tạp chí physics for you tháng 1 năm 2016

Volume 24
Managing Editor
Mahabir Singh
Editor
Anil Ahlawat
(BE, MBA)

No. 1

January 2016

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CONTENTS

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Physics Musing Problem Set 30

8

Core Concept

12

Thought Provoking Problems

22

PMT Practice Paper

25

JEE Accelerated Learning Series
Brain Map

31
46

Ace Your Way CBSE XI

57

JEE Workouts

64

Ace Your Way CBSE XII

68

Exam Prep 2016

75

Physics Musing Solution Set 29

81

Live Physics
You Ask We Answer

83
84

Crossword

85

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Physics for you | january ‘16

7


P

PHYSICS

MUSING

hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment
the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material.
In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed
solutions of these problems will be published in next issue of Physics For You.
The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those who
send atleast five correct solutions will be published in the next issue.
We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.

30
single oPtion correct tyPe

1. Two batteries of emf e1 and e2 having internal
resistance r1 and r2 respectively are connected in
series to an external resistance R. Both the batteries
are getting discharged. The above described
combination of these two batteries has to produce a
weaker current than when any one of the batteries is
connected to the same resistor. For this requirement
to be fulfilled
e2
r
r
(a) e must not lie between 2 and 1
r2 + R
1
r1 + R
r
r +R
e
(b) 2 must not lie between 2 and 2
r1 + R
r1
e1
r
e
r
(c) 2 must lie between 2 and 1
r2 + R
e1
r1 + R
e2

r
+R
r
(d)
must lie between 2 and 2
.
e1
r1
r1 + R
2. A particle of charge per unit mass a is released from

origin with velocity v = v0i in a magnetic field

3 v0
B = − B0 
k for x ≤
2 B0a

3 v0
and B = 0 for x >
2 B0a
π 

The x-coordinate of the particle at time t  >
 3B0a 
would be

8

(a)

π 
3 v0
3 
+
v t−
2 B0a 2 0  B0a 

(b)

π 
3 v0

+v t −
2 B0a 0  3B0a 

(c)

3 v0 v0 
π 
+ t −
2 B0a 2  3B0a 

(d)

3 v0 v0t
+
2 B0a 2
Physics for you | January ‘16

3. An ideal choke takes a current of 10 A when
connected to an ac supply of 125 V and 50 Hz. A
pure resistor under the same conditions takes a
current of 12.5 A. If the two are connected to an
ac supply of 100 V and 40 Hz, then the current in
series combination of above resistor and inductor
is
(a) 10/ 2 A
(c) 10 2 A

(b) 12.5 A
(d) 10 A

4. When an athlete runs with some acceleration, he
leans forward. The line joining his centre of mass
to his foot which is in contact with the ground
makes an angle q with the vertical. The coefficient
of friction between his foot and the ground is m. His
foot does not slip on the ground. His acceleration
is
(a) mg
(b) mgtanq
(c) gtanq
(d) (gtanq)/m
5. In a regular polygon of n sides, each corner is at
a distance r from the centre. Identical charges
are placed at (n – 1) corners. At the centre, the
magnitude of intensity is E and the potential is V.
The ratio V/E is
(a) rn

(b) r(n – 1)
(c) (n – 1)/r
(d) r(n – 1)/n
Solution Senders of Physics Musing
1.
2.
3.
4.

set-29
Amatra Sen (WB)
Manmohan Krishna (Bihar)
Meena Chaturvedi (New Delhi)
Naresh Chockalingam (Tamil Nadu)
set-28

1. Azhar Qureshi (Bihar)
2. Amatra Sen (WB)
3. Preeti Puri (Haryana)



subjective tyPe

6. In figure, a long thin wire carrying a varying current
I = I0 sin wt lies at a distance y above one edge of a
rectangular wire loop of length L and width W lying
in the X-Z plane. What emf is induced in the loop?
X

I

W
y

L

Z

Y

7. A wire is wrapped N times over a solid sphere of
mass m near its centre, which is placed on a smooth
horizontal surface.
A horizontal magnetic field of

induction B is present. Find the
m R
angular acceleration experienced

B
by the sphere. Assume that the
I
mass of the wire is negligible
compared to the mass of the
sphere.

