directly.
For option (a), we can find A and if we know initial velocity and initial position. Option (c) cannot given the values
A of and so it not the correct condition.
22.(ABCD)Period of oscillation changes as it depends on mas and becomes three times. The amplitude of oscillation does not
change, because the new object is attached when the original object is at rest. Total energy does not change as at
extreme position the energy is in the form of potential energy stored in spring which is independent of mass, and hence
maximum; KE also does not change but as mass changes the maximum speed changes.
23.(BD) The maximum extension x produced in the spring in case (a) is given by
F
F kx or x
k
The time period of oscillation is:
T 2
mass
m
2
force constant
k
In case (a) one end A of the spring is fixed to the wall. When a force F is applied to the free end B in the direction, the
spring is stretching a force on the wall which in turn exerts an equal and opposite reaction force on the spring, as a result
which every coil of the spring is elongated producing a total extension x.
In case (b), both ends of the spring are free equal forces are applied at ends. By application of forces both the cases are
same.
Thus. The maximum extension produced in the spring in cases (a) and (b) is the same. In case (b) the mid-point of
spring will not move, we can say the block connected with the springs whose lengths are half the original length of
spring. Now, the force constant of half the spring is twice of complete spring. In case (b) the force constant
= 2 k. Hence, the time period of oscillation will be
T 2
m
2k
;
T
2
T
Hence, the correct choices are (b) and (d)
24.(ACD)
The only external horizontal force acting on the system of the two blocks and the spring is F. Therefore,
acceleration of the centre of mass of the system is equal to F / m1 m 2 .
APP | Simple Harmonic Motion
79
Solution | Physics
Vidyamandir Classes
Hence, centre of mass of the system moves with a constant acceleration. Initially there is not tension in the spring,
therefore at initial moment m2 has an acceleration F / m2 and it starts to move to the right. Due to its motion, the spring
elongates and a tension is developed. Therefore, acceleration of m2 decreases while that of m1 increases from zero,
initial value.
The blocks start to perform SHM about their centre of mass and the centre of mass moves with the acceleration
calculated above. Hence option (b) is correct.
Since the blocks start to perform SHM about centre of mass, therefore the length of the spring varies periodically.
Hence, option (a) is wrong.
Since magnitude of the force F remains constant, therefore amplitude of oscillations also remains constant. So option (c)
also wrong.
Acceleration of m2 is maximum at the instant when the spring is in its minimum possible length, which is equal to its
natural length. Hence, at initial moments, acceleration of m2 is maximum possible.
The spring is in its natural length, not only at initial moment but at time t T , 2T , 3T , ..... also, where T is the period of
oscillation. Hence, option (d) is wrong.
25.(ACD)
When the block is released suddenly, it starts to move down. During its downward motion the rubber cord
elongates. Hence, a tension is developed in it but the block continues to accelerate downwards till tension becomes
equal to weight mg of the block.
After this moment, the block continues to move down due to its velocity and rubber cord further elongates. Therefore,
tension becomes greater than weight; hence, the block now retards and comes to an instantaneous rest.
A lowest position of the block, strain energy in the cord equals loss of potential energy of the block. Suppose the block
comes to an instantaneous rest when elongation of the rubber cord is equal to y. Then
1 2
2mg
ky mgy y
and 0
2
k
Hence, block will be instantaneously at rest, at y = 0 and at y = 2mg/k.
In fact, the block oscillates between these two values.
Since the rubber cord is elastic, tension in it is directly proportional to elongation. Therefore, the block will perform
SHM.
It amplitude will be equal to half of the distance between these extreme positions of the block or amplitude is
1 2mg mg
l
2
k
k
k
m
Since k and m are not given in the question, it cannot be calculated. Hence option (d) is not correct.
Hence, option (b) is correct. The angular frequency of its SHM will be equal to ;
26.(AB) When point of suspension of pendulum is moved upwards, geff g a, geff g and as T 1 / g eff , hence T,
decreases, i.e., choice (a) is correct.
When point of suspension of pendulum is moved downwards and a 2 g , then T decreases, i.e., choice (b) is also
correct. In case of horizontal acceleration
geff g 2 a 2 , i.e. geff g
i.e., again, T decreases
27.(BC) The maximum potential energy of linear harmonic oscillator is equal to the total mechanical energy at extreme positions
of the oscillations hence option (C) is correct. The maximum kinetic energy of the oscillator is (1 / 2)kA2 100 J hence
option (B) is also correct.
APP | Simple Harmonic Motion
80
Solution | Physics
Vidyamandir Classes
28.(BCD)
Due to the Pseudo force on block (considered external) its mean position will shift to a distance mg/K above
natural length of spring as net force now is mg is upward direction so total distance of block from new mean position is
2mg/K which will be the amplitude of oscillations hence option (C) is correct. During oscillations spring will pass
through the natural length hence option (D) is correct. As block is oscillating under spring force and other constant
forces which do not affect the SHM frequency hence option (B) is correct.
a
29.(BC) Both will oscillate about equilibrium position with amplitude tan 1 for any value of a.
g
If a g , motion will be SHM, and then
Time period will be 2
30.(BD) When x
l
a2 g 2
2
A
4
A KA
F1 Kx K
4 4
2
and
1 A KA2
U1 K
2 4
32
1
1 m2 A215 m2 A2 15
K1 mv12
2
2
16
32
and
1 A
kA
U2 k
4U
2 2
2
2
A 15
A
V1 A2
4
4
When x = A/2
F2 kx
kA
2 F (Magnitude)
2
2
2
3
4
1
4
A
A2 A
v1
KE2 mv22 K1 0.8 K1 k
2
5
2
5
2
31.(AC) Let O be the mean position and a be the acceleration at a displacement x from O.
At position I, N – mg = ma
N0
V2
At position II, mg – N = ma
For N = 0 (loss of contact), g a 2 x
Loss of contact will occur for amplitude xmax g / 2 at the highest point of the motion.
APP | Simple Harmonic Motion
81
Solution | Physics
Vidyamandir Classes
32.(AB) The time period of simple harmonic pendulum is independent of mass,
so it would be same as that T 2 1 / g . After collision, the combined
mass acquires a velocity of v0 / 2 as a result of this velocity, the mass
(2m) moves up and at an angel 0 (say) with vertical, it stops, this is
the extreme position of bob.
From work-energy theorem, K Wtotal
2
v02
2m v0
1 cos 2 sin 2 0
2mgl (1 cos 0 ) ;
2 2
8 gl
2
v
v
;
If 0 is small, sin 0 0 0 0
sin 0 0
2 4 gl
2
2
2 gl
0
So, the equation of simple harmonic motion is 0 sin(t )
33.(AD) Statement (a) is correct. At any position O and P or between O and Q, there are two accelerations – a tangential
acceleration g sin and a centripetal acceleration v 2 / l (because the pendulum moves along the arc of a circle or
radius l), where l is the length of the pendulum and v its speed at that position. When the bob is at the mean position O,
the angle 0 , therefore sin 0 ; hence, the tangential acceleration is zero. But at O, speed v is maximum and the
centripetal acceleration v 2 / l is directed radially towards the point of support. When the bob is at the end points P and
Q, the speed v is zero, hence the centripetal acceleration is zero at the end points, but the tangential acceleration is
maximum and is directed along the tangent to the curve at P and Q. The tension in the string is not constant throughout
the oscillation. At any position between O and end point P or Q, the tension in the string is given by T mg cos
At the end point P and Q, the value of is the greatest, hence the tension is the least. At the mean position O,
0 and 1 g which is the greatest ; hence tension is greatest at the mean position.
34.(BD) Initially,
1 2 1 2
kA mv0 A
2
2
Next, mv0 = 2mv
A =
A
2
APP | Simple Harmonic Motion
m
v0
k
v = v0/2
f=
1
2
1
1
v
kA2 2m 0
2
2
2
2
k
f
2m
2
82
Solution | Physics
Vidyamandir Classes
35.(AC)The situation is similar as if a block of mass m is suspended from a vertical spring and a constant force mg acts
downwards. Therefore, in this case also block will execute SHM with time period T = 2
At compression x
F = kx
;
m
k
F
k
x
This is also the amplitude of oscillation.
Hence A = F/k
At mean position speed of the block will be maximum. Applying work energy theorem
1 2 1 2
2Fx kx 2
kx mv
v=
2
2
m
K3 K 4
36.(B) K eq K1 K 2
70 N / m
K3 K 4
37.(C) It starts the motion from x = 0 in +ve direction
F. x =
;
x
F
k
; Vmax
F
mk
38.(C) 39.(B)
w
K
800
kx
kx0
400 20 rad / s
m
2
When car is accelerated let elongation is x0
a
extreme
K x0 mg ma
800 x0 20 20
mg
mg
x0 5 cm
equilibrium
When acceleration ceases let elongation of spring in equilibrium position k x = mg
800 x = 20
Hence amplitude = x0 x 2.5 cm
x = 2.5 cm
Initially the block is at right extreme position. Hence initial phase = / 2 .
40.(A) At x = 0, magnitude of displacement from means position 2m
41.(B)
Time period
a 2 x m / s
v 2
2
2
2 m / s 2 2 2 1 rad / s
2
2 sec
v
v 0
dv
dx
v 2
2 x
2
2 2
2
vdv
v 0
2 2
2
2
42.(B) As A is at its negative extreme at t = 0
3
So x 3 2sin 2t
2
x 3 2cos (2t )
1
2
m v 2
2 x dx
0
2 v 02 4 J
43.(D) As B is at its equilibrium position and moving towards negative extreme at t = 0
So
y 4 2sin(2t )
y 4 2sin(2 t )
APP | Simple Harmonic Motion
83
Solution | Physics
Vidyamandir Classes
44.(B) Distance between A and B x 2 y 2 (3 2 cos 2t )2 (4 2sin 2t ) 2
4
3
9 4 cos2 2t 12 cos 2t 16 4sin 2 2t 16sin 2t 29 20 cos 2t sin 2 t
5
5
29 20 sin(2t 37)
Maximum distance 29 20 49 7cm
;
Minimum distance 29 20 9 3cm
45.
[A – q s ; B – p r; C – q s ; D – q s ]
When u is min equilibrium is stable and particle performs SHM. When u is max equilibrium is unstable.
46.(2)
When block is displaced x from its mean position than extra elongation
in the spring will be 4x and hence
Frest 16kx
f
1 16k 1 4 22
2
2 m
2
7
47.(8)
When the liquid in left vertical arm is displaced threw a distance x then
liquid in tilted arm also move a distance x along the tube.
Difference of height H x x cos
Now use can calculate Pexce which will provide restoring force.
48.(8)
When displacement of the ring is , then extension in spring = (2a + x0)
Energy of system,
1
1
d
E = k (2a x0 )2 mga I 2
where =
2
2
dt
1
11
E = k (2a x0 )2 mga ma 2 ma 2 2
2
22
k
2
dE
d
d
3
d
= k (2a + x0). 2a
mga
+ ma 2 2
dt
dt
dt
2
dt
As
dE
3
d 2
0 , k (2a + x0) 2a – mga = ma 2
dt
2
dt 2
3
d 2
4ak + 2akx0 – mga = ma2 2
2
dt
49.(2)
8 k
d 2
3m
dt 2
8k
3m
=
d
The effective acceleration of a bob in water = g = g 1 where d & D are the density of water & the bob
D
D
respectively.
= specific gravity of the bob.
d
Since the period of oscillation of the bob in air & water are given as T= 2
T/T =
g
=
g
& T = 2
g
g
g (1 d / D)
d
1
= 1
= 1
g
D
s
APP | Simple Harmonic Motion
84
Solution | Physics
Vidyamandir Classes
Putting T/T = 1/2,
50.(6)
E=
1
m2 A2
2
we obtain:
E=
1
m(2f)2 A2
2
Putting E = K + U we obtain, A =
51.(2)
1/2 = 1 – 1/s
1
2 25 /
1 1
s 2
A=
s=2
2E
m
1
2f
2 (0.5 0.4)
0.2
A = 0.06 m.
The loss of potential energy by the gravity = gain in potential energy in the spring
mgx =
1 2
2mg
kx x
2
k
At equilibrium f x 0
;
m
mg kx0 = 0
x0 =
y
mg
k
So, required ratio = 2 : 1
52.(4)
The bob will execute SHM about a stationary axis passing through AB. If its
effective length is l then :
l
l
T 2
;
l
2l
g
sin
g g cos
53.(3)
g
;
2l
2
g
2 0.2 2
s
10
5
2
For small angular displacement of cylinder.
The energy of system angular displacement is
1
1
1
E k (2R)2 k ( R)2 Mv 2 I 2
2
2
2
2
Where v is the velocity of centre of mass and is the angular velocity
of cylinder. Since E is constant.
dE
d
d
dv
d
0
4kR 2
kR 2
Mv
I
0
dt
dt
dt
dt
dt
5kR 2 MR 2 I 0
Compare it with a 2
54.(5)
T 2
;
Thus
5kR 2
( MR 2 1)
2
10k
3M
where I
or
MR 2
2
T 2
5kR 2
3
MR 2
2
3M
10 K
By phasor diagram
Angle traced = 53° + 37° = 90°
0
T
Time taken
T 5s
360
4
55.(4)
600 A + 300 = 2ma
450A + 150 – f = ma
f mg 150 A
56.(1)
When block is displaced down by y, then level of liquid in beaker also rises.
APP | Simple Harmonic Motion
85
Solution | Physics
Vidyamandir Classes
A
A
y A y
4
4
So,
yA
upthrust y 3dg
34
4y A
A
dha y
3d g
4
3 4
4g
a
y
h
57.(9)
58.(8)
;
T 2
2f
mR
f
aCM
m
3f
aPOC
A(10)2 cos(10)t
m
For no slipping
100 A 3g 9
h
h
4g
g
A 9cm
xu m
2 2k
After collision at the left end, the motion of the dumbbell can be conceived as superposition of translation of mass
centre of the dumbbell and oscillations about the mass centre. For minimum value of l, the right ball of the dumbbell
must collide with the right most ball at the end of the first half period of the oscillations.
l
m(k1 4k2 )
59 . T 2
k1k 2
( 2)r
60. 2
g
APP | Simple Harmonic Motion
86
Solution | Physics
Vidyamandir Classes
WAVE MOTION
1.(B)
f2
f1
240
260
T2
T1
T1 507 N ;
507 10000 V
507
T 2 507 10000 V
V 0.0075 m3
2.(B)
4.(D)
3.(B) I A2
Compare with the standard equation of the wave.
6.(B)
Beats are produced due to the difference in apparent frequency of the two tuning forks.
7.(A)
At 4 sec V 40 m / s
8.(C)
x
x, t A sin t
c
9.(D)
f
Apply equation for Doppler’s effect.
5.(D)
320
1000 889 Hz
320 40
x 11 2 2 (optical path)
f1
340 17 19
f 2 340 34 18
10.(D)
IA
A2 f 2
A A
IB
AB2 f B2
11.(C) Let y A sin kx wt 0
Also kx1 wt wt k
At t 0, y 0 at x 0 0 0
dy
dt
12.(C)
A wcos kx wt V1 at x x1 and t 0 but kx1 w
C A
f max f 0
C
and
C A
f min f 0
,
C
x1
x
V1t
eq is y Asin
Acos
Acos
x1
V1
k
m
990 330 10
330 10
990
= 960 Hz
1020 Hz ;
330
4
330
c
350
13.(A)
2 2 32 3
1.211 m f
289 Hz
2
1.211
14.(C) Amplitude as function of x taking open and (antinode) as reference point is
2
2a cos kx 2a cos
. x 2 a cos at x = 10 cm if 40cm
amplitude is zero
2
15.(D) Along perpendicular bisector, path difference is zero, but phase difference between L1 and L2 is , So, waves interfere
destructively.