8. A stone is dropped from a balloon going up with
a uniform velocity of 5.0 m s–1. If the balloon
was 50 m high when the stone was dropped, find

its height when the stone hits the ground. Take
g = 10 m s–2.
9. A body weighs 98 N on a spring balance at the
north pole. What will be its weight recorded on
the same scale if it is shifted to the equator? Use
g = GM/R2 = 9.8 m s–2 and the radius of the earth
R = 6400 km.
10. A wheel of radius r and moment of inertia I about
its axis is fixed at the top of an inclined plane of
inclination q as shown in figure. A string is wrapped
round the wheel and its free end supports a block of
mass M which can slide on the plane. Initially, the
wheel is rotating at a speed w in a direction such
that the block slides up the plane. After some time,
the wheel stops rotating. How far will the block
move before stopping?


M


nn

10

Physics for you | January ‘16



Superposition of Waves

When two or more waves of similar kind, simultaneously
arrive at a point then the resultant disturbance (or
displacement in case of particles) at the point of meeting
is given by a vector sum of the disturbances produced
by each of the arriving waves.
After superposition, the waves pass through as if they
did not encounter each other, hence there is no change
in the properties of either of the arriving waves after
superposition.
If the disturbances are produced along same line, vector
sum becomes algebraic sum.
Let us see what we understand from this.
Consider two disturbances travelling in opposite
directions meet each other as shown.
1 cm s–1

1 cm s–1

1cm
1 cm 1 cm 1 cm 1 cm 1 cm 1 cm

We are going to draw the shape of the resultant waveform at
(i) t = 2 s (ii) t = 2.5 s (iii) t = 4 s.
To do this, imagine individual waves travelling as if
they are not meeting each other and then we apply
superposition.
t=2s
2 cm

1 cm 1 cm

t = 2.5 s

t=4s

1 cm s–1

1 cm 1 cm

1 cm s–1

1 cm 1 cm

Now, you see, that during superposition, the waves
individually appear to loose their identity, but
after superposition we can clearly see, they were always
there!
Applications of Superposition of waves





• Interference
• Standingwaves
• Beats

Interference

Beforewebeginwiththetopic,weneedtounderstand

what are coherent sources. These are such sources
for which the phase difference is independent of
time. This clearly is possible only if the frequency
of both the waves are identical else (w1 – w2)t will
come out to be time dependent expression.
Coming back to interference; when waves of similar
kind from two or more coherent sources simultaneously
arrive at a point then the resultant intensity at the
point of superposition is different from the sum of
intensity of the arriving waves and is dependent on
the phase difference between the arriving waves which
is directly dependent on path difference between the
arriving waves.
To put this in simple words, let me put a simple
statement, imagine two coherent sources of light made
to interfere. There might be a situation where
light + light = darkness
Isn't it opposite to our common sense? Yes, it is since
we expect more bright light when two light sources
are made to superimpose. But this logic is true only
for non-coherent sources.
Let us see this through an
example. Consider waves
S1
from two coherent sources
S 1 and S 2 meeting at a
point P travelling different
S2
distances.

Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata

12

physics for you | January ‘16

x1
x2

P



The displacement equations for the two sources are say
S1 : y1(x1, t) = A1sin(wt – kx1)
S2 : y2(x2, t) = A2sin(wt – kx2)
Clearly, the phase difference between them is
Df = (wt – kx1) – (wt – kx2)
= k(x2 – x1)
2p
\ Df = (Dx )
l
where Dx = x2 – x1 = path difference between the
arriving waves.
Hence from phasor analysis
of SHM, we now understand A2
that the resultant amplitude of

oscillation can be found out.

\ AR = A12 + A22 + 2 A1 A2 cos(Df)
where AR = resultant amplitude of oscillation.
Now we understand that A1 and A2 are fixed for wave
but if we change the location of sources from the point P,
x1 and x2 changes, hence Dx changes, hence Df changes.
Therefore AR becomes dependent on Dx.
Now, considering intensity of waves, as we know,
I ∝ A2
2
2
2
\ from, AR = A1 + A2 + 2A1A2cos(Df)
we have ,
I R = I1 + I2 + 2 I1I2 cos(Df)
where IR = intensity of the resultant wave due to arriving
waves of intensities I1 and I2.
Now extreme cases of interference may arise
€ Constructive interference: When the waves arrive
such that crest of one wave coincides with the crest
of the other, the waves are said to be in phase and
in such case the amplitude and hence intensity of
resultant wave is maximum.
A1


A1 + A2 = AR

A2

This clearly is possible only if one wave is shifted

with respect to the other wave by an integral
multiple of complete wavelength (l).
 2p 
\ Dx = nl ⇒ Df =   (nl) = n(2p)
 l 
\ AR
= A1 + A2
max

IR

max

14

=

(

I1 + I2

)2

physics for you | January ‘16

A1
+

AR = A1 – A2



AR

A1

+

€ Destructive interference: If the waves arrive such
that crest of one wave coincides with the trough
of the other, the waves are said to be out of phase
and in such case the resultant amplitude and hence
intensity of wave is minimum.