16.(AC) Only these two satisfies,
17.(BC) y
0.8
4 x 5t
2
0.8
2
y = f (x + vt) (wave moving in ve x direction) v
5
16 x t 5
4
Distance moved by wave = (speed of pulse) (time)
Distance moved by pulse in 2 seconds = 5/4 2 = 2.5 m
APP | Wave Motion
5
d2 y
1 d2 y
dx 2
v 2 dt 2
87
5
m/s
4
Solution | Physics
Vidyamandir Classes
18.(ABCD)
19.(BC)
Ieq = I1 + I2 + 2 I1 I 2 cos
For maximum & minimum
cos = 1
Imin = 0 and Imax = 4 I
2 x 2 x
3
=3m
120 t = 2nt
n = 60 Hz.
I
T = 648 N
m
v = ns = 180 m/s v =
2
Amplitude at a distance x = 2a sin x = 4.2 cm
20.(AC) |V p |
dy
dt
2
x 3t 2 1
2
=
02 1
2 x 3t 3 speed of particle at t = 1 and x = 3
2 3 3 3 0
Speed of wave =
Coeff of t
coeff of x
3
1
3 cm / s
300 0
App. frequency of B as heard by A = 500
437.5 Hz
350 50
21.(ABC)
2
f
2
22.(AC)
2
2
300 25
App. frequency of B as heard by O = 500
468.75 Hz
350 50
350 25
App. frequency of A as heard by O is 500
464.28 Hz
350
Diff = 468.75 464.28 4.47
300 50
App. frequency of A as heard by B = 500
428.57 Hz
350
2
path diff (p) p
and
p 1m
2
4
2 Hz , at x = 20 cm, t = 4 sec y 0.15 sin 4 4 0.15 sin
5
3
2
23.(BCD)
24.(ABD)
It is a known fact as well as experimentally and analytically verified that wave speed depends on the
properties of the medium and is same for the entire wave. The particle velocity is given by
vP
y
A cos(kx t )
t
Where symbols have their usual meanings. It is clear form above expression that vP depends upon amplitude and
frequency of wave which are wave properties and are having different values for different particles at a particular
instant.
25.(ACD)
y f ( x vt )
Particle velocity ;
vP
To find velocity of wave
d
( x vt ) 0
;
dt
APP | Wave Motion
dy
vf ( x vt )
dt
dx
v
dt
88
Solution | Physics
Vidyamandir Classes
26.(AD) v
T
For equilibrium Mg mg sin 30 T
M = m/2
100
Mg
9.8 10
3
M (9.8)
9.8 10 3
;
100
M (1000)
M 10kg and m 20kg M
27.(CD) Since the first wave and the third wave moving in the same direction have the phase angles and ( ) , they
superpose with opposite phase at every point of the vibrating medium and thus cancel out each other, in displacement,
velocity, and acceleration. They in effect, destroy each other out. Hence we are left with only the second wave which
progresses as a simple harmonic wave of amplitude A. the velocity of this wave is the same as if it were moving alone.
28.(ABD)
If P divides AB in ratio 1:4, then the fundamental frequency corresponds to 5 loops, one loop in AP and 4 loops
in PB which corresponds to 5th harmonic of 1 kHz. Hence fundamental = 5 kHz.
If P be taken at midpoint, the third harmonic will have three loops in each half of the wire AB. Hence total number of
nodes (including A and B) will be 5 + 2 = 7.
If P divides AB in the ratio 1:2, the fundamental will have three loops, corresponding to the frequency of 3 kHz. For this
string to vibrate with the fundamental of 1 kHz, the tension must be (T/9).
The wire AB will be symmetry, vibrate with the same fundamental frequency when P divides AB in the ratio a:b or in
the ratio b:a.
29.(BC) At any point on line AB, the phase difference between two waves is zero and hence waves will interference
constructively.
Along CD, the phase difference changes and waves interference constructively and destructively and, hence sound will
be loud, faint and so on.
y A sin[t k ( AC )]
30.(CD)
k ( BC )
2
;
yB sin[t
;
BC AC n 15, 35,55, 75.....
4
For maximum intensity at C
k ( BC AC )
2n k
2
31.(AD) In both case (A) and (D) the source and observer are relatively at rest, thus neither of them is approaching or separating
from each other. Effectively, it is the medium that moves in each of these cases. The received (apparent) frequency
differs from the emitted frequency if and only if the time required for the wave to travel from the source to observer is
different for different wavefronts. With a uniform steady motion of the medium, past the observer and source, the
transit time from source to observer is the same for all wavefronts. Hence is follows that apparent frequency is equal to
the true emitted frequency. Thus there is no Doppler effect. In case (B) and (C), Doppler effect will be observed as the
source and observer have a relative speed and so they will approach or recede from each other.
32.(BC) As time increases, the source and detector are relatively approaching each other up to t t0 , where t0 is the instant
when the source and detector are located perpendicular to direction of motion.
d cot 0
d cot 0
v0 t0
;
t0
2
2v0
For
t t0 , f ap f 0
APP | Wave Motion
;
For
t t0 , f ap f 0
89
Solution | Physics
Vidyamandir Classes
33.(ACD)
If intensity at points is I, then energy density at that point is E = I/v, where v is wave propagation velocity.
It means that E I , Hence, the graph between E and I will be a straight line passing through the origin. Therefore (a)
is correct and (b) is wrong. Intensity is given by :
I 22 n2 a 2 pv
Hence,
E 2 2 n 2 a 2
It means that E n2
Hence, the graph between E and n will be parabola passing through origin, having increasing slope and symmetric about
E-axis. Hence, option (d) is correct.
Particle maximum velocity is
u
1
u0 a 2na
na 0
;
Hence, E u02
2
2
It means that graphs between E and u0 will be a parabola, have increasing slope and will be symmetric about E-axis.
Hence. Option (C) is also correct.
34.(ACD)
4
For observer O1 ,
V Vs V V / 5 4V
f
f
5f
For O2 , there is change of medium. Hence, at the surface of water, keeping frequency unchanged.
V
4V
a w
w 4 a
Velocity of wave relative
f
to observer
w
16V
5f
V
5 21V 5 f 21 f
w
5 16V
16
4v
35.(BC) As the support is rigid, the wave is reflected in opposite phase hence at the support destructive interference takes place
and node will be obtained. Due to nodes and antinodes at different positions, intensity of wave varies periodically with
distance.
36.(C) 37.(A)
38.(A)
Mass per unit length of the string is
m Ad (0.80 mm 2 ) (12.5 g / cm3 )
(0.80 10 6 m2 ) (12.5 103 Kg / m3 ) 0.01 Kg / m
Speed of transverse waves produced in the string
v
T
64
80 m / s
M
0.01 Kg / m
The amplitude of the source is a = 1.0 cm and the frequency is n 20 Hz . The angular frequency is 2n 40s 1 .
Also at t = 0, the displacement is equal to its amplitude, i.e., at t = 0, y = a. The equation of motion of the source is
therefore.
y (1.0cm) cos [(40 s 1 )t ]
APP | Wave Motion
. . . .(i)
90
Solution | Physics
Vidyamandir Classes
The equation of the wave travelling on the string along the positive X-axis is obtained by replacing t by (t – x/v)] in
equation (i). It is, therefore,
y (1.0 cm) cos[(40s 1 ) {t ( x / v)}]
(1.0 cm) cos[(40 s 1 )t {( / 2) m 1}x]
. . . .(ii)
The displacement of the particle at x = 50 cm at time t = 0.05 s is obtained from equation (ii)
y (1.0cm) cos[(40 s 1 ) (0.05 s )
{( / 2)m 1}(0.5 m)]
(1.0 cm) cos[2 ( / 4)]
1.0 cm / 2 0.71 cm
39.
40.
[A – q s ; B – p r ; C – p r; D – q s ]
Situation (i)
direction same ;
Situation (ii)
direction opposite ;
frequency different
frequency same
[A – q ; B – p ; C – r ; D – r]
f f
(A)
f 1 2
2
(B)
A 2 A
(D)
A1 A 2
Ratio
2
A1 A 2
2
(C)
41.(1)
Beat frequency f 2 f1
The general equation of a wave travelling along the positive x-direction is y = f(x ct) where c is the wave velocity.
At t = 0 , y
1
1 x2
Now , at t = 1s : y
(given) and the equation of the wave reduces to y = f(x)
1
2
=
1
2
f (x 1)
1 (x 1)
2 2x x
Comparing the equation with y = f(x - ct) at t = 1s, we get c = 1 m/s.
42.(2)
Let f1 and f2 be the frequencies of tuning forks P and Q,
Then | f1 f2 | =
Apparent frequency for O corresponding to signal directly coming from
v
O
Q = f2
v vq
7
2
Q
Q
v
2vq v
Apparent frequency of the echo = f2
f2 = f2 2
2
v vq
v v q
Since, f2 = 5
(given),
f2 = 163.5 Hz.
Now, f1 = 163.5 3.5 = 167 or 160 Hz.
When P is filed, its frequency will increase, since it is given that filed P gives greater number of beats with Q. It implies
that f1 must be 167 Hz.
APP | Wave Motion
91
Solution | Physics
Vidyamandir Classes
43.(4)
Required beat frequency = |f1 f2|
Where, f1 = apparent frequency for the motorist corresponding to the
signals directly coming to him from source, and f2 = apparent
frequency for the motorist corresponding to the signals coming to him
after reflection.
Vm
Vb
Wall
V Vm
Now, f1 = f
V Vb
Where f is the frequency at which signals from sources are incident on wall.
V V Vm
V Vm
V
f = f
f2 = f
f
V Vb
V Vb V
V Vb
Hence, the beat frequency = |f1 f2| =
44.(2)
2Vb V Vm f
V
Vb2
.
The new position of the 8 kg block
T sin = 6400 N; T cos = 4800 N
Squaring and adding T =
(64)2 (48)2
Now velocity of transverse wave =
T
=
= 80 N
80
20 10
4
= 2 10-2 m/s= 2 cm/s.
46.(4)
mid point will be having maximum displacement (Antinode) when string will vibrate in 3rd harmonic (in second
harmonic mid point will be position of node)
yR = y1 + y2 = 3Acos(t kx) + A cos(3t 3kx)
yR = 4 A sin3(t kx)
AResultant = 4 A
47.(3)
Ymax Y1(max) Y 2(max)
48.(7)
C V1
C V2
V1
V2
Δf f 0
f0
f 0 1 f 0 1
C
C
C V1
C V2
45.(3)
2
V1 V2
49.(2)
99
50
Velocity of the wave,
2
2V1
2V2 2
2
f 0 1
1
V C
C
C
2m / s
T
(16 105 )
V
2000 cm / s
0.4
20
0.01 s
200
Time taken to see the pulse again in the original position 0.01 2 0.02 s
Time taken to reach to the other end
50.(4)
19.2 10 3 kg / m
From the free body diagram
T 4 g 4a 0 ;
T 4(a g ) (2 10) 48 N T
Wave speed :
v
T
48
50 m / s ;
19.2 10 3
APP | Wave Motion
So n = 4
92
Solution | Physics
Vidyamandir Classes
51.(2)
Given that
x 40cos (50 t 0.02 y )
particle velocity
dx
(40 50) { sin(50t 0.02 y)}
dt
vp
Putting x 25 and t
1
s
200
1
v p (2000 cm / s ) sin 50
0.02 (25)
200
52.(4)
0.8
y
2
2
(3 x 24 xt 48t 4)
53.(2)
0.8
amax 2 A g
Amin
54.(3)
g 2
4 2 F
0.8
2
3[ x 8 xt 16t 2 ] 4
x 4t x vt
3( x 4t )2 4
10 2 m / s
;
v 4m / s
2 v
, v
F
2
2 103 m 2 mm
4F
f T for strings.
On increasing the tension by 1%
f 101T
1
f
1.01T
1
(1 0.01) 2 1
f
200
T
;
f
1 1
Beat frequency, f f f
f
Number of beats in 3s 1 30 30
55.(7)
f0 fc 2
1
1
V
2 or V / L 8
2
L
4
L
f 0 f c
56.(1)
In the second case,
V V
7V
7
(8) 7
L 8L 8 L 8
Intensity is given by I
p02
2v
Here v and are same for both. And also given that I is same both. So pressure amplitude is also same for both.
57.(7)
Intensity from a point source varies with distance as I
1
r2
Let at distance r1 10m , intensity is I1
Then given 20 10 log
I1
I0
(i)
Let for r r2 , sound level be zero. Then intensity at that point should be I 2 I 0 .
APP | Wave Motion
93
Solution | Physics
Vidyamandir Classes
2
I1 r2
I1 r2
I 2 r1
I 0 r1
From, Eqs. (i) and (ii), we get
2
And
r
20 10 log 2
r1
58.(8)
2
(ii)
r
20 20 log 2
r1
r2
10 10r1 7 m
r1
The observer will hear a sound of the source moving away from him and another sound after reflection from the wall.
The apparent frequencies of these sounds are
v
v
f1
f , f2
f
v
u
v u
v
v
2uvf
2uf
8
Number of beats f 2 f1 =
f 2
2
v
u
v
u
v
v u
59.(4)
Here I1 1.0 10 8 W / m 2
r1 5.0 m, I 2 ?, r2 25m
1
We know that I 2
r
I1r12 I 2 r22
60.(5)
;
I2
I1r12
r22
1.0 108 25
4.0 1010 W / m 2
625
We can see from figure, in a stationary wave two successive of equal amplitude, if separated by equal distances then
this distance must be . Thus we have
4
15cm
Or
60 cm
4
Thus there are four loops in 120 cm length of string.
This corresponds to 3rd overtone oscillations. As
shown in figure if we consider origin O is at a node
then the amplitude of a general medium particle, at a
distance x from O can be given as
R A0 sin kx
Where A0 is the maximum displacement amplitude. First point from the origin where amplitude is 3.5 mm is at
distance
7.5 cm . Thus we have
8
2
3.5 A0 sin
7.5
60
APP | Wave Motion
or A0
3.5
3.5 2 mm 5mm
sin( / 4)
94
Solution | Physics
Vidyamandir Classes
ELECTROSTATICS
1.(C)
2.(C)
3.(B)
A
q
4m
kq kq
=
+ Vind.
5
4
1
Vind. = –9 × 103
= – 0.45 kV
20
B
3m
C
S
If another shell is kept upside down over it complete a sphere, net field should become zero.
qE
Net downwards acceleration on body of mass m = g
= anet
m
If E = uniform electric field in downwards direction
2v
2v
If it hits after time t =
=
anet
qE
E
g m
planet
at maximum height vf = 0
v2f = v2i – 2anet h In uniform field (Gravitation + Electric field time to reach highest point = t / 2 ]
qE
v2 = 2 g
h
m
V between ground and highest point : V = (E) h
2v
2v
q
qE
2v
anet =
g
E g
m
t
t
m
t
v
m 2v
E=
- g and h = (average velocity) × t
h = t
q t
2
m 2v v
mv
gt
So V = - g t =
v
q t
q
2
2
4.(C)
If q denotes the charge in the medium from R to r, then E
K (Q q )
2
r
q can be calculated by integrating the charge dq in a thin shell of radius r, thickness dr. This gives q = 2r2 – R2)
2
i.e,
E
2
K (Q 2 ( r R ))
r
2
Electric field will be uniform if coefficient of r is made zero
5.(D)
dV
E=
=
20 x 20 3a x
dx
=
ln x
2a
a
ln 3a x
2a
a
2
r
dV
2 0
2k
R2
2
0
Q 2R 2
r
dx
dx
x
3a x
= 2 ln 2 ln 2 = 2 ·2 ln 2 = V
0
A
– VB
q0 ln 2
0
For earthed conductor [the inner shell], V = 0
1 q q '
Here V =
where q is charge that would appear on inner shell as it is grounded q = –q/3
4 0 3r r
Hence, the charge flown to earth = 0 – (–q/3) = q/3
6.(B)
2 0
kQ
work = (VA – VB) q0 =
APP | Electrostatics
95
Solution | Physics
Vidyamandir Classes
7.(B)
1 =2
Enet E
8.(A)
1
+
+
+
(q,) +
P+
+
+
+
E 11 0
r 2 R cos
r
2
1
E
dq
E11
dq
dV k
r
dq 2 r dr
/2
2 2 R cos 2 R sin d
r
2 R cos
R
.
0
The system can be treated as a system of two dipoles having dipole moment p = qa each. If you think a little, you
V dV 4k R
9.(B)
dV k
d
0
sin d =
will realize that the dipoles are perpendicular to each other. Obviously, the net dipole moment is qa 2 .
10.(D)
Qx
E x
2
4 0 R x
F
Qq
4 0 R3
2
3/ 2
Qx
F
–q
O
4 0 mR3
T 2
Qq
(Towards mean position)
x
+Q
(for x < < R)
40 R 3
11.(B) Charge distribution is as shown :
(Q
Q1 Q2
1
Q1
1 = 2 = 3
Q1 2 )Q2 Q3
Q1 Q1
4 R
On solving we get
2
=
Q1 Q2
4 2 R
;
2
=
Q1 Q2 Q3
4 3R
2
Q1 : Q 2 : Q 3 1: 3 : 5
3
2
1
12.(C)
13.(A) Field at P due to outside changes =
K 2Q
2 R
r
3r
, 0
,
2
2
14.(A)
r
30 r
60
r
p
0, 0, 0
p
2
KQ
4 R
2
7 KQ
16 R 2
towards O
p cos
using V r,
4 0 r 2
r
3r
p cos 30
p cos 60
net potential at ,
, 0 is
2
2 2
4 0 r
4 0 r 3
APP | Electrostatics
96
3p
2 4 0 r 2
p cos 60
4 0 r
2
p
3 1
8 0 r
2
Solution | Physics
Vidyamandir Classes
15.(B)
f
Kq1q2
4 r 2
2
and
f K q1 q2
where q1 and q2 be initial charges.
3
4 0 r 2 4
4q1q2 3q12 3q22 6q1q2 ; Let
q1
1
x 3x 2 10 x 3 0 x 3 or
q2
3
16.(AC) If forces due to both the fields cancel out then (A). If forces due to both fields are along the same line and the
particle is given the initial velocity along the same line, then also (A).
In any other case, the net force acting on the particle will be uniform. From basic 2-D motion, we know that if
acceleration is uniform but not parallel to initial velocity, the trajectory of the particle is a parabola.
17.(AD) Net charge on the surface of conductor is zero. Hence, potential at the center, due to charges on the surface of
conductor is zero (as the center is equidistant from all points on the surface). Hence (A).
At point B, total potential is same as that at center since volume of a conductor is an equipotential volume.
Subtracting the contribution of point charge q from the total potential will give the potential at B due to charges
induced at the surface of sphere.
18.(ABD)For (A), use the fact that potential at center of sphere, due to Q2 will be exactly cancelled by charge –Q2 induced
on the surface of the cavity. Hence (A).
Potential at any point at surface of cavity is equal to potential at the centre of sphere. Also, electric field inside
cavity is known. So, potential at any general point inside cavity can be found be performing the line integral of
electric field from the surface of cavity upto that point. This will solve (B).
Potential outside the sphere cannot be found easily as the charge on the outer surface will be non-uniform. Hence C
is incorrect as the given answer for uniform Q2 + Q3.
(D)can be calculated by making the net potential at the center of sphere zero and finding new charge on sphere.
19.(AB) x0 deformation in equilibrium state.
2Kx0
·q
0
x0
q
2 K 0
Springs are connected in parallel Keq = 2K
Angular frequency =
2K
m
20.(AD) We know that for constant r, electric field due to a dipole is maximum at axial point and minimum at equatorial
point. Magnitude of electric field continuously decreases from 2KP/r3 to KP/r3 as we move from an axial point to an
equatorial point, keeping r constant. In the present case, if E = 10N/C at any point, then it cannot be less than 5N/C
or more than 20N/C at any point.
21.(BCD)Even if the metallic sphere is neutral, it will still be attracted towards the positive charge due to induced charges.
22.(BCD)
(A)
(B)
(C)
(D)
23.(BC) E =
V=
is incorrect by conservation of charge.
is correct. The net field inside the sphere is zero only because the electric field due to induced
charges cancels the external electric field at all points inside the sphere.
is a basic property of conductors
As stated in part (B), the electric field lines are eliminated within the spherical region.
kq
kq 2
6kq
Enet = 3E cos = 3 × 2
2
a
a
a 3
3kq
a
APP | Electrostatics
(B)
(C)
97
Solution | Physics
Vidyamandir Classes
24.(AC) Potential due to a uniformly charged spherical shell is given by:
V
KQ
for r R
r
and
V
KQ
for r R
R
Q2c
a bc
Q Q2
Q1 = y + x + y – Q2 – x y = 1
2
Q2+x
xa + xb + (Q2 + x) c = 0 x =
25.(AB)
y x
–x x
–x
y
–Q2–x
Q2 ca
Potential different between A & B is V =
.
(a b c) S 0
Similarly, p.d between C & D depends upon Q2
26.(B) In part B, difference is due to the mass of electrons. Part (C) will not be true for a curved electric field line.
27.(ABD)
Ex = 10V/m, E Ex
28.(ABCD) (A) Net charge enclosed by Gaussian surface is zero.
(B) Net charge enclosed by Gaussian surface is zero.
(C) As the change is displaced towards sphere amount of negative charge induced on right half of sphere will
increase hence net flux through right hemispherical closed Gaussian surface increases.
(D) Same reason as (C) and hence charge distribution on outer surface of sphere will change.
+
––
––
–
+
29.(AD) Uniform charge on outer surface
+
q
therefore potential is
4 0 R
–
–
q ––
+
+
+
+
30.(ABC)
VA = VB work = 0
+
+
q +
31.(ABC)
If moved slightly along x-direction, say towards left, the attractive force of left wire will be more than the attractive
force of the right wire. Hence equilibrium will be unstable in this direction.
If moved in y-direction, the force will remain zero.
If moved in z-direction, electron will be attracted back towards the wires.
32.(ABCD)
and
Charge on a1 = (r1 )
Ratio of charges =
r1
r2
E1, Field produced by a1 =
as r2 > r1
Therefore E1 > E2
V1 =
Charge on a2 = (r2 )
K [ (r1 ) ]
r12
=
KQ
r1
;
E2, Field produced by a2 =
KQ
r2
i.e. Net field at A is towards a2.
K .(r1)
= K
r1
V1
k r2
k V1 V 2
r2
33.(B) At C and E, the electric fields will be subtracted and at D they will be added.
APP | Electrostatics
98
Solution | Physics
Vidyamandir Classes
34.(AD)
35.(ACD)
36.(BC) Position A is position of equilibrium and B is critical position so
tension and speed during motion is maximum at A and minimum at B.
VMIN VB geff
2 g
VMAX VA 5geff 5 2 g
TMAX TMIN 6mgeff 6 2mg
37.(AD) By conserving energy of the first ball,
0
q2
40
0
q2
40
1
1
1
1
.....
r1N
r12 r13 r14
And for the second ball,
1
1
1
.....
r
r
r
2N
23 24
K K1 K 2
q2
4 0
K1 0
K2 0
1
d
[It can be observed K1 K 2 ]
q 40 dK
;
curved
38.(ABCD)Flux through two circular surfaces is
q
2
1
2 0
4R
3 R 2 4 R 2
q
5
0
q
q
4 q
0 5 0 5 0
39.(AB) Consider a small element AB, is very small. Then
AB R (2)
Change on AB is dQ
dQ.q
2T sin
4 0 R
2T
Qq
4 0 R
2
2
Q
Q
2 R
2 R
Qq
4 0 R2
or T
2
Qq
2
8 0 R 2
40 (ACD)At steady state electric force is balanced by centrifugal force.
At r radial distance
M 2 r 2 eEr
Now integrate E
APP | Electrostatics
dV
m2 R 2
and get potential at centre
dr
2e
99
Solution | Physics
Vidyamandir Classes
41.(A) Let us assume that the cavity is filled with equal and opp. charge density and .
C1C2
Ep
C1P PC2 =
i.e. parallel to line joining C1 and C2
2 0
2 0
a
42.(A) | E p |
| C1C2 |
2 0
2 0
For Q. 43 – 47
Let us consider forces acting on bead P as shown in Figure. These forces are :
(i)
Weight mg vertically downwards
(ii)
Tension T in the string
(iii)
Electric force between P and Q given by
F
(iv)
C2
C1
qq
1
. 1 2
40 2
Normal reaction N1
The net force along the string is (T F ) . Bead P will be in equilibrium, if the net force acting on it is zero.
Resolving forces mg and (T F ) parallel and perpendicular to plane AB, we get, when the bead P is in
equilibrium,
and
and
mg cos 60 (T F ) cos
. . . .(i)
N1 mg cos30 (T F ) sin
. . . . (ii)
For the bead at Q, we have
mg sin 60 (T F )sin
. . . . (iii)
N2 mg cos 60 (T F ) cos
. . . . (iv)
43.(C) Dividing Eq. (iii) by Eq. (1), we get
tan tan 60 or 60 which is choice (C)
44.(C) Using 60 in (iii), we have
mg sin 60 (T F ) sin 60
Or T F mg
qq
1
. 1 2 mg
40 2
. . . . .(v)
So the correct choice (C)
45.(C) From Eq . (ii) we have (since T F mg )
N1 mg cos30 mg sin 60 2 mg cos 30 3mg
46.(A) From Eq. (iv) we have
N2 mg cos 60 mg cos 60 mg
;
;
Which is choice (C)
Thus is the correct choice is (A)
47.(D) When the string is cut, T 0 . Putting . Putting T 0 is Eq. (v), we get
mg
qq
1
. 1 22
4 0
The right hand side of this equation should be positive which is possible if q1 and q2 have opposite signs. Thus, for
equilibrium the beads must have unlike charges. The magnitude of the product of the charges is
| q1 q2 | (4 0 ) mg 2 ,
which is choice (D)
APP | Electrostatics
100
Solution | Physics
Vidyamandir Classes
48.
[A q,r ; B-p,s,t ; C- p,q,t ; D- p,s,t]
for by symmetry
49.
kQ
kq q
kQ
&E= 2 &U= 1 2
r
r
r
v=
is | | to the axis
is as to the axis.
[A pq ; B - s ; C- s ; D – r]
For small '' separation of balls = land hence, T = mg
1
q2
=T =
q 2 3 , q , T 2
2
2 40 l 2
Tsin
In satellite
T=
50.
[A q ; B p ; C- s ; D- r]
V0 =
51.(9)
1
q2
(geff = 0) T 2 ,
4 0 2l 2
K 2 q
Kq
Kq
+
=
(Q, R)
R/2
2R
R
V0 (due to inner surface charge) = –
Kq
(P, R)
R
When outer surface is grounded charge '–Q' resides on the inner surface of sphere 'B'
Now sphere A is connected to earth potential on its surface becomes zero.
Let the charge on the surface A becomes q
kq kQ
–
=0
a b
q=
a
Q
b
Consider the figure. In this position energy stored
2
1 a
1 a
Q2
Q
Q (Q )
E1 =
+
+
80a b
80 b 40b b
when 'S3' is closed, total charge will appear on the outer surface of shell 'B'. In this position energy stored
2
1 a 2
1 Q
E2 =
80b b
Heat produced (Q) = E1 – E2 =
52.(2)
Q 2 a(b a )
= 1.8
8 0b3
So,
5Q 5 1.8 9
Takes all three bodies as system electrostatic force
(1) Apply law of conservation of momentum in direction y
(2) Apply law of conservation of energy,
So
m A v A m B v B mC vC 0 vC 2 v A (as VA = VB)
A
C
APP | Electrostatics
VC
B
VA
VB
101
Solution | Physics
Vidyamandir Classes
Change in electrostatic P.E. = Increase in KE
1
1
kQ 2 kQ 2 1
–
= mAvA2 + mBvB2 + mCvC2
2
2
2
2
53.(4)
arˆ
E
r2
V =
54.(2)
55.(3)
56.(9)
dV E d r
1 1
V a
r 2 r1
In uniform electric in vertical direction if (+ve) charge feels extra acceleration in downward direction, then (–ve)
charge will feel acceleration in upward direction.
;
v = 0, h = height
;
;
– (5 5) 2 = –2gh
v = 0, h = h
F
v2 – u2 = 2 g E h ;
0 – (13)2 = – 2 g
uq– = u(say)
v = 0 h = ht
m
Let
;
F
v2 – u2 = 2 g E h
m
–u2 = 2 g
FE
m
FE
h
m
h ; u = 9 m/sec
At a distance r from the center, take a thin shell of thickness dr. A charge dq = dv
will be trapped inside this shell. Now apply Gauss’ law on this volume (dv) to obtain the required result.
Q
If it was complete sphere, then total f flux
0
if position is changed, then f lux through the two hemisphere is also interchanged 1
L
59.(3)
So VC = 2 m/s
R
kq kQ
qR
0
Q=–
+
q
x
R
x
+
+
.
dQ qR dx 103 12
Q
= 2 =
= 0.5 × 10–3 = 0.5 × 10–6
dt
2 2
x dt
x
Net Electric field t P
E E1 E2 due to vertical wire
1
2
E1 =
in + y direction , E2
due to vertical wire in x direction.
2 0 y
2 0 x
/ 2 0 y
E
1
x
x
angle of electric field with x direction tan = 1 = 1
= 1 1
3; 1 =3
E2
2 / 20 x
2 y
2 y
2
3
uq+ = 13 m/sec
58.(1)
;
a a 2a
280
a a
=
=2×
= 16 V
= –
r2 r1
5 7 35
35
vuncharged = 5 5 m / sec
v2 – u2 = –2(g) h
57.(6)
[vA = vB, VC = –2vA
L
L
Net YE dq YE dY E Y dY
0
0
0
2
Net
EL
2
Net 3EL2 3E
I
2M
2ML2
APP | Electrostatics
102
Solution | Physics
Vidyamandir Classes
60.(8)
The bowl exerts a normal force N on each bead, directed along the radial line or at 60.0° above the horizontal.
Consider the free-body diagram of the bead on the left with the electric force Fe applied :
Ey N sin 60 mg 0
Fx Fe N cos 60 0
ke
N
ke q 2
R
2
mg
sin 60
N cos 60
mg
mg
tan 60
3
1
9.0 109 Nm2 / C 2
40
1/2
mg
q R
k 3
e
Thus,
61.(5)
Assume ' ' and ' ' in the cavity then
V
V
3 K 4 3
. R
2 R 3
3
4 R
K .
3 2
R
2
VC V V 2K R 2
V
5R 2
12 0
dq
A dx A
qin
3
1
dx
[area under the curve] =
0
0
0
0
0
4
Flux will be maximum when maximum length of ring is inside the sphere.
62.(0.75)
63.(2)
K R 2 5K R 2
3
3
This will occur when the chord AB is maximum. Now maximum
length of chord AB= diameter of sphere. In this case the arc of ring
inside the sphere subtends an angle of
R
charge on this arc =
.
3
64.(1)
at the centre of ring.
3
E .dA
R
R
3
Solving ans 2
0
3 0
;
E
dV
E 4r 2
0
k r n 4 r 2 dr
0
4k r x 3
0 x 3
k
n 3 0
r n 1
;
n 1 2
n 1
APP | Electrostatics
103
Solution | Physics
Vidyamandir Classes
65.(2.5) In the remaining three quadrants, put three more quarter sheets to convert this given arrangement to that of infinite
sheet. Now contribution from all the four quarters to the z-component will be same. Hence due to a quarter E.F. at
1 z
z ˆ
point (0, 0, z) will be, E
k.
kˆ
4 2 0 z
8 0 z
Hence potential difference between points (0, 0,1) and (0, 0, 2) will be,
2d
v2 d vd
E.dl
;
d
2d
v2 d vd
d
z ˆ
dl dzkˆ; E
k
8 0 z
z ˆ ˆ
k .dz k
8 0 z
8 0
2d
d
z
z
vd v2d
;
dz
d
8 0
Substitute the value of σ and d
66.(4.8) E net at M 2 E1 2 E cos
2K
2K 1
2
.