A2

This clearly is possible only if one wave is shifted
with respect to the other by an odd multiple of half
wavelengths.
l
\ Dx = (2n + 1)
⇒ Df = (2n + 1)p
2
\ AR = A1 – A2
min

I Rmin =

(

I1 − I2

)2

Hence, generalising,
A1 – A2 ≤ AR ≤ A1 + A2

(

)2

(

)2

I1 − I2 ≤ I R ≤ I1 + I2
Special case:
If the arriving waves are of identical amplitude and
hence intensity, then
A1 = A2 = A0 (say)
I1 = I2 = I0 (say)
\ AR
= 2A0, IR
= 4I0
max

and AR

min

max

= 0, IR

min

= 0.

Now, this is what I was talking about when I said
light + light = darkness!
In general at any point in such case,
 Df 
 Df 
AR = 2 A0 cos  , I R = 4 I0 cos2  
 2 
 2 
Let us now see an application of interference.
Q1. Two coherent sources
S1 and S2 are symmetriR
cally placed with
S1
S2
respect to centre at
D
3
a separation 3l as
R >> 3
shown. D is a detector
which can measure
the resultant intensity

at all points on circumference. Find





(i) the number of maximas and minimas detected
in one complete rotation.
(ii) the angular position of the 1st maxima after
starting.
A(x = 0)
Soln.: (i) In such
x = 
question analyse one
x = 2
quadrant, since all
quadrants are identical.
S1
S2
B (x = 3)
Let us see how.
3
At point A,
S1A – S2A = 0
\ Dx = 0, since the same path length for waves
arriving from S1 and S2.
At point B, S1B – S2B = 3l
\ Dx = 3l
Hence A and B are locations of maximas but in
between A and B, there must be other locations

where Dx would gradually increase from Dx = 0 to
Dx = 3l.
\ In between 0 and 3l, there would be locations
of Dx = l and Dx = 2l too.
Hence total maximas
= 2 × 4 (each quadrant) + 1 × 4(each point on
x and y axis)
= 12
l
For minimas, Dx = (2n + 1)
2
Hence between each successive maxima, there
would be minima too, so,
l 3l 5l
Dx = ,
,
in each quadrant.
2 2 2
\ Total minimas = 3 × 4 = 12
(ii) Given : R >> 3l hence, from all points on
circumference S1 and S2 will appear to be close by
points.
\ Dx = 2l
∆x = 2λ
sθ
o
⇒ 3lcosq = 2l
c
3λ θ
θ

2
S1 θ S2
⇒ cos q =
3
∆ x = 3λ
3λ
2
⇒ q = cos −1  
3

Standing Waves/Stationary Waves

When two waves of similar kind, same frequency
and amplitude travelling from opposite directions
superimpose we get standing waves. Now to understand
better what this actually means, let us start by considering
two such waves.
16

physics for you | January ‘16

y1(x, t) = Asin(wt – kx)
y2(x, t) = Asin(wt + kx)
Hence, considering superposition,
yR(x, t) = y1(x, t) + y2(x, t)
= A[sin(wt – kx) + sin(wt + kx)]
= A[2sin(wt) cos(kx)]
= 2Acos(kx) sin(wt)
= Axsin(wt)
where Ax = 2Acos(kx)

This resultant wave clearly shows that it is not a
travelling wave atleast.