2
a
a 5
5
2
2
8 K 8 K 48 K
substituting values Ans is 4.8
a
5a
5a
The value of flux is maximum through surface AB GH, because charge in front of this surface is maximum, and
67.(9)
flux is =
68.(4)
9q
.
240
Work done by magnetic force is zero so from work energy theorem
1
1
m p vB2 m p v A2 q v
2
2
and simultaneously there is no change of velocity component along the direction of perpendicular to electric.
v A sin vB sin
After solving v 16 Volt
69.(20) For maximum angular velocity, rotation is equal to 90º.
WEF KE
1
(qE ) R (mR 2 ) 2
2
70.(4)
;
2(q ) E
mR
Ql
2
2R
mR
QlE
mR 2
substituting values Ans is 20
The force experienced by an electric dipole placed in a non-uniform electric field is given by, F p
E
where
l
E
is directional derivative of E along the dipole moment. Here, dipole is placed along x-axis, so
l
E
E
corresponding to component of
is along x-axis.
l
u
E
E
ˆ
ˆ
ˆ
E p
6 xi 6 yi 0k
p(6 4iˆ 6 0 ˆj )
F | p |
x
x x component
APP | Electrostatics
104
F 24 piˆ
Solution | Physics
Vidyamandir Classes
DC CIRCUIT AND CAPACITOR
1.(B)
2.(B)
V2
P=
B
V2 V2 V2
....
P1 P2 P3
Equivalent circuit.
R1
R2
R1
R2
G
A
R4
R4
R3
R3
D
+
3.(D)
C
–
(1680 + r) I = 20 (2930 + r) I = 30 2 × 2930 + 2r = 3 × 1680 + 3r r = 820,
4.(D)
r
r
r
(r)
2r
A
I 8mA
Resistors whose resistances are written in brackets are parallel.
2r
(r)
r
r
2r
r
r
2r
A
r
r/2
C
r
D
r
r
2r
2r
2r
2r
r/2
B
r
A
r/2
2r
C
r
B
D
r
4r
r
C
C
A
B
2r
A
B
r
r
r
r
D
4r
5.(A)
E = j [j = current density]
j=
I
πr 2
r = [a+
6.(A)
4r
[r = radius of cross section at distance 'x' from left end]
b a x ]
Hence, E =
l
Resistance of each part =
Vl 2ρ
πR ( al b a x ) 2
R
2n
R
R
For 'n' such parts connected in series, equivalent resistance, say R1 = n = . Similarly, equivalent resistance,
2n 2
R
1 R
say R2 for another set of n identical resistors in parallel would be
=
n 2 n 2n2
For getting maximum of R1 & R2, they should be connected in series & hence, Req = R1 + R2 =
APP | DC Circuit and Capacitor
105
R
1
1 2
2
n
Solution | Physics
Vidyamandir Classes
7.(A)
9.(C)
I 0.1Ω
5 103 A
I 50mA
8.(C)
0.2 0.3 0.5
V A VB voltage drop across capacitor + voltage drop across resistor
18
3
.1 0 R 3
3 R
Q
16
iR 11 i.3 103 i 5mA
C
4
Power delivered by capacitor P iV (5mA)(4V ) 20mW
11
2E
5R
2
2
q
Steady state p.d. across capacitor : V0 = E q0 = CV0 =
EC t = 0 =
3
3
i
10.(D) Initial charging rate = initial current in the line of capacitor =
2
3 EC
2E
5R
=
5
RC
3
11.(A) Suppose that the inner sphere is at a higher potential than outer sphere. Let the current be i.
Consider a thin shell of thickness dr at a distance r from the centre. Let voltage across it be dV. Then, applying
V iR
dr
dr
1
Edr i
2
2
σ4πr
4πr kE
C
i
i
where C
E
r
4π k
4πkr 2
dV
E
dr
C
dV
C (nr )ba Va Vb
r
dr
For thin shell: dV i
Using,
V Va Vb C ln(b / a )
V 2 4 k
i
i
n(b / a )
4 k
[np(b / a)]2
P.d. across C is zero
charge = 0
V
12.(D)
13.(D) Let n1 : no. of capacitors be connected in parallel, n2 : no. of such parallel combination connected in series.
n2
n1 8 F
1000
4 and
16 F n1 8
n2
250
Total no. of capacitors required = 32
14.(B)
Charge on C = sum of charges at 4&5 = 2 CV 2 2 10 40 C
APP | DC Circuit and Capacitor
106
Solution | Physics
Vidyamandir Classes
inside dielectric, field
15.(B)
0
0
1
1
Force between A & M =
A.
2 0
(Using Eq = F)
2 1 A
2 0
16.(B) Current is obviously constant, by charge conservation. Using i = n0eAvd, we can say that if A is non-uniform,
vd will be non-uniform. Similarly, since vd depends on electric field, electric field will also be non uniform.
17.(AC) Current is obviously constant, by charge conservation. Using i = n0eAvd, we can say that if A is non-uniform,
vd will be inversely proportional to A. Similarly, since vd increases with electric field, electric field at A will be
more than B.
18.(A) Mark the voltages as shown in figure:
By KCL, sum of all currents emerging from point P is zero
i.e.
x 8 x 6 x 0 x 4
0 . Solving this, we can get x.
2
5
4
3
19.(ABC)
Let the point where jockey touches wire be called S. Then the direction of current shown in figure
indicates
that voltage across QS is less that E2. This can happen if:
1.
E1 is too low
2.
r is too high ( if r is too high, it will take up more voltage and less will be left for QS).
20.(BD) Let the currents be as shown in the figure:
KVL along ABCDA
– 10 i – 2 + (2 – i)1 = 0 i = 0
10
A
B
2
i
Potential difference across S = (2 – i)1 = 2 × 1 = 2 V.
21.(ABCD)
2-i
1
D
2
C
After redrawing the circuit
(a)
4 = 5A
(b)
From loop (1)
– 8(3) + E1 – 4(3) = 0 E1 = 36 volt
(c)
From loop (2)
+ 4(5) + 5(2) – E2 + 8(3) = 0
E2 = 54 volt
(d)
From loop (3)
– 2R – E1 + E2 = 0
R=
E2 E1 54 36
=
=9
2
2
APP | DC Circuit and Capacitor
107
Solution | Physics
Vidyamandir Classes
22.(AC) Equivalent diagram is as shown. If P is moved 2cm right, then R1 = 12 Ω , R3 = 3 Ω
R1
R
R1
= 2 (Hence wheat stone will be balanced.)
R3
R4
R
5
10
If S is moved left by cm, then R3 =
and
3
3
20
R
R
R4 =
hence 1 = 2 (hence wheatstone’s bridge will be balanced.)
3
R3
R4
R R2
R R2
R
R
1
1
AC
CB
L/4
3L / 4
4R
5R
R2
and R1
3
3
10 5
3 2
eq
1 1
3 2
23.(AB)
24. (AC)
5/6
1V
5/6
6
req
5
i
R1 R2 3R and
R2
P
R3
G
Q
S
R4
R R1
R
2 R R1 2 R2
2 / 3
/3
1
0.5 A
4 6
5 5
4
0.4 V
;
5
3i1 10 1.0 0.4 0
4
0.4 V
5
i1 ve
0.4 2i2 5 0
i2 ve
P.d. = 0.5
0.5
25. (AB) If P is opened effective resistance would increase A is correct & C is wrong. As drop across R1 is reduced
B is correct. Battery is ideal hence D is wrong.
26.(AB) The effective emf of secondary is 0V
A & B are correct and D is wrong. There will be current in 1V battery C is wrong.
27.(ABC)
B2 and B3 are in series,
i2 i3 ;
Also, V2 V3 V1
P1 P2 P3 P4 P2 P3 or i22 R2 i22 R3 R2 R3
V2 V3
V2
V1
V2 V2 V2
V2
1 1 1 and P3 3 1
2
R1
R1
36
R3
4 R3
V12
V2
R3 9Ω R
36
4 R3
2
P1 P4 I12 R1 I 42 R4
APP | DC Circuit and Capacitor
or
I4
2
36 I 4 R4 R4 RΩ
3
108
Solution | Physics
Vidyamandir Classes
Voltage output of battery = V1 V4
V2
V2
P1 4 1 V1 12V & P2 4 4 V4 4V
36
4
28.(ABCD)
i1
VA VB
10
5
A
3
5
8
i2 i1 1 1 A
3
3
6
6
G
VE VD V A VB 10
i1 i3
5
V1 V4 16V
i3
10
3
1A
R
1A
1 F 2
H
10
5A
in loop F E D C
3 3
2 3i3 4 i1 i3 3 2 i1 i3 0
3
i1 + i3
2 + i1 + i3
2A
A
i3
4 3
A
i2
E
i1
6
C 4 D
B
2 51V in loop F G H C
1 4 2 3 2 i1 i 3 0 1 29 V
R
VA VG
1
2 1
1
22
29.(ABCD)
(1)
(2)
(3)
4 x 10 y 6( I x) 0
3( x y ) 7( I X y ) 10 y 0
6( I x) 7( I x y ) 2 I 49 0
259
5
A 4.98 A
y 4.98
0.34 A
52
74
5(4.98 0.34)
I
7.73 A
3
x
I x y 2.41A
I x 7.73 4.98 2.75
30.(ACD)
At t = 0, capacitor is uncharged and there will be no voltage across it. Hence it can be short-circuited in
solving the circuit. After a long time, no current flows through the branch containing capacitor. i.e. rest of
the circuit elements will be in series.
31.(ACD)
At steady state : I(3) + I(2) = 15
I=3
KVL
C D E a b C
(V/C) – I(3) +
q
q
–7+ =0
11
5
q
q
+ = 7 + 3 × 3 = 16
11 5
q = 55 C
q
q
55
KVL : a b
Va – 7 + = Vb Va – Vb = 7 – = 7 –
= –4V
5
5
5
P.d. across C1
q
55
=
= 5V ;
11 11
P.d. across C2
q
= 11V
5
P.d. across terminal = 15 – I(2) = 15 – 3 × 2 = 9V
APP | DC Circuit and Capacitor
109
Solution | Physics
Vidyamandir Classes
32.(AB) We know that the capacitance of an empty capacitor increases k times if a dielectric is inserted in it. Therefore, in
this case, the capacitance of combination will increase upon insertion of a dielectric. Also, by Q = CV, charge
supplied by battery also proportionately increases for keeping V constant.
33.(ABC)
At t = 0, capacitor will have no voltage across it. Hence A. Voltage across capacitor will gradually increase
with time. Hence B. C can be calculated by the usual charging equation for capacitor.
34.(ACD)
Q = CE
Wbattery = Uf – Ui + H
1
CE2 + H
2
1
1
H = H1 = CE2 – CE2 = CE2
2
2
E · CE =
When switch is closed all the energy stored in capacitor is
released as heat (H2)
Hence H1 = H2 =
35.(ABC)
1
CE2
2
2i1 + 3(i1 + i2 – i3) + 4(i1 + i2) – 10 = 6
i2 + 2i3 = 3
3(i1 + i2 – i3) – 6i3 = 0
36.(AC) Charge flown through battery = C1V 6 20 120 C
After closing S2 , common potential VC
Final C2 C2VC 3
120 C
9 F
C1
120
40 C
9
2
2
120C
+
–
initial
0
C2
C1
120 q
+
–
C2
+ q
–
final
2
120 80
40
J 0.53 m J
Heat produced =
2 6 2 6 2 6
37.(ABCD)
Q1 Q2 Q
and
Q1 2Q2
;
Q2
Q
3
APP | DC Circuit and Capacitor
Q1 2Q2
C
C
2Q
Q1
3
and potential at point A is
2Q
3C
110
Solution | Physics
Vidyamandir Classes
38.(ACD) i
E
5R
Equation of charging of capacitor
q CEe t /
q ECe t /5 RC
t = 5 RC ln2
at
q
EC
2
U cap H 2 R H 3R
Q 2 (Q / 2) 2
H 2 R H 3R
2C
2C
3Q 2
H 2 R H 3R
8C
( H i 2 R and i is same in series,
2
H 3R
H R)
2
3 3Q
9Q
5
8C
40C
9
CE 2
40
39.(ABCD)
Uncharged capacitor behave like zero resistance
36 36
I
12 amp
Req 3
IC
6
12 8 amp
9
In steady state, capacitor behave like a large resistance
36
I
9 amp
4
40.(ABCD) Initially capacitor behave like zero resistance.
Pinitially
V02
3
P
2 2 0
2 V0
3 P0
Finally capacitor behaves like large resistance.
Pfinal
41.(ABCD)
V02
P
0
V02 2
2
P0
Potential difference across two plates must be equal to .
So potential difference across any point on plate-1 and any point on plate-2 will be constant
So electric field will be same.
42.(ABCD)
Whenever a charged capacitor is connected to other charged or uncharged capacitor we have to distribute
charged by applying Conservation of Charge and equality of potential.
APP | DC Circuit and Capacitor
111
Solution | Physics
Vidyamandir Classes
43.(ABD)As Q1 Q2
or C1 E1 C2 E2 Hence (A) is correct.
t
q Q0 e RC
Slope
t
dq
1
Q0e RC
dt
RC
Slope
1
RC
(B) is correct
R1C1 R2 C2
As Q1 Q2 Hence (D) is correct.
R1C1 R2 C2 but C1 and C2 are not known hence (C) is incorrect
By energy conservation A is correct.
44.(ABCD)
B is correct as Current is max at t = 0 . Just before steady state voltage across C is same as battery hence C is
correct.
45.(ABC)
ABCD
Initial charge (before filling the dielectric slab) = 10 × 10 = 100 C
Final charge (after filling the dielectric slab) = 10 × 30 = 300 C
Increase in charge = 200C.
46.(A) Reading of C = V {i in that branch = 0} ; Reading of A =
47.(D) VB = iRammeter = 0
V
V
Ratio =
R
V/R
R
Vcap. = 0
48.(C)
V
R
L
R
2V
at t = 0 inductor is open circuit i
V
2R
V
V/2
O
3V/2
O
2V
49.(A) 50.(A) 51.(C)
E2 E1 iR1 0
;
i
V 3V
2 2
E1
E2 i x R2 E2 0
ix
p.d. = 2V –
E1 E2
R1
APP | DC Circuit and Capacitor
E2
C
R1
–
+
x
i
112
R2
i–x
Solution | Physics
Vidyamandir Classes
VA E1 E2 VB
VA VB E1 E2 constant .
1 2
1 R2 2 R1
After long time, VC 1 I R1 (current through capacitor = 0) 1
R1 =
R1 R2
R1 R2
1 R2 2 R1
Hence QC VC . C C
R1 R2
52.(A) Potential gradient in AB when both k1 and k 2 are open
E2 Pot.gradient AJ
2
125
E1RAB
2
R1 RAB AB
125
volts / cm.
31.25 0.5 volts
53.(A) Equivalent diagram when both k1 and k 2 are closed is as shown
54.(A)
V AJ VCD
E2 E 1 .
2V
0.5
5 r 7.5Ω
10
5 r
2 1
i1
C
r
Q2
2 A 0 / d
60 60
26
R
i
q2
q2
R
dq1
1
0
t
3q1 q C dt
q1
0
Q
1 e
3
3t / RC
r
D
i2
5Ω
R
J 300 J
q1
9Ω
c
KVL :
A 0
C
d
J
0.5V
D
q2 d
A 0
b
Q
Q
S
q1
q1
1Ω
C
5Ω
55.(C) Before closing switches
56.(A) At time t
B
9Ω
1Ω
0.5V
Energy stored =
A
J
A
AJ
E
AJ 2 AB 12.5 cm
AB
E1
2V
2q1d
A 0
a
iR 0
. . . .(i)
q1 q2 Q
dq1
i
dt
Q q1 2q1
dq1
R
0
C
C
dt
dq1 3q1 Q
R
dt
C
R
3q1 Q
t
n
3
Q C
. . . .(ii)
. . . .(iii)
3q1
Q
1 e 3t / RC
201 e C
500t
57.(D)
q2 60 20 40C at t
58.
[A-qs ; B-p ; C-p ; D - t]
From charge conservation, current through any cross section remains constant. Current density is current/Area of
cross section. Only in option A cross sectional area is constant. Hence Q fits only with A.
APP | DC Circuit and Capacitor
113
Solution | Physics
Vidyamandir Classes
Similarly, from i = n0eAVd, drift velocity depends on the ratio of current and cross sectional area. So, the above
argument applies in part (S) as well.
59.
[A - p r s ; B – p r s ; C – r ; D - q r]
Q
C
(A)
a
C0V0
C0
d
c
b
(B)
d
0
a
V0
c
b
(C)
(D)
C0V0
2
C0V0
2
C0V0
2
C0
2
C0
V0
2
C0V0
2
b
0
a
c
KC0V0
KC0V0
d
KC0V0
a
C0V0
KC0V0
b
C0
2
C0
2
KC0
K 1
C0
2
C0
2
C0
2
KC0
K 1
KAε 0
d
KC0
KC0
KC0
K 1
C0
KC0
KC0V0
K 1
KC0V0
K 1
d
c
KC0
K 1
KC0
K 1
V
V0
V0
V0
V0 ( K 1)
2K
0
V0
V0
V0 ( K 1)
2K
0
V0
V0
(K + 1)V0
V0
V0
V0
V0
60.
[ A p q ; B s ; C r ; D q]
It is easy to see that the capacitor will initially drive the current in the same direction as battery (Anti-clockwise).
The capacitor will discharge after some time. Then it will get charged in the opposite direction, the direction of
current remaining anti-clockwise throughout.
Graph between charge and current can be drawn from the KVL equation around the circuit.
61.
[A-p, q]
62.(1)
Assuming potential at B 0
I2
63.(5)
30 5
1
25
[B-q, s]
;
[C-q, s]
I1
[D-q]
30 25
1
5
The equivalent resistance at 0°C is
APP | DC Circuit and Capacitor
114
Solution | Physics
Vidyamandir Classes
R10 R20
… (i)
R10 R20
The equivalent resistance at tC is
RR
R 1 2
… (ii)
R1 R2
But R1 R10 (1 t )
… (iii)
R0
R2 R20 (1 2t )
And R R0 (1 eff t )
… (iv)
… (v)
Putting the value of (i), (iii), (iv), (v) in eqn. (ii),
64.(2)
Let x be the number of electrons striking the surface per unit time.
I
F = PA = nmv = mv
q
I=
65.(1)
PAq
mv
;
I=
9.1 1 10 4 1.6 10 19
= 2 Ampere
9.1 10 31 8 10 7
Current in circuit
I=
66.(2)
5
eff
4
;
1
A = 0.2 A which will pass through 10 and 20 in both the cases.
5
Taking potential at A to be zero, potential at B = 3V and potential
at B' = 3V and potential at C = 6V
Let VD be potential of point D then sum
of charges reaching point D is zero
VB VD VB ' VD
(V VD )
+
+ C
=0
RV2
RV1
RV3
3 VD 3 VD 6 VD
+
+
=0
R
R
R
RV RV RV R
1
1
3
12 – 3VD = 0
; VD = 4 volts reading of V3 = 2 volts.
67.(3)
Req Rin
68.(3)
4 12
3 (Max power delivered when internal resistance = External resistance)
4 12
(1)
x(9 r ) y (14 r )
(2)
xr 7.5
APP | DC Circuit and Capacitor
115
Solution | Physics
Vidyamandir Classes
yr 5
(3)
By (1), (2), (3)
9 x 7.5 14 y 5 24z zr
So,
9 x 14 y 2.5 ;
So,
7.5
24 z zr 9
7.5
r
So,
24 7.5 (9 r) 7.5(9 r )
5
14 5
r
r (24 r )
(24 r )
(9 r )(7.5) 24
1
(24 r) r
14 y 9x 2.5
9
7.5 1
r
7.5(9 r )
z
24 r
r (24 r )
;
14 (9 r)(7.5)
5 1
r
r
;
70 5r 67.5 7.5 r 2.5 2.5 r r 1
x 7.5 y 5 24 z z 14(5) 5
5(14 r ) 7.5(9 r )
25z 75 z 3
;
Voltage across V3 zr 3
69.(0)
Since potential difference across AD and across CD is same, so A & C are on same potential. Therefore energy
stored in the capacitor at given instant is zero.
b
70.(1)
dR
dr ; R
dr
2 Lr
2 Lr
a
71.(3) Current through branch containing capacitor is
dq
I 3e t amp. At t 0, I 3A
dt
C
72.(2.5)
5
4
3
C
1,3
2,5
C
4
C
2
q1
1
q1
45
2, 4, 5
Again
C
q2
5CV
2
q2
1
1, 3
C
3
2
V
C/2
5CV
3
V
2
5
5 5 5CV
2
Work done by battery V q2 q1 CV 2
= 30 10 J 2.5mJ
6
6
2 3
APP | DC Circuit and Capacitor
116
Solution | Physics
Vidyamandir Classes
4C
C
11C
Ceq 3
C
7C
7
3
73.(1.57)
74.(3)
1 1 1
Ceq
8 12 8
1
24
3
3 2 3
75.(1.33)The network is equivalent to
Therefore equivalent capacitance
=
76.(3)
2C C 4C
2C C
3
[2C series C] // [C series 2C]= 2
When switch is closed capacitor behaves as conductor and finally behaves like a very large resistance hence
potential drop across resistors will be zero at steady sate. Equivalent capacitance will be 3C and X =3.
77.(2)
Let a be the side length of square and be the position where galvanometer gives zero deflection. To have zero
deflection bridge is to be balanced.
RAB
RBX
RAD RDC RCX
[ RDC and RCX is in series]
400
a tan
100
a
200 500 400 (a a tan )
a
1
400 tan
2 500 400(1 tan )
Solving tan 3 / 4 ,
t be the time taken from start, t[ is radian]
APP | DC Circuit and Capacitor
117
Solution | Physics
Vidyamandir Classes
37
t
180
360
5
78.(22.5)
R
3
R3
t = 74 sec = 2 × 37 sec.
On solving reading of voltmeter is v 22.5 volt
79.(1.4) Use result of Req. for cross symmetry
2
1
1
R1 R
1
R1R2 R2 R3 R3 R1
R2 2 R
1
1
2
Req
R3 R
R1 R2 R3
80.(3)
1 5
2
2
1
2 2
2
2
1
2R
R R 2 5 . 2R 5
2R
7
7
Req
7R
2 R2 7
2R
2R
During charging for 1 Req C
Req R3 R
1 RC
During discharge 2 Req C
Req
Req.
7R
5
R1 R2
3R
R3
R1 R2
2
3
RC
2
1 2 2
n 3
2 3 n
2
APP | DC Circuit and Capacitor
118
Solution | Physics
Vidyamandir Classes
MAGNETIC EFFECT OF CURRENT
1.(C)
Bilt Blq
mv i l B t Bilt v
M
M
0 v 2 2 gh
2.(B)
B x
0 I 2
I I dx
, dF BI1dx 0 1 2
2 x
2 x
F
3.(A)
R
4.(C)
The plane of motion of the particle is the z-x plane. It is a case of uniform circular motion a .v 0
5.(C)
Magnetic moment M =
6.(C)
0 I i
k
B
4 a 2 2
2
7.(B)
8.(D)
B 2l 2 q 2
M2
2 gh
0 I1I 2 2 a dx 0 I1I 2
=
ln 2
2
2 a x
mv
qBR
qBr qr
, 2R r V
V
0 ni
qB
M
2M 2M
B
0 I1
2
qL
q 2
1
mR 2 M qR 2
5
2m 2m 5
0 I2
k
210
2 2 10 2
2m 2 1
T
2s ;
qB
11
j
R
mv 11
1m
qB 11
1 E y q 2 1 1 1 2 2
t
m
2 M
2 1
2
z = 2 R = 2m
x = 0, y
2
,2 m
2
Co-ordinates will be 0,
9.(A)
Force on PS F I
10.(A)
R
mv
qB
mv
2 qB
Force on PQR F' I 2 R B
2 R B
F' 2 F
. Hence particle will enter in the region where magnetic field is absent. The return path will be
C
identical to path ABC. So required time t 2 t AB t BC
O
m mv / 2 qB
= 2
cos V
qB
Also sin
mv / 2qB
mV / qB
1
2
V
R
M
from triangle BMO
B
A
m
m
m
Hence, t 2
4
4qB q B 2 q B
mv
2 qB
mv
2 qB
11.(B) Speed can change only due to work done by electrostatic force.
Hence
qEZ
1
2
mv 2
APP | Magnetic Effect of Current
10q Z
1
2
mv 2 V
119
20 q z
m
Solution | Physics
Vidyamandir Classes
12.(D)
B
0 I1
, let the current in PQ be i2 dF Bi2 dx =
2
net anticlockwise torque on PQ is dF . x
0
For equilibrium
13.(C)
0i1i2
2
mg
2
i2
0i1i2
2 x
0 i1i2
2
x
dx
i2
0i1i2
2
i1
P
dx
B
Q
mg
0i1
mV
is decreasing as v is decreasing
qB
it enters at P charge is negative bending shows the direction of force
R
i1
14.(B)
I 2
I
; i2
2
2
B1
0i1 2r
i
and B2 0 2 .
2r
2
2r 2 z
B1 and B2 are in opposite direction, but have same magnitude
15.(A)
dF
dx
B1
B3
16.(BCD)
KI
KI
KI / a
, B2
a
4
4a 2
KI
a/ 2
0 I 2
8 2 r
net field is zero.
2 1 KI
4
a
2 KI
B3 B1 B2
a
m B
U m . B
17.(BD) Electric field along the axis is non zero due to P.D. along axis.
18.(BCD) Force on ab will be stronger than bc.
19.(ACD) K .E qV
1
mV 2
2
R
mV
qB
B
1
q
0 2 dx
2
R
20.(AD) Use symmetry and Ampere’s law
21.(AD) d B
0 2 xdx d
R 2 x 2
22.(ABC) ArcAB
;
mV
r
3
3qB
I T m
Time ' t '
2 3 6 3qB
APP | Magnetic Effect of Current
120
Solution | Physics
Vidyamandir Classes
23.(ABC)The particle will move along an arc which is part of a circle of radius r
mv
Bq
From the figure we can see r = R
R
mv
r / 2 r
; T
Bq
V
2v
24.(CD) For cylinder B
rR
mV
m
T
Bq
2 Bq
0ir
i
;ra 0 ;ra
2
2 r
2 R
We can consider the given cylinder as a combination of two cylinders. One of radius ‘R’ carrying current I in one
direction and other of radius R/2 carrying current I/3 in both directions.
At point A: B
25.(ABC)
0 I / 3
I
0 0
2 R / 2
3 R
At point B: B
0 4 I / 3 R
0 I
0
2
3 R
2
2 R
y is speed of light x and z also have same dimension
26.(ABC)
F Il B .
27.(AD) Normal force of the rail on the wire = Bil
100
3
. 1 . 30 N
5
4
force of static friction at t = 0 is 20 N
max force of friction at t = 0 is Biintial l. 2
But weight = 2 g 20 N
100
. 1 normal force is decreasing
5 0.5t
friction is also decreasing max. value of force static
When max. frictional force reduce to weight of the rod, it stats moving
Normal force at time t is Bil = 2
200
3
20
5 0.50t 4
30 20 2t t 5 sec
28.(BC) Consider the solid cylinder as super position of solenoids.
29.(ABC)
30.(ACD)
L
, if we double radius and cross-sectional radius, then resistance will be halved.
A
Due to sheet
R
0 k 0 (2bJ )
0 JB.
2
2
The slab is symmetrical under translation in y so field is independent of y.
B
APP | Magnetic Effect of Current
121
Solution | Physics
Vidyamandir Classes
Symmetric under rotation by 180° around Z axis, so y component of field is odd function of x. consider the ampere
loop shown in diagram
2 Bh 0 (2 xhJ )
B 0 Jx
31.(AD) I
2
Ma 2
a
I 0 a 2 B0
2 M 2
2
12
Ma 2 Ma 2
I 0 a 2 B0
2
6
I 0 a 2 B0
32.(ABCD)
(A)
2
Ma 2
3
3I 0 B0
2M
We have F q E (V B )
F q aiˆ ( xiˆ yiˆ) biˆ ciˆ dkˆ
F q iˆ(a yd ) ˆj ( xd ) kˆ( xc yb)
(B)
for c 0; d 0; y 0
F qaiˆ
(C)
So particle will move along straight line with increasing speed
for c 0; d 0; y 0
F q[(a yd ) i yb kˆ]
| F | q (a yb)2 ( yb)2
(D)
So particle will moves along helix of varying pitch
Here a 0, v B 0
And v is perpendicular to B so particle will move in circular both.
33.(AB)
34.(AC) T
2m
2
Bq
B0
APP | Magnetic Effect of Current
122
Solution | Physics
Vidyamandir Classes
At t
T
; velocity of particle is v0iˆ v0 kˆ
B0 2
Speed will always remains constant v0 2
2V0
2
At t
T ; displacement is equal to pitch, x V0T
B0
B0
2 2V0
2
T ; distance = speed T
B0
B0
Arc
6.28
35.(ABC)
;
R
3
Radius
2
R3
I
B 0
4R
40
Solving B
107 T
9
36.(A) First particle will travel along parabolic path OA. Let
qE
time from O to A is t. a y
m
2
3mv
3mv
x
(2v cos 60)t0
t0
qE
qE
At t
qE 3mv
0
m qE
Hence at point A, velocity will be purely along x-axis and it will be 2v cos 60 v.
v y u y a yt0 2v sin 60 ,
37.(B) Now magnetic field is switched on along y-axis. Now
its path will be helical as shown below with
increasing pitch towards negative y-axis. r
x x0 r sin (2v cos 60)t0
38.(C) z-coordinate : z (r r cos )
mv
qB
mv
sin t
qB
v 3
mv mv qB
sin
t
qE qB
m
mv
qB
1 cos
t
qB
m
39.(C) In triangle PMC
MP
cos 53
MC
3
R
5
4 R
12 = 8R
3
R m (R is the maximum radius of half – circle)
2
mu
Rmax max umax 3 m / s
qB
APP | Magnetic Effect of Current
123
Solution | Physics
Vidyamandir Classes
40.(B)
R
mu
24 m
qB
Let, MPQ
By geometry
CPO (37 )
In CPO
OC
OP
sin(CPO ) sin(PCO )
20
24
sin(37 ) sin(180 37)
5
5 6
1
sin(37 )
sin(37 )
3
2
7
qB
rad .
180
m
2rad / sec.
t
7
sec.
360
41.(D) Since there is no current passing through circular path, the integral
B d along the dotted circle is zero.
42.(B) Let segment OB = OC and arc BC is a circular arc with centre at origin. Since the shown closed path ABCA
encloses no current, the path integral of magnetic field over this path is zero.