The quantity, Ax = 2Acos(kx) is constant for a location
and the value is position dependent whereas
y(x, t) = Axsin(wt) indicates that the displacement
along y-axis of a particle located at x varies simple
2p
harmonically with time period, T = .
w
Hence, sin(wt) being same for all particles, each
executesSHMwithsametimeperiod.
Hence, we conclude that Ax indicates the amplitude
of oscillation at a location x.
Ax = 2Acos(kx)
\ –2A ≤ Ax ≤ 2A
But amplitude indicates magnitude of maximum
particle displacement. Hence –2A is same as 2A
when we talk of amplitude.
\ 0 ≤ |Ax| ≤ 2A
Hence at all locations the amplitude varies between
0 to 2A.
Imagine what do we mean when we say amplitude
of oscillation is zero?
Amplitude indicates maximum displacement and
maximum displacement itself is zero indicates
particles are not oscillating at all, i.e. they are stationary
at their respective points. It is due to these particles

that the name stationary wave is coined.
The locations where amplitude of oscillation is
(i) maximum (Ax = ±2A) are said to be antinodes.
Ax = ±2A = 2Acos(kx)
cos(kx) = ±1 ⇒ (kx) = np
p
 p 
⇒ x = n  = n
k
 2 p / l 
l
3l , ...
l
⇒ x = n   = 0, , l,
2
2
2
(ii) minimum (Ax = 0) are said to be nodes.
Ax = 0 = 2Acos(kx)



⇒ cos(kx) = 0

p
⇒ kx = (2n + 1)
2
l
l 3l 5l
⇒ x = (2n + 1) = ,

,
, ...
4
4 4 4
Let us try to draw the envelope of the particle's
displacement (for standing waves in a stretched
string) with these locations of nodes and antinodes.
Antinodes



Nodes

2A
2A
x=0
x=
4

x=
2

x=
x = 
4

x = 
4

x = 

2

The separation between two consecutive nodes or
l
antinodes is said to be loop length which is .
2
The bold lines drawn above indicate the boundary
within which the particles are confined to move.
How would standing waves in stretched string appear?
As below,

\ f1 = 2f0


All the particles located between two consecutive
nodes oscillate in same phase, i.e., they reach their
extreme ends and mean position together. Particles
located on opposite sides of a node always move
opposite to each other, hence are 180° out of phase
with respect to each other.
Hence, unlike travelling wave, the energy does not
propagate from one end to the other, it gets confined
between a loop.
Application of standing waves
Standing waves in stretched string fixed at both ends:
The fixed ends will always
be a displacement node.
The smallest frequency
l
with which standing

waves can be set up is
said to be fundamental frequency (f0).
18

physics for you | January ‘16

T
m
where, T = tension in string
m = mass per unit length.
Since velocity is only medium dependent, v = f l
suggests that for minimum frequency, wavelengthl should
be maximum. Hence keeping the two extreme ends
fixed (nodes) the largest wavelength that can be drawn
by inserting an antinode in between is shown here.
\ v = f0l0
v
v
⇒ f0 =
=
l0 2l
v

⇒ f0 =
l= 0
2l
2
The frequencies higher than the fundamental frequency
with which standing waves can be set up in the system are
said to be overtones, whereas all integral multiples of

fundamental frequency are said to be harmonics.
1st overtone
The wavelength decreases hence frequency increases.
v v
v
f1 =
= = 2 
 2l 
l1 l
Velocity of wave, v =

1st

2nd

overtone =
harmonic
Similarly,fornth overtone,

1 = l

fn = (n + 1)f0 = (n + 1)th harmonic
v
= (n + 1)  
 2l 
Standing waves in organ pipe:
The open ends of the organ pipe behave as displacement
antinode whereas the closed end behaves as displacement
node.
Open organ pipe

Closed organ pipe
It has both ends open. It has one end closed.
Fundamental mode

l=

0
2

v
\ f0 =
l0
v
⇒ f0 =
2l

l=

0
4

v
\ f0 =
l0
v
⇒ f0 =
4l


1 w2 − w1

\ fb =
=
= f2 − f1
Tb
2p

1st overtone

\ fb = f2 – f1
1 = l

31

=l

4
v
\ f1 = × 2 = 2 f0
v
2l
\ f1 =   × 3 = 3 f0
 4l 
\ 1st overtone = 2nd
st
\1 overtone = 3rd
harmonic
harmonic
Generalising
th
nth overtone

n overtone
th
= (n + 1) harmonic = (2n + 1)th harmonic
v
v
= (n + 1)  
= (2n + 1)  
 2l 
 4l 

Beats

When two sound waves of frequency close to each
other (but not equal) superpose at a point, then at the
point of superposition the phase difference keeps on
changing with respect to time.
Df = (w1 – w2)t
Hence the amplitude of resultant wave will also keep
on changing at a location as time passes by.
ButeverytimeDf= n(2p) we get to hear a maxima since
waves become in phase and everytime Df= (n + 1)p we
get to hear a minima.
So, the appearance is a sound of alternating intensity,
alternating between a certain maximum and minimum
intensity.
This phenomenon is beats and the number of maximas
heard per second is beat frequency.
To find beat frequency (fb)
Df = n(2p)
⇒ (w2 – w1)t = n2p

 2p 
⇒ t = n
 w2 − w1 
2p
2(2 p)
3(2 p)
=
,
,
, ....
w2 − w1 w2 − w1 w2 − w1
Tb
Tb
The time gap between two successive maximas is
beat period (Tb).
2p
\ Tb =
w2 − w1

Therefore beat frequency is just the difference of
two frequencies of the source.
Applications:
Q2. An aluminium rod having a length 100 cm is
clamped at its mid point and set into longitudinal
vibrations of fundamental mode. Given r= 2600 kg m–3
and Y = 7.8 × 1010 N m–2 for aluminium. Find the
frequency of sound emitted.
Soln.: Velocity of wave
v=

7.8 × 1010
Y
=
= 30 × 103 m s −1
r
2600

30 × 103
v
\ f0 = =
≈ 2740 Hz.
2l
2 ×1
Q3. A tuning fork of unknown frequency produced
4 beats per second when sounded with another
of frequency 254 Hz. It gives the same number
of beats per second whenever unknown tuning
fork is loaded with wax. Find its frequency before
waxing.
Soln.: Remember, two terms with tuning forks
(i) Waxing : Here wax is added to the prongs due
to which frequency of wave decreases.
(ii) Waning : Here the prongs are filed due to which
frequency of wave increases.
Coming back to the question, having 4 beats/second
gives two possible situations
254
+4
258

–4
250

waxed

waxed
f < 250

f < 258
Here the difference with
254 will initially decrease
as long as f > 254 but if
f < 254 then it might
happen that it reaches
f = 250 Hz and again we get
fb = 4 Hz, hence 258 Hz
is the required frequency.

Here difference with
254 will clearly be
greater than 4. Hence
fb > 4 which is not possible.

physics for you | January ‘16

19


Q4. For a certain organ pipe three successive frequencies
are observed at 425, 595, 765 Hz respectively.

Taking speed of sound in air to be 340 m s–1, find
the fundamental frequency.
Soln.: Note the ratio,
425 : 595 : 765 = 5 : 7 : 9
\ Odd multiples ⇒ closed organ pipe
\ 5f0 = 425 ⇒ f0 = 85 Hz

Q7.
vw
A

fa
T
324 18
=
=
=
fa′
T′
289 17
4 + ft 18
⇒
=
⇒ ft = 50 Hz
1 + ft 17
\

20

physics for you | January ‘16

B

A vehicle is moving towards a large building with
a speed v0 and blows a horn of frequency f0 which
travels with a speed vw. Find the number of beats
detected by A who is moving with the vehicle and B
who stands in between vehicle and building.
Soln.: Using method of images, we create a virtual
source behind the building.
vw

vw
A

f0

f0

B

v0

v0

S

⇒ ft = 205 Hz

Soln.: v ∝ T

\ Decrease in temperature
⇒ decrease in velocity of wave
⇒ decrease in frequency of air column (fa)
Two situations were possible.
fa – ft = 4 or ft – fa = 4
The second equation suggests that if fa decreases,
fb > 4 (increases) which is a contradiction.
Hence, fa – ft = 4
... (i)
Again on decreasing temperature,
fa′ – ft = 1
... (ii)

f0
v0

Q5. A tuning fork vibrating with a sonometer having a
wire of length 20 cm produces 5 beats per second. The
beat frequency does not change if the length of wire is
changed to 21 cm. Find frequency of tuning fork.
1
Soln.: For a sonometer wire, f s ∝
l
Hence if length increases, fs decreases.
Hence we conclude that initially fs was greater than
ft (frequency of tuning fork) and later it became
smaller by same amount.
\ fs – ft = 5
ft – fs′ = 5
f

l ′ 21
5 + ft 21
\ s = =
⇒
=
f s′ l 20
ft − 5 20
Q6. A tuning fork and an air column whose temperature
is 51°C produce 4 beats in one second when sounded
together. When the temperature of air column is 16°C
only one beat is detected per second. Find frequency
of tuning fork.