B
Hence
A
B d
C
B d
B
A
B d 0
C
Because B is perpendicular to segment AC at all points,
A
therefore
B d 0
C
B
Hence
A
B d
B
C
0 I OB() 0 I
1
B d
tan 1
2 2
2
2
43.(C) Consider two points P and Q lying on dotted circle and equidistant from origin O. We draw a circular arc QP with
centre origin O. The path integral of magnetic field, that is, B . d along the dotted circle between two points P
and Q is also is equal to path integral
APP | Magnetic Effect of Current
B d along the arc QP whose cenre is at origin.
124
Solution | Physics
Vidyamandir Classes
Therefore the path integral of magnetic field
B d along the dotted circle between two points P and Q
0 I OP () 0 I
2 OP
2
The value of will be maximum when chord OQ and chord OP will be tangent to the dotted circle, that is,
Hence the required maximum value
0 I
.
6
[A – s ; B - p ; C – q ; D – r]
45.
A-p, q; B-p, s; C-p, r; D-p
46.(9)
Energy density of electric field, u
47.(5)
Bz
48.(3)
Use F q V B
44.
.
3
1
0 E 2
2
R
mV
qB
Energy density of magnetic field. uB
1 B2
2 0
0 I
10
107
105 T
4 R
10 102
I
0 I
2 R 2
R2
Bp 0
4 R 2 x 2 3/2
2 R 2 x 2 3/2
Bc
2
0 I
2R
0 I 0 I
R
2
16R 16 R R x 2 3/2
2R R 2 x 2
0 I
4R 2 R2 x 2
R 3x
0 I
l
B 4
2
l
l2
4 x 2 2 x 2
4
4
49.(4)
B1
50.(5)
Fm qvB, and directed radially outward.
l2
4 x 2
4
,
N mg sin qvB
1/2
8R3 R 2 x 2
mv 2
R
3
k 3
(N will be max at 90 )
2mgR
mg qB 2 gR 3mg qB 2 gR
R
The magnetic field due to ring in x-y plane is
I
B1 0 kˆ
2R
the magnetic field due to ring in y-z plane is
I
B2 0 iˆ
2R
and the magnetic field due to ring in x-z plane is
I
B B1 B2 B3
B3 0 ˆj
2R
N max
51.(7)
APP | Magnetic Effect of Current
125
Solution | Physics
Vidyamandir Classes
i
B 0 kˆ iˆ ˆj
2R
Hence, X = 3 and Y = 4
52.(2)
30 I
2R
B
3 0 I
4 R
=
To enter region 2; Radius in region I should be greater than `d
R
mV
d
qB
;
V
qBd
m
;
V
1.6 10 19 0.001 5 102
9 1031
;
V
8
107
9
To come out of region 2; diameter 2R d
2mV
d
qB
;
V
qBd
2m
;
V
1.6 10 19 0.002 5 1012
2 9 1031
8
V 107 ms 1
9
53.(3)
8
Hence for both region V 107 ms 1
9
The magnitude o f magnetic moment is
M iA 10 (10 102 )2 Am2 10 102 0.1 Am2
The normal on the loop is in x z plane. It makes 60° angle with x-axis.
M M cos 60iˆ M sin 60 ˆj
0.1
M
iˆ 3 ˆj
2
Balancing torque
54.(4)
mg
;
M
3 ˆ
M iˆ
Mj
2
2
M (0.05) iˆ 3 ˆj Am 2 ; X 3
2R
R 2
I
B
2
2R
R 2
(R) g
I
B
10
2
I 4
55.(1)
Pitch will be
P V| | T
and
:
4
I
B
T
2R 2mV
V
qBV
2 m
qB
2m
T
qB
V 2 R
p
2
V
2
p 2R
p
R
2
56.(5)
B d 0inet
APP | Magnetic Effect of Current
126
Solution | Physics
Vidyamandir Classes
r
r
B 2r 0 J 0 2r 2 dr
a
0
57.(7)
J
I
2
(b a 2 )
B d 0 Ienclosed
B (2r ) 0 [ J (r 2 a 2 )]
2 10 7
58.(2)
5.5
;
B
0 I r 2 a 2
2 r b2 a 2
(0.21)
(3)
;
7T
1.1 10
Magnetic field is non zero only in the region between the two solenoids, where B 0 n2i2
2
n2i 2
B2
0 22
2 0
2
The energy per unit length = energy per unit volume × area of cross section where B 0
59.(3)
Energy stored per unit volume
0 n22 i22
n2 i 2
[(r22 r12 )] 0 1 1 [(r22 r12 )]
2
2
(Since n1i1 n2i2 )
E is parallel to B and v is perpendicular to both. Therefore, path of the particle is a helix with increasing pitch.
Speed of particle at any time t is
v vx2 v 2y vz2
. . . (i)
2
qE
Here, v 2y v z2 v02 and vx2
t and v 2v0
m
Substituting the values in eq. (i), we get
t
60.(4)
3mv0
qE
A and P will have same momentum in magnitude and they will move in opposite direction. They will move in the
circle of same radius and the same centre but in opposite directions. If they meet after time t then
At P t 2
t
t
2
2
2eB
A P 2eB
4m
( A 4) m
4( A 4)m
2eB 4m( A 4)
; A At
eBA
4m
eBA
2( A 4) 48
n 48
A
25
APP | Magnetic Effect of Current
127
Solution | Physics
Vidyamandir Classes
EMI & AC CIRCUIT
1.(B)
BVl L
di
0 ;
dt
At maximum current
di
0
dt
1
1
mV02 LI m2
2
2
Conservation of energy
I m V0
2.(A)
Answer follows from concept of motional emf and F q V B
3.(C)
Let ‘O’ be the instantaneous centre of rotation.
V 0
m
L
x V1
x l V2
V2 x l
V2 x V1x V1l
V1
x
V1l
V V V V V
;
1 2 1 2 1
V1l
l
V2 V1
1
2
Emf = B l x x 2
2
x
4.(D)
Fb BIL
Br
Induced current: I
2
/2
R
B r
B r3
Fb B
r
2R
2R
2
2
To maintain constant angular velocity: F(r) = FB r / 2 F
FB B 2 r 3
2
4R
4E
E L
L i i
R
R 4
5.(C)
Flux through a closed circuit containing an inductor does not change instantly
6.(B)
2
2i 2 a
The magnetic flux in inner loop due to current in outer loop is BA 0
a
i cos t
0 2b 0
4 b
e
7.(D)
8.(A)
d a 2 i0
0
dt
2b
d
dq
R
q
dt
dt
r
i
0 An
2r
b
9.(A)
sin t
Induced emf
q
0 nAi
R
2rR
b
0 I
vdx
2 x
a
Bvdx
a
Induced emf =
0 Iv b
E2
ln Power dissipated =
2
R
a
APP | EMI and AC Circuit
128
Solution | Physics
Vidyamandir Classes
E2
Also, power = F. V F
VR
10.(A) W = (L) F = L ×ILB = L
1 0 IV b
F
ln
VR 2
a
2
L2 B 2V
1J
R
11.(A) The growth of current is given by i i0 1 e t /
di i0 t /
e
dt
Energy stored in the form of magnetic field energy is: U B
Rate of increase of magnetic field energy is: R
This will be maximum when
dR
0
dt
1 2
Li
2
dU B
di
Li
dt
dt
e t / 1 / 2
Substituting: Rmax
12.(C) Resultant voltage = 200 volt
Since V1 and V3 are out of phase 180°, the resultant voltage is equal to V2
13.(C)
i1rms
Erms
X c2 R12
130
10 A ;
13
i2 rms
Erms
Li02
V2 = 200 volt
13 A
X L2 R22
Power dissipated = i12rms R1 i 2 2 rms R2 102 5 132 6 W = power delivered by battery = 500 + 169×6 W
14.(B) For constant speed I B cos 30 m g sin 30
Z
B V cos 30
B cos 30 mg sin 30 ind B v cos 30 (Side view)
R
B
30
V
2mg R
3v
B
30
I B
30
I
mg
2
15.(B) When K1 and K3 are closed charge on capacitor and current in inductor is given by
1
q c 1 et / RC 2 2 1 e t / 0.5 2 4 1 et = 4 1 at t 1 sec
e
Rt
2t
1
4
1 e L 1 e 2 2 1 e t = 2 1 at t 1 sec .
R
2
e
Now when K1, K3 are opened and K2 is closed then it is an L – C ckt and Let q0 be maximum charge.
From conservation of energy for L-C ckt
I
42 1
q02
1
e
q2
q02
1
L I 2
2C 2C 2
22
22
16.(A) I = 10 A, V = 1000 volt, VR = IR = 1000 V it, Vc = 200 V
2
1
1
2 22 1
2
C
2
1
q0 4 2 1
e
2
V 2 VL VC VR2
2
10002 VL 200 1000 2
APP | EMI and AC Circuit
V2 200 V
129
(Resonance condition)
Solution | Physics
Vidyamandir Classes
17.(C)
ind B V 2 1 2 4 volt.
I2
I1
4 3I1 2 I1 I 2 0
4V
4 6 I 2 2 I1 I 2 0
Solving I1
18.(C)
ind
2
3
A, I2
1
3
6
A
2 1
F I1 I 2 B 1 2 2 N
3 3
;
2
3
I1 + I2
is maximum when velocity is maximum.
L
max B d x Vmax =
x
B d x A sin K x
dx
0
L
2 L B A
2
cos K x
= BA .
K
, L
K
2
0
19.(C) Let it takes time Δt to switch off the magnetic field
20.(B)
B R 2
Δt
Included emf =
Force experienced
BRq
2t
Field produced
Δt
velocity gain
BRq
adt 2m
B R 2
BR
2 R
ΔT
2Δt
5gR
q 2 5g
m B R
24.(ABC)
XC
0
I
R
tan 2
I1 X C
1
1
I 2 I12 I 22 V12 2 2
R
X
C
V02 V12 V22 2V1V2 cos 90
I2
1
1
2
2
R
XC
I 2 X L2 2
I
IX L
1
1
2
2
R
XC
R
2
R X C2
R2 X 2
2 X L RX C R
V02 I 2 2 C 2 X L2
R X
R 2 X C2
C
R2
2
2
I2 2
X
2
X
X
X
C
L
C
C
2
R X C
X C2 2 X L X C 0
XC 2XL
1
2 L
C
21.(ACD)
2
1
2 LC
2
VR2 VC VL 130 VR 50V , VR2 VL2 100
22.(CD) i 5sin t 53
23.(ABCD) xL xC
VC
; VL IX L
I
25.(A) Dependence of current, whether it will increase or decrease with change in L or C depends on the frequency.
APP | EMI and AC Circuit
130
Solution | Physics
Vidyamandir Classes
26.(ABD) Rate of work done by external agent is:
dw BIL.dx
=
=BILv and thermal power dissipated in the resistor = eI = (BvL) I
dt
dt
27.(BD) Equivalent circuit is E B
a2
2
a2
2
2
B a 4
I
2 5r
E B
t
28.(AC) i
t
R
E
1 e L
R
29.(AD) i
30.(ABCD)
AB cos AB cos t ,
31.(AC) B x
0 I
2 x
2a
d Badx ,
I
20 xadx
a
a ln 2
M 0
2
R
E
1 e L
R
d
AB sin t
dt
0 Ia
ln 2 MI
2
Apply len’z law for direction of induced current.
32.(AC) Not required
33.(ABD)
i t 4 1 etl1 ,
PB Vi 32 1 e t
PR i 2 R
34.(AC) Len’z law
8
80V
0.1
dB
36.(BCD)For r R ,
E 2 r r 2
dt
dB
For r > R,
E 2 r R 2 .
dt
Also apply lenz’ law for direction of E
dq
di q
37.(ABC) i
4t
Va 1 4i Vd
dt
dt 2
1
38.(AC) Va Vb Vc Vb Bl 2
2
35.(BCD)
BA 8wb ,
av
39.(A) The equivalent diagram is:
The induced emf across the centre and any on the circumference is:
1
B r 2
e B l 2
2
2
APP | EMI and AC Circuit
131
Solution | Physics
Vidyamandir Classes
40.(AC) I upper
I lower
20
100 2
20
50 2
;
;
4
4
I I12 I 22
V100
20
100 2
ahead of voltage
behind voltage
1
10
0.3 A
100 10 2
41.(AD) In right ring emf induced in ABC part and CDA part will be same. Simplified diagram may be
So there will be current in each ring and so magnetic force on them will be non-zero.
42.(ABC)(A)
(B)
(C)
t 0
iL 0
V
V
2
t
2V
V
i
i2
3R
3R
After opening
v
iA
3R
VA
VL VA VR 2 R
VA
43.(AC) X C
V
3
V
2V
3R
3
V
3
2
XL
5
Z R2 ( X L X C )2 ,
R
0.8
Z
Now solving get answer.
44.(AC) r
1
50 ,
LC
Band width
f r 25 Hz
R
R
r
50 1
250 ; Q
1
L
band width 250 5
25
(2 1 ) (2 1 )2 42r
APP | EMI and AC Circuit
;
;
(250 )2 4(2500 2 ) ;
132
2 1 250 and 12 2r
72500 2 10 725
Solution | Physics
Vidyamandir Classes
45.(CD) iR
Ldi Mdi Ldi Mdi
0
df
dt
dt
dt
iR 0
i
R
2
R
R
46.(CD) If source AC or DC, LC oscillations will take place. So charge on capacitor will not be constant at any time.
Time constant
Leff
0 & Power delivered by battery is constant P
rR
47.(ABCD)
E.2r
d
0nCt r 2
dt
rR
d
0 nCt R 2
dt
The line charge will produce radially electric field which is perpendicular to induced electric field
48.(ABCD)
At t 0 capacitor behave like conductor & inductor behave like insulator. At steady state capacitor behave like
insulator & inductor behave like conductor
49.(C) As the loop enters the region it will experience a magnetic force in a direction opposite to gravity but still
gravitational force may be greater than or less than the magnetic force hence speed may increase or decrease. Once
it gets completely inside the magnetic field then speed will increase after that instant.
E.2r
q
di
L 0
C
dt
2
d q
q
d 2q
d2x k
2
0
q
0
compare
this
equation
with
x0
dt 2 m
dt 2 LC
dt 2
50.(ABC) For given situation
q x, i v, L m, C
1
k
Solving equation
q q0 cos t & i q0 sin t
According to given conditions
q 2 1 2 q0 cos 2 t 1 2 2
cot 2 t 1
Li
Lq0 sin t
2C 2
2C
2
3 5 7
3
5
7
t , , , ........
t
,
,
,
......
4 4 4 4
4 LC 4 LC 4 LC 4 LC
51.(B)
52.(C)
53.(B)
E 2
R R2 dB
2
4 dt
dB
6t 2 24
dt
dB
6t 2 24 ;
dt
APP | EMI and AC Circuit
F qE
E 2 r r 2 .
dB r dB
dt 2 dt
dB
30 T / s
dt t 1s
Apply Lenz’s law for direction of E
133
Solution | Physics
Vidyamandir Classes
54.(D) 55.(C) 56.(A)
When connected with the DC source R
When connected to ac source I
Using I rmsVrms cos
12
3
4
V
Z
2
Vrms
cos
Z
12
2.4
2
3 2 L2
2
Vrms
R
1
R L
C
2
L 0.08 H
24W
2
57.
[A- q, s; B-p, r, s; C -p, r, s; D-q, s ]
(A)
Due to current carrying wire, the magnetic field in lop will be inwards w.r.t. the paper. As current is
increased, magnetic flux associated with loop increase. So a current will be induced so as to decrease
magnetic flux inside the loop.
(B)
option in (B) will be opposite of that in (A)
(C)
when the loop is moved away from wire, magnetic flux decreases in the loop. Hence the options for this
case shall be same as in (B)
(D)
when the loop is moved towards the wire, magnetic flux increased in the loop.
58.
59.
60.