Building



S

For both A and B we have two sources S and S′, but
for B, the frequency of waves detected from both of
them would be identical and equal to
 vw 
fappB = 
f0
 vw − v0 
hence no beat is heard by him.
SpeakingofA, he hears two frequencies
(i) f1 = f0 (since there is no relative motion between
him and S)

v +v 
(ii) f2 =  w 0  f0
 vw − v0 
\ fb = f2 – f1
2v0 f0
v +v

=  w 0 − 1 f 0 =
vw − v0
 vw − v0 
nn

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physics for you | january ‘16

nn
21


By : Prof. Rajinder Singh Randhawa*

THERMODYNAMICS
1.

A large tank A of water is kept at a constant
temperature q0. It is connected to another body B
of specific heat c and mass m by a conductor of
length l, area of cross-section A and thermal
conductivity K. Initial temperature of body B is q1.
Find the variation of temperature q of the body B
at time t.

K
l

A

2.

1

t=0

B

One end of a rod of length L and cross-section
area A is kept in a furnace at temperature T1. The
other end of rod is kept at a temperature T2. The
thermal conductivity of the material of the rod
K and emissivity of the rod is e. It is given that
T2 = TS + DT, where DT < < TS, TS is the temperature
of surroundings. If DT ∝ (T1 – TS), find the
proportional constant.
Furnace
T1

3.

by w2. What is the ratio of
4.

Tank of water

0

in the vessel which is maintained at a temperature T.
The mean square of velocity of the molecules of B
type is denoted by v2 and the mean square of the
x-component of the velocity of A type is denoted

Insulated
L
Insulated

N molecules each of mass m of gas A and 2N
molecules each of mass 2m of gas B are contained

v2

?

An ideal monoatomic gas is confined in a cylinder
by a spring-loaded piston of area of cross-section
8 × 10–3 m2. Initially, the gas is at 300 K and
occupies a volume of 2.4 × 10–3 m3 and the spring
is in its relaxed state as shown in figure. The gas is
heated by a small electric heater until the piston
moves out slowly by 0.1 m. Calculate the final
temperature of the gas and the heat supplied (in J)
by the heater. The force constant of the spring is
8000 N m–1, atmospheric pressure is 1 × 105 N m–2. The
piston is massless and no friction exists between
the piston and the cylinder.

Heater

TS
T2

w2

Piston

5.

Rigid
support

Figure shows three isotherms at temperatures
T1 = 4000 K, T2 = 2000 K, and T3 = 1000 K. When
1 mole of an ideal monoatomic gas is taken through
paths AB, BC, CD and DA, (a) find the change in

*Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh, Ph. 09814527699

22

physics for you | January ‘16


internal energy DU, (b) find the work done by the
gas W.
A

P

B

VA

1.

V

C T = 2000 K
2
T3 = 1000 K
VB

B

K (T1 − TS − DT )
= 4σeTS3 DT
L
K (T1 − TS ) 
K
=  4σeTS3 +  DT

L
L
\ DT =

The rate of increase in thermal energy of the

tank B is
dU B
dq
= mc
dt
dt
dU
dQ
B because no work is done in change
Since
=
dt
dt
in volume, we have

=
3.

− KAt

The mean square velocity of gas molecules is given
3kT
.
by v 2 =
m
3kT
... (i)
For gas A, v 2A =
m
For a gas molecule,

( v 2x = v 2y = v z2 )

2
2 3kT
For gas B, v B = v =
2m
Dividing eqn. (ii) by eqn. (iii), we get
kT
w2
2
= m =
2
3kT 3
v
2m

or (q0 − q) = (q0 − q1 )e mlc

Given that T2 = TS + DT

.

 3kT 

 kT
w2 = v 2x =  m  =
 3  m

q − q −KA
dq

KA
=
t or ln 0
=
t
∫
q0 − q1 mlc
q1 q0 − q mlc

t

Rate of heat conduction through rod = rate of heat
lost from right end of the rod.
KA(T1 − T2 )
... (i)
= σAe(T24 − TS4 )
L

(4σeLTS3 + K )

or v 2x =

q

2.

K

v3
3

From eqn. (i), we get

 q − q  or dq = KA dt
dq
=K 0
A
 l 
q0 − q mlc
dt
Integrating both sides, we get

− KA

(4eσLTS3 + K )

v 2 = v 2x + v 2y + v z2 = 3v 2x

mc

or q = q0 − (q0 − q1 )e mlc

K (T1 − TS )

Comparing with the given relation DT ∝ (T1 – TS),
Proportionality constant

At t



0

(as DT << TS)

Substituting in eqn. (i), we have

m
c

= TS4

\ T24 − TS4 = 4(DT )(TS3 )

solutions
Let the temperature of the body B be q, in time
dt, let a heat dQ flow through the conductor
rod.
dQ q0 − q
Then
=
KA
dt
t
A

4

DT 

T24 = TS4 1 + 4

TS 


T1 = 4000 K
D

4

 DT 
\
= (TS + DT )
1 + T  ,

S 
Using binomial expansion, we have
T24

4.

... (ii)
... (iii)

Since F = kx = 8000 × 0.1 = 800 N
Pressure exerted on the piston by the spring is
physics for you | January ‘16

23


F

800
DP = =
= 1 × 105 N m −2
−
3
A 8 × 10
The total pressure of the gas inside the cylinder
is
P2 = P1 + DP = Patm + DP
P2 = 1 × 105 + 1 × 105 = 2 × 105 N m–2
Since the piston has moved outwards, then increase
in volume of the gas is
DV = A × x = 8 × 10–3 × 0.1 = 8 × 10–4 m3
The final volume of the gas,
V2 = V1 + DV
= 2.4 × 10–3 + 8 × 10–4 = 3.2 × 10–3 m3
Let T2 be the final temperature of the gas, then
P1V1 P2V2
PV
=
⇒ T2 = 2 2 T1
T1
T2
P1V1

T2 =

2 × 105
1 × 105

×

3.2 × 10 −3

2.4 × 10 −3

× 300

= 800 K
Let the heat supplied by the heater be Q, then
Q = W + DU

= 3000 × 8.314 = 2.49 × 104 J
DUBC = CV (TC – TB)
3
= × R(2000 − 4000) = –2.49 × 104 J
2
DUCD = CV(TD – TC)
3
= × R(1000 − 2000) = –1.25 × 104 J
2
DUDA = CV(TA – TD)
3
= × R(2000 − 1000) = 1.25 × 104 J
2
\ Total change in internal energy
DU = 0
(b) Work done by the gas during path AB under
constant pressure is
P(VB – VA) = R(TB – TA)

= 8.314 × 2000 = 16.628 × 103 J
No work is done during BC and DA.
Work done on the gas during CD is
R(TC – TD) = 8.314 × 1000 = 8.314 × 103 J.
.. . Net work done by the gas during the cycle is
(16.628 × 103 – 8.314 × 103) J = 8.314 × 103 J.

V2
x
kx 
Now, W = ∫ PdV = ∫  Patm +  Adx

A
V
0

nn

1

kx 2
= Patm Ax +
2
kx


 P = Patm + A and dV = Adx 
Substituting the values, we get W = 120 J.
Also, U = nCV DT
Number of moles of gas,

PV
n=
RT
3
For monoatomic gas, Cv = R
2
PV  3 
3 PV
\ DU =
DT
 R  DT = ×
RT  2 
2 T
3 1 × 105 × 2.4 × 10−3
= ×
× 500
2
300
= 600 J
\ Heat supplied by the heater
Q = 120 + 600 = 720 J
5.

24

(a) The change in internal energy is given by
DUAB = CV(TB – TA)
(for n = 1 mole)
3
= × R(4000 − 2000) = 3000R

2
physics for you | January ‘16

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PRACTICE PAPER

CLASS
XI

single oPtion correct
This paper contains 45 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which only

one is correct. (Mark only One Choice).
Marks : 45 × 4 = 180

Negative Marking (–1)

1. A body of mass 7m initially at rest explodes into two
fragments of masses 4m and 3m. If the momentum
of the lighter fragment is p then the kinetic energy
released in the explosion will be
11 p2
5 p2
7 p2
9 p2
(a)
(b)
(c)
(d)
24 m
14 m
16 m
24 m
2. A particle of mass m describes a circle of radius r.
4
The centripetal acceleration of the particle is .
r2
The momentum of the particle is
4m
2m
4m
2m