[A - p; B - r; C - s; D - s ]
A - s; B - p; C - s; D - s
A - s; B - q; C - p; D - p
61.(6)
62.(3)
Use maxima-minima
When current is maximum, vL 0
63.(2)
64.(5)
65.(0)
66.(5)
67.(8)
68.(3)
69.(8)
XL
X L 100 3 X C
R
V
20
I= R
5A
R
4
tR
t
V
V RC
L
I I L I C ; I L 1 e ; I C e
R
R
ind BV 1
di
10e 4t 4 40e 4t
dt
di
VL L
dt
BlV
i
F ilB ,
R
tan 60
B 0
E
70.(5)
d
dt
cos
t C/K
i
E
R
B. area C kt a 2
1 d
R dt
q idt
1
, 45 , tan 1 L R
2
APP | EMI and AC Circuit
ZR
1 d
. dt
R dt
;
134
2
2 2
d
0 C a 2 C a 2
= 8 coulombs
R
R
R
R
3
rad/sec = 500 rad/sec.
L 6 10 3
Solution | Physics
Vidyamandir Classes
71.(4)
e Bv
I
Bv
R
F BI
v
dv B 2 2v
ds
mR
0
B2 2v
R
B 2 2
mR
dv
v0
s
ds
0
s 4m
8
3 5 12 Volt
10
q CE (1 et / )
72.(2)
e B v
73.(5)
E t / 12 2
e
2
R
4 3
Total magnetic flux through a super conductor is equal to zero.
i
self ext 0
24 6 12(1 e t / ) et /
2
3
self ext
Li B0 r 2
B0 r 2
L
AC ammeter shows rms current
So, when both currents are flown simultaneously, AC ammeter gives
i
74.(4)
I AC 62 82 10 & DC ammeter gives,
I DC 6
Difference in readings 10 6 4
75.(4)
di
vB
dt
Amperes force law opposes velocity
L
iB m
dv
dt
Figure
di
m d 2v
dt
B dt 2
So
d 2v
dt 2
B 2 2
B
v 0 i.e.,
mL
mL
APP | EMI and AC Circuit
135
Solution | Physics
Vidyamandir Classes
76.(3)
I
5
R 1
R 4
20
2
2
5 (20 L)
77.(4)
L
2
3
H
4
P V1i1 V2i2
10 103 25 V2
V2
104
25
Now
78.(5)
I rms
V1
n1
8 104
V2
V
n2
1 25
Erms
Z
Ev
1
R L
C
2
0.2 amp
2
H I 2 Rt 2 10
t
20
(0.2)2 100
or
(0.2)2 100 t 20
5sec
79.(2)
When velocity is maximum then net force acting on the conductor is zero.
80.(2)
zRL R 2 2 L2
1
zRLC R 2 L
C
cos RL
2
R
2
R 2 L2
According to question
1
(L )2 L
C
APP | EMI and AC Circuit
2
C
1
22 L
136
Solution | Physics
Vidyamandir Classes
RAY OPTICS AND WAVE OPTICS
1.(B)
3
2
sin c
c 60
cos 60
R
Rd
d = R.
2.(A)
Coordinate of image formed by lens f
(f, 0)
For lens 2 :
u 2 f , h0 d
1 1 1
v u f
v2f
v 2f
m
1
u 2 f
hi d
Coordinate of image 3 f 2 f , d d
3.(C)
For
y 0 For
x3
1 1 1
v u f
1 1 1
x 3 15
5
x
2
x0
y 1
5
2
Equation of line passing through (0, 1) & , 0
y 0.4 x 1
Fringe-width
4D
d
4.(D)
The effective distance of the screen = 2D + 2D = 4D
5.(C)
Iˆ Nˆ rˆ nˆ
6.(B)
The necessary and sufficient condition for all the rays to pass around the arc is that the ray with least angle of
incidence should get internally reflected.
From the figure, it becomes obvious that the ray with least angle of incidence is the
one which is incident almost grazingly with the inner wall.
For this ray, if ‘‘ be the angle of incidence
sin =
Rd
where d is the diameter of tube .
R
APP | Ray Optics and Wave Optics
R
137
Solution | Physics
Vidyamandir Classes
But
C
sin sinC
d
1
d
1 R
R
1
Rd 1
R
4 3/ 2
R
1/ 2
d 1
R
1-
R 12 cm
7.(B)
Hence the least radius required is 12 cm.
There is no change of phase in the transmitted ray due to difference in refractive index. In the reflected ray, phase
change occurs when a ray is reflected from an optically denser medium.
8.(C)
d 2n 1 D
2
2d
9.(D)
AB 2 cm
n 0, 1, 2,.....
From refraction at air-water boundary,
1
2
1
10.(C)
11.(B)
2 sin or 30
3 cos 2
cos
sin c
2
1
2
3
2
1
y
1 cos 2 C 2
1
1 nˆ pˆ
3
2
cm.
( is the angular position of the point)
2
BC
5
D
2
2
2
2
2
2 3
4 2
1
5
25 1
1
;
2
3 3
5
12.(C) For required condition the light ray should be parallel to principal axis after refraction through curved surface of
lens.
3/ 2
1
3/ 2 1
x = 40 cm
x
20
13.(B) For min , i should be maximum and maximum possible value of i 90 .
Hence if for i 90 , min C then light ray does not cross the curved surface for any value of i.
Hence from Snell’s Law.
1 sini sin 90
sin 90 sin 90
cos 1
1
1 sin 2 C
i
APP | Ray Optics and Wave Optics
1
cos
1
1
cos C sin ce C
1
2
138
1
2
1
1
2
2 min 2
Solution | Physics
Vidyamandir Classes
14.(A) Optical path difference = S2 O n2 S1O t n2 tn3
t n2 n3 or t n2 n2 which ever is positive
Wave length in vacuum = n11
15.(C)
I I max cos 2
16.(A)
phase diff. =
2 optical path diff.
Wavelength in vaccum
=
2 t n2 n3
n11
x
I max I max cos 2
x
x
1
cos
2
x
D
. . . . . . for smallest x x
3
3 3d
9
r d sin s 1 t mm n
16
18.(BC)
17.(AC)
D
d
19.(ABCD)
20.(BD)
There is a dark fringe at O if the path different ABO AO ' O
2
D
2d 2 d 2
d min
2D
D 2
2
The bright fringe is formed at P if the path different ' AO ' P ABP
2 D2 d 2 2D
2
D D2 x 2 D 2 d 2 D2 x d
Given, d d min
Solving, x d min
x 2 d 2 2 xd
x2 d 2
2 D 2D
2D
D
2
21.(ABCD)x at O = d [path difference is maximum at O]
So, if d
7
, O will be minima
2
d , O will be maxima
d
5
, O will be minima and hence intensity is minimum.
2
If d = 4.8, then total 10 minimas can be observed on screen, 5 above O and 5 below O, which correspond to
3
5
7
9
x , , ,
,
2
2
2
2
2
APP | Ray Optics and Wave Optics
139
Solution | Physics
Vidyamandir Classes
22.(AD) If the amplitude due to two individual sources at point P is A0 and 3A0 then the resultant amplitude at P, will be :
2
A A02 3 A0 2 A0 3 A0 cos
13 A0
3
Resultant intensity, I 13 A02
cos i
cos i
t1 sin i 1
t 2 sin i
23.(BC) Shift d 1
2
2
2
2
n
sin
i
n
sin
i
1
2
3
4
i 37n1 t1 4.5 cm n2 , t2 2 cm
2
3
0.8
d 1
2
3
2
0.6
2
9
0.8
0.6 1
2
2
4
2
0.6
3
2 0.6 1.129 0.39 1.5 cm d d1 d 2
24.(AC) From the geometry of prism : 1 = 60, r = 30
5
4
5 1 4
sin r sin 2 sin 2
3
3
3 2 3
5
5
sin 2 2 sin 1 .
8
8
Then apply Snell’s law :
Total internal reflection at the point P is only possible if P > m
25.(BCD)
Apply Snell’s law : 2 sin i 1 sin r sin i k sin r
From the given graph, angle of deviation decreases and becomes zero at k k 2
Hence, 1 r i
(By geometry)
6
at k = k2, r i 0 means, k2 = 1.
when k , r 0, by the Snell’s law, 2 r i i
k1 = must be less than k2 from the given graph.
3
26.(BD) There are two possibilities 2 = 90 (According to the question) = 45
dy
dy
Slope of the tangent = tan
tan 45 1
dx
dx
Now,
dy 2 L
x
cos (From the given equation)
dx L
L
dy
x
x 1
dy
2cos
cos Since,
1
dx
L
L 2
dx
APP | Ray Optics and Wave Optics
140
x 2
,
L 3 3
x
L 2L
,
3 3
Solution | Physics
Vidyamandir Classes
From the displacement method : h0 h1 h2 9 4 6 cm. Hence, option (C) is correct
27.(BCD)
h1 9 3
h0 6 2
m1
;
m2
h2 4 2
h0 6 3
From the displacement method :
m1 m2
f
d
f
3 2 d
2 3 f
D 2 d 2 902 d 2
4D
4 90
… (i)
… (ii)
On solving equation (i) and (ii) we get, d 18 cm, f 21.6 cm
Hence, option (D) is correct.
For the position of object :
x2 x1 d 8, x2 x1 D 90
x1 36 cm, x2 54cm . Here x1 and x2 be the position of object for two positions of the lens.
Hence, option (B) is correct.
D = 96 cm
m1
4 , also m1 m2 = 1
m2
28.(ABCD)
Also
V1
u1
m1 2, m2
1
2
m1 V1 2u1
Also V1 + u1 = 96
3u1 = 96
u1 = 32
V1 = 64
u2 = 64, V2 = 32
Distance between two positions = u2 u1 32 = L
Focal length =
D 2 L2
4D
962 322
4 96
= 64/3 cm
For shorter image, V = V2 = 32 cm
29.(AC) sin 1
n
n1
then will also greater then sin 1 3 if n3 n1
n2
n2
If n3 n1 then
C
D
r
A
B
At AB, n3 sin r n2 sin
sin r
n2
. sin
n3
n2
n n
. sin sin1 1 1
n3
n2 n3
n
r sin 1 1 TIR at CD
n3
1 1.4 1
1
1
. . for biconcave f = 24 cm
f 1
16
24
1
1
1
For concavo convex converging,
1.4 1 f 120 cm
f
16 24
30.(BCD)
APP | Ray Optics and Wave Optics
141
Solution | Physics
Vidyamandir Classes
Power =
1
5
D
1. 2 6
1 1. 4 1
1
1
f 76.8 cm
f 1.6 16 24
1
1
1
1
3 1
3/ 2 1
31.(AC)
1
1
;
f air 2 R1 R2
f water 4 / 3 R1 R2
From these two equations we get,
f water 4 f air 4 f
In medium of r. i 1.6,
In air object was inverted, real and magnified. Therefore, object was lying between f and 2f. Now the focal length
has changed two 4 f. Therefore, the object now lies between pole and focus. Hence, the new image will be virtual
and magnified.
32.(AC) The intensity of light is I () I 0 cos 2
2
2
2
Where
( x ) ( d sin )
(i)
For 30
c 3 108
300 m and d = 150 m
v
106
2
1
(150)
2 4
300
2 2
I
(Option A)
I () I 0 cos 2 0
4 2
(ii)
For 90
2
Or
and I () 0
(150)(1)
300
2 2
(iii)
For 0, 0 or 0
2
(Option C)
I () I 0
33.(AD) Final image is formed at infinity if the combined focal length of the two lenses (in contact) becomes 30 cm or
1
1 1
30 20 f
f 60 cm
i.e., when another concave lens of focal length 60 cm is kept in contact with the first lens.
Similarly, let be the refractive index of a liquid in which focal length of the given lens becomes 30 cm. Then
1 3 1
1
1
20 2 R1 R2
1 3/ 2 1
1
1
30
R1 R2
From equations (1) and (2), we get
9
8
APP | Ray Optics and Wave Optics
……(1)
……(2)
142
Solution | Physics
Vidyamandir Classes
34.(BD) 21t n
n
640 3
n
240 n
21
2 4
t 240 nm, 480 nm,....
t
35.(BD) The condition for getting maxima is d sin m. The wavelength of electron will be given as h / mv.
The distance between successive maxima will increase if becomes smaller. For this either d should be increased
or should be decreased. For the latter we must increase the voltage V.
36.(B) Focal length of plano-convex lens
f 20cm
1
2
feq
f
f eq
20
10 cm
2
37.(B) As mass of lens = mass of particle
m
kx0
x
2
2 0 .2
k
gk
g
Time period T 2
T
, lens will come to same position (mean).
2
T
Distance travelled by particle in time
2
1
1
2
u gt 2 10 0.1 0.05m
2
2
Velocity of particle v p g t 10 1 1 m/s
At time t
u = 5 cm
Velocity of lens v = 10 m/s
1 1 1
v u f
1
1
1
v s 10
Location of image
v 10 cm
2
v
Velocity of image : vi / l vo / l
u
2
10
vi / l 10 1
5
38.(C)
39-41. 39.(B) 40.(C) 41.(B)
sinc
1
1
3
r1 30 tan c
cos c
30
2
Ceiling
r2
2
300cm
3
cm = 15 2 cm
APP | Ray Optics and Wave Optics
vi / l 36 m/s
30cm
2
c
143
r
1
r1
c
Water surface
c
Solution | Physics
Vidyamandir Classes
r 10 3 r1
(Hence shadow of ring will be formed on roof)
Radius of shadow = r2 r 300 tan 2 = 10 3 300 tan 2
10 3
1
Now from shell’s Law 3 sin 1 sin 2 tan 1
30
3
Hence radius of shadow = 10 3 300 3 310 3 cm
3
1
2
sin 2
tan 2 3
rmax r1 for which light will come out of liquid surface and shadow of ring will be formed on ceiling
42.
[A-p, q] [B-q, r, s] [C-p, q, r] [D-q, r, s]
44.
[A-q, t] [B-p, r, t] [C-p, q, s, t ] [D-p, r]
45.
[A-p, r] [B-q]
43.
[A-p, t] [B-r, t] [C-s, t ] [D-s, t]
[C-p, s] [D-p, r]
At centre intensity will be maximum for both wavelengths.
d
n n 0, 1, 2 .......
D
n D
= 0.2 mm, 0.4 mm,.......for 4000 A = 0.4 mm, 0.8 mm,.......for 8000 A
d
For maxima:
For minima: 0.1mm, 0.3 mm,.......for 4000 A = 0.2 mm, 0.6 mm,.......for 8000 A
46.(2)
Ray diagram is shown
AOB 2r
1
AOB why?
2
1
AMB 2r r
2
2
In AMB
; ir rir
2
sin i
Now
sin i 3 sin 2i
sin r
2
sin i
sin i
cos 2i
2 3 sin 2 i sin i 3 0
sin r
Now, AMB
sin i
3
1
,
.
2
3
Rejecting the negative sign
We get i = 60°.
47.(4)
48.(4)
1
1
w
… (i)
fair
50 u
1
1
1
and
fw
40 w u
4
For plane surface up
3
… (ii)
f w 4 fair
For curved surface up
APP | Ray Optics and Wave Optics
144
Solution | Physics
Vidyamandir Classes
1 1
v u
R
49.(8)
sin c
v
;
25
, u 4
8
;
R 25 cm =
1
m.
4
2
3
At face AC, i is 60°
i c
50.(4)
51.(16) In one case image is virtual (u = -10cm)
In another case image is real (u = -40cm)
v1
uf
10 f
u f 10 f
;
v2
40 f
40 f
In both situations, sign convention is opposite.
v2 v1
v2
10 f
40 f
10 f 40 f
f 16cm
;
52.(10) hI1 hI 2 h02
4hI21 h02
h1 1
h0
2
1 (45 x )
2
x
x 90 2 x
x 30cm
1 1 1
30 15 f
53.(30) m1m2
f 10cm
f
f'
f f'
xf
f x
d f ' xf d ( f x) f '( f x)
x f
xf dx xf ' 0
54.(8)
55.(1)
;
f f ' d 30cm
1
1
1
1
1
1
( rel 1)
and
2
f
f meq
fm'
f
R1 R2
From the information given, f1st lens 30cm
From lens formula (for second lens)
1 1 1
1
1
1
v u f
22 x 30 x 30
APP | Ray Optics and Wave Optics
145
Solution | Physics
Vidyamandir Classes
56.(3)
Given d 4 3cm sin 30
17
8
sin sin
16
17
AB d sec 30 8 cm
4
8 4
t
15 t
t 7.5cm 75 mm
tan
57.(8)
1.5 102
n
nD n 750 10 9 1
d
0.4 10 3
6 10 6
7.5 10
7
60
8 minima
7.5
I
58.(209) I I 0 cos2 , here I 0
2
4
2
1
cos
2 2
2 3
3
L
2
627
2
L
nm
3
3
59.(4)
(1 1)t1 10
60.(3)
I 4I 0 cos2
;
L 209 nm
3
now (1 1)t1 (2 1)t2
2
t2 4 m
2
Case 1 : 0 I 4 I 0
Case 2 : I
Now,
3I
4 I 0 cos 2
4
2
cos
3
;
2
2
2 6
3
( 1)t 2 ( 1) t 2
6000
;
;t
;t
2000 Å
3
6( 1)
3
APP | Ray Optics and Wave Optics
146
Solution | Physics
Vidyamandir Classes
MODERN PHYSICS
1.(C)
2.(A)
3.(A)
4.(A)
5.(A)
Energy released = (80 × 7 + 120 × 8 – 200 × 6.5) = 220 MeV
1
1
1 1
27.2 13.6 Z 2
ΔE 13.6Z 2 2 2
4 36
n1 n2
27.2
4
8
By dividing we get,
12
x 5.95eV
4.25 x 18
3
1 1
4.25 x 13.6 Z 2
9 36
Maximum energy is liberated when transition is from n 5 to n 1 and minimum energy is liberated when transition
is from n 5 to n 4.
E1
E1 52.224 E1 ( )54.4 eV
52
1.224eV
Q ne
It
I
or,
n
t
t
e
3
(3.2 10 ) (1)
2 1016
19
(1.6 10 )
P h / λ and K
and
E1 E1
9
9
2
E1
54.4
2
400
400
5
4
P2
h2
2m 2mλ2
For X-ray photons, it is also maximum energy
So,
6.(A)
7.(B)
hc
h2
λ 0 2mλ2
or,
1
2 1
13.6 3
w 4
2
42
3
1
2 1
13.6 3
wV
2
52
4
Solving V = 0.85 eV.
dN
At time t;
t2 N
dt
d2 N
dN
2t
2
dt
dt
8.(D)
1
13.6 1 13.6 z 2
9
9.(A)
No. of neutrons
3.2 10 11
2mλ2 c
h
(w = work function)
d2N
2 0
d t
1
1
2
9
x
100 106
λ0
0 2t0 t02 N0
n2 9
z2
8
n2
N0
t02 2t0
2
z 3 and n 9
2.5 7.8 1018
10.(A) Let the radius of the n th Bohr orbit be r and let the velocity of the electron in this orbit be v
APP | Modern Physics
147
Solution | Physics
Vidyamandir Classes
Angular momentum of the electron, L mvr
nh
2π
Also,
mv 2 KZe2
2
r
r
Solving, we get
v
2πKZe 2
nh
h
nh 2
mv 2πKZme 2
For the first excited state in the Hydrogen atom, Z 1 and n 2 ,
So, de Broglie wavelength of the electron,
λn
11.(BC) K 20.4 eV
λ
no excitation of hydrogen atom
h2
πKZme 2
collision will be elastic
12.(AB) im intensity, eVs hυ o
13.(ABC)
K max E W
TA 4.25 WA
Given TB TA 1.50 4.70 WB
On solving, WB WA 1.95 eV
λ
1
h
as λ
K
2 Km
λB
KA
λA
KB
1/ 2
TA
2
TA 1.5
On solving, TA 2eV
WA 4.25 TA 2.25 eV
WB WA 1.95 4.20 eV
TB 4.70 WB 0.50 eV
14.(AB) | F |
dU Ke2
…… (1) ;
dr
r4
Ke2
r
4
mv 2
r
…… (2) ;
mvr =
nh
2
…… (3)
Solving (2) and (3) : T.E = KE + PE
Total energy n6
–3
Total energy m .
15.(ACD)
P.E. 2 K .E. and
APP | Modern Physics
rn
n2
2
148
Solution | Physics
Vidyamandir Classes
Two or more lighter nuclei combine to form a heavier nucleus in fusion reaction.
16.(ABC)
17.(BC)
18.(ACD) c
a
E1 5RH / 36 5
E2 3RH / 4 27
λ1 1
1
,b
λ2 c
a
19.(ACD)
20.(ABCD)
Under normal conditions total energy, potential energy and kinetic energy in ground state and first excited
state are –13.6 eV,–27.2 eV, 13.6 eV, –3.4 eV, –6.8 eV and 3.4 eV respectively. If potential energy in ground state is
taken to be zero, then kinetic energy will remain unchanged but potential and total energies are increased by 27.2 eV.
Therefore, the new values are 13.6 eV, 0, 13.6 eV, 23.8 eV and 3.4 eV respectively.
21.(AD) Two or more lighter nuclei are combined to form a relatively heavy nucleus to release the energy
1
22.(AC) R R0 A 3
16
For O , R R0
1
(16) 3
1
128 3
For 54 X , R ' R0 (128)
;
;
R'
R 2R
16
23.(BC) (1) Due to emission of particles mass will almost remain unchanged.
128
1/ 3
V'
4 3
R 8V
3
(2) No. of particles decayed 3 10 22 , so charge 3 10 22 1.6 10 19 4800 C .
24.(ABCD)
(A)
(B)
(C)
(D)
Maximum potential will be equal to the stopping potential which depends on and nature of
material.
KQ
RV
Q
R
K
Since V and K are constant, maximum positive charge appearing depends on R.
As the sphere gets charged (which goes on increasing), it applies a force on the emanating
electrons thus reduces the velocity of emanating electrons.
Initially the sphere in uncharged, thus KEmax of emanating electron is independent of radius of
V
sphere.
25.(BC) Energy of incident photons, E
hc 1242
6 eV
λ
207
Cut-off potential is given, VC 4 V
Therefore, kinetic energy of the fasting moving photoelectron, K max eVC 4 eV
Work function of plate A, 0 E K max 2 eV
Therefore, longest wavelength of light that can cause emission from plate A,
hc 1242
m
621 nm
0
2
Number of photons striking plate A per second,
APP | Modern Physics
149
Solution | Physics
Vidyamandir Classes
N ph
4
sin 30
Energy received at the plate per second 60 20 10
1
Energy per photon
100e
6 e
Number of electrons emitted from plate A per second,
Photoelectric current 5 10 4
e
e
N ph
Therefore one electron is emitted per
photons, i.e., 20 photons.
Ne
Ne
26.(ACD)
K
1 2
mv ,
2
r
mv
,
eB
h
mv
Radius of the orbit is proportional to n2
1
Ionization energy 13.6 2 0.85 eV
4
27.(ABC)
λ
n4
hc
1242 16
97.4 nm
1 13.6 15
1
13.6 2 2
4
1
Energy of photon h 6.6 1034
6.78 1027 kg m/s
Speed of light
λ 97.4 109
As the electron can at the most lose as much energy as it gained in the first transition, it can only emit a photon of
wavelength higher than 97.4 nm.
pa Momentum of photon =
28.(BC)
(1)
is a β - decay process, so the mass that converts to energy is just the mass of the parent nucleus minus the mass
of the daughter nucleus.
(2)
is a β - decay process, so the mass that converts to energy is the mass of the parent nucleus minus the mass of
the daughter nucleus plus the mass of two electrons.
29.(AD)
30.(ACD)
1
hc
1
13.6
n 1 2 n2
λL
p L = Momentum of emitted photon =
1
hc
13.6
n 1 2
λM
λ L is proportional to
n n 1
2
2n 1
h
2n 1
p L is proportional to
2
λL
n n 1
λ M is proportional to n 1
… (1)
and
2
31.(B) Ground state energy (in eV) is E1.
Given condition, E2 n E1 204 eV
1
E1 2 1 204 eV
4
n
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150
E2n En 40.8 eV
Solution | Physics
Vidyamandir Classes
1
3E
1
E1 2 2 21 40.8 eV
n
4n
4n
By dividing equation (1) and (2), we get,
… (2)
1
1 4n 2
5
3
4n2
On solving, we get, n 2
32.(C)
4
E1 n 2 (40.8)eV 217.6eV
3
33.(B)
E1 (13.6) Z 2 217.6 eV
Emin E2n E2 n1
or,
(Put n 2)
Z 4
E
E1
2
4n
(2n 1) 2
(Put n 2)
E1 E1 7 E1
10.58 eV
16 9
144
34.(B) Decay constant for the decay of A into X, λ1
log e 2
T1
Decay constant for the decay of A into Y, λ 2
log e 2
T2
dN
λ1 N λ 2 N λ1 λ 2 N
dt
This means that the effective decay constant when both decay processes are going on simultaneously is λ = λ1 λ 2
If the instantaneous number of nuclei A is N, then
So, effective half-life, T
log e 2
TT
1 2
λ
T1 T2
35.(A) Let the instantaneous number of nuclei A, X and Y present be N A , N X and N Y .
Then,
dN X
λ1 N A
dt
dNY
λ2 N A
dt
and
Dividing the two equations, we get
dN X
λ
1
dNY
λ2
λ 2 dN X λ1dNY
NX
λ
1
NY
λ2
N X T2
NY
T1
36.(B) Maximum energy = Binding energy of products – Binding energy of reactants
Emax 41.5 40 8.5 40 1320 keV
37.(D) Let the velocity of the Ca-40 nucleus and the beta particle (electron) after the disintegration be v1 and v2 respectively
Conserving momentum,
APP | Modern Physics
1
v2 0
1800
40 v1
151
v1
v2
72000
Solution | Physics
Vidyamandir Classes
1 1 2
v2
2 1800
2
So, ratio of kinetic energies,
72000 5.18 109
1
KCa
40 v12
2
38.(A) The energy liberated in the decay of a K-40 nucleus into a Ca-40 nucleus is,
Kβ
Q m K-40 m Ca-40 931.5 MeV
Here m K-40 and m Ca-40 denote the mass of a K-40 atom and a Ca-40 atom respectively.
m Ca-40 m K-40
Therefore,
39.
[A – p ; B – r; C – r ; D – r]
40.
[A – s ; B – r; C – p; D – q]
41.(1)
r
Q
39.9640 0.0014 39.9626 u
931.5
P
Since, r
Bq
P
q
1
re
2
Pα 1 Pe
2 2 1
Given rα
42.(6)
Pα Pe
h
Since, λ
p
or,
n 1
1
P
λα λe
λ
So,
or,
Given λ A N A λ B N B
λ t
ln 2
ln 2 λ Bt
)
(4 N 0e A ) ( N 0 )
(e
TA
TB
e(λ A λ B )t 8
(λ A λ B )t ln8 3(ln 2)
ln 2 ln 2
t 3ln(2)
2
1
43.(7)
x and y are number of α- decays and β- decays respectively
92 2 x y 85
or,
Similarly,
44.(4)
2x y 7
238 4 x 210
x7
… (1)
… (2)
1
Ephoton 13.6 1 eV 13.0 eV
25
(Momentum conserved)
E / c mv
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152
Solution | Physics
Vidyamandir Classes
E
(13)(1.6 1019 )
4 m/sec.
mc (1.67)(1027 )(3)(108 )
v
45.(6)
λ1 ( Z 2 1) 2
λ 2 ( Z1 1) 2
[Since,
1
( Z 1) 2 ]
λ
1 ( Z 2 1) 2
4 (11 1) 2
On solving Z 2 6
46.(6)
When electron jumps from nth state to ground state, number of possible emission lines
Here, number of possible emission lines
n( n 1)
.
2
( n 1)(n 2)
10 (given)
2
On solving, n 6
47.(8)
a v2 / r
Z2
(1/ Z )
a
So,
Thus, a Z 3
3
3
a1 Z1 2
8
a2 Z 2 1
48.(2)
The shortest wavelength of Brackett series is corresponding to transition of electron between n1 4 and n2 .
Similarly, the shortest wavelength of Balmer series is corresponding to transition of electron between n1 2 and
n2 .
13.6 13.6
16 4
Thus, we have ( Z 2 )
49.(8)
50.(8)
dN
dt
n2
3
n 2
6
n2
2
n 2
hr
1
N N0 e
2
.6
Z 2
N 0 2 3
N0
N
8
N N 0e t
n
dN
dt
n N 0 t
From graph equation is : n
or,
n
dN
dt
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1
2
dN
dt
t6
. . . . (i)
4
4 3
6 4
t 4
. . . . (ii)
153
Solution | Physics
Vidyamandir Classes
Comparing (i) and (ii)
1
and n N 0 2
2
N
51.(9)
N0
P
P
N0
N
N0
N 0e
t
e
t
1
e 2
4.16
e2.08 8 (using log table)
Let the velocity of the alpha particle before the collision be u1
Let the velocity of the alpha particle and the deuterium nucleus after the collision be v1 and v2 respectively.
52.(4)
53.(1)
Conserving linear momentum,
4u1 4v1 2v2
Newton’s experimental law,
v2 v1 u1
u1
and
3
Solving, we get
v1
So,
1
4 u1 2
K
2
9
2
K' 1
u1
4
2
3
1
1
13.6 2 2
n
1
12.7
v2
4u1
3
1
1
n
2
1
9
2
n
136
n2
n2 15.11
n4
Force
54.(1.18)
127
136
136
9
2 Power received cos i 2 3 0.2 cos i cos i 109
3 108
Speed of light
Wavelength of K-alpha line is proportional to
N
1
Z 12
2
Cr 25
1.18
Fe 23
Therefore,
55.(25.69)
56.(7)
hc
1242 9
25.69 nm
1 1 13.6 4 8
13.6 4 2 2
1 3
(Mass + energy) of the system will remain conserved. Thus (5 + mass energy of A) + (3 + mass energy of B) = (KE of C
+ mass energy of C + excitation energy) [5 3 (35 34.99) 930 8.3]MeV KE of C.
(17.3 – 10.3) MeV = KE of C.
57.(159) Radiation T 4
So T2 2T1 and by Wein’s displacement law
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1
T
154
Solution | Physics
Vidyamandir Classes
So 2
f
1
hc
3000Å; by Einstein’s photoelectric equation
eVs
2
hc
hc
1
1
eVs
(13.6eV )12 2 2 4.14 2.55
3000 Å
2
4
1.59eV 1.59
or 159
100
58.(255) Energy of the photon
hc 1240
51 10.2 eV
6200
Since six spectral lines are obtained, thus transition is from ground state to n = 4.
Also since the atom is not hydrogen, thus only possible atom is He .
[Photon of
6200
nm corresponds to transition from 4 to 2]
51
Thus E in collision = 51eV
For minimum kinetic energy of neutron, collision must be perfectly inelastic.
1 2
vrel 51eV
2
14 2
mv 51eV
25
1 2
mv 63.75 eV
2
[m = mass of neutron, v = velocity of neutron]
59.(4)
The given wave is superposition of 3 waves with frequency, 0, 0 and 0 ; max (0 )
Emax hvmax h
(0 )
2
hvmax KEmax 4eV
60.(160) P 700 103 1.6 1019
10 10 3 ;
dN
dt
dN 102
1
1012
N 0
dt 10 14 7 16 11.2
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;
155
ln 2
14 86400 1012
N0
160 1015
14 86400
11.2 ln 2
Solution | Physics