(c)
(d)
(a)
(b)
r
r
r
r
3. Figure shows a block of mass m1 on a smooth
horizontal surface pulled by a
m
string which is attached to a
block of mass m2 hanging over a
frictionless pulley which has no
m
mass. The blocks will move with
an acceleration
 m + m2 
(a) g
(b)  1
g
 m2 
1

2

 m2 
(c) 
g
 m1 + m2 

 m1 
(d) 
g
 m1 + m2 

4. Two blocks m1 and m2 are in contact over a
frictionless table; m1 = 2.0 kg, m2 = 1.0 kg.
In the first case a horizontal F = 3 N
m1
m2
force of magnitude 3 N
is applied to block m1. In
Case 1
the second case this force
F=3N
is applied to block m2. The
m1
m2
forces of contact between
Case 2

the blocks in the first and the second cases, respectively
are
(a) 3 N, 3 N
(b) 2 N, 2 N
(c) 1 N, 2 N
(d) 2 N, 1 N
f1
5. A uniform ladder is in

equilibrium against a rough
A
N1
wall as shown. Points A and
B respectively are the point
N2 C
of contact of ladder with wall
B

and with the ground. Point C
is the CM of the ladder.
f2
W
1. Torque due to friction f2 about point A is not
zero.
2. Torque due to friction f2 about point B is zero.
3. Torque due to weight is zero about A, B and C.
4. Torque due to f1 and N1 is not zero about A.
(a) Only 1 and 3 are correct
(b) Only 2 and 3 are correct
(c) Only 3 and 4 are not correct
(d) Only 1 and 4 are correct
6. The moment of inertia of an annular disc (a disc
with concentric cavity) of mass M radius R and
cavity radius r about an axis passing through its CM
and normal to its plane will be
1
1
2
2

(b) M R2 − r 2
(a) M R + r
2
2
1
1
2
2
2
2
(c) M 2R + r
(d) M R + r
8
4
7. A body is moved along a straight line by a machine
delivering constant power. The distance moved by
the body in time t is proportional to
(a) t1/2
(b) t3/4
(c) t3/2 (d) t2

(
(

)

)

(
(

)
)

Contributed by : K. P. Singh, KP Institute of Physics, Chandigarh, 09872662552
Physics for you | January ‘16

25


8. A body of mass 3 kg is under a force which causes
t2
in metre, with
displacement in it, given by s =
3
time t in seconds. What is the work done by the
force between time t = 0 and t = 2 s ?
(a) 8 J
(b) 5.2 J (c) 3.9 J (d) 2.6 J
9. The linear momentum p of a body varies with time
as p = 5a + 7bt2 where a and b are constants. The
net force acting on the body for one dimensional
motion varies as
(a) t2
(b) t–1
(c) t–2
(d) t
10. A body is acted upon by a force proportional to
square of distance covered. If the distance covered
is denoted by x, then work done by the force will

be proportional to
(a) x
(b) x2
(c) x3
(d) x–2
11. The potential energy function along the positive
x-axis is given by U (x ) = −ax + b , a and b are
x
constants. If it is known that the system has only
one stable equilibrium configuration, the possible
values of a and b are
(a) a = – 1, b = 2
(b) a = – 5, b = 1
(c) a = 1, b = – 2
(d) a = 5, b = – 3
12. Neglecting the friction and weights
of the pulley, which one of the
T
following is the force F required to
T
F
B
lift a 100 N load in the system of
T
T
pulleys as shown in the figure?
A
(a) 20 N
(b) 25 N
(c) 30 N

(d) 35 N
W
2

2

1

1

13. A body is falling freely under the action of gravity
alone in vacuum. Which of the following quantities
remain constant during the fall?
(a) Kinetic energy
(b) Potential energy
(c) Total mechanical energy
(d) Total linear energy
14. A motor drives a body along a straight line with a
constant force. The power P developed by the motor
must vary with time t as shown in figure
(a)

P

(b)

(c)

P

t

26

17. A block of mass 5 kg is resting on a smooth surface.
At what angle a force of 20 N be acted on the body
so that it will acquire a kinetic energy of 40 J after
moving 4 m ?
(a) 30°
(b) 45°
(c) 60° (d) 120°
18. The moment of inertia of a body about a given axis
is 1.2 kg m2. Initially, the body is at rest. In order
to produce a rotational kinetic energy of 1500 J, an
angular acceleration of 25 rad s–2 must be applied
about that axis for a duration of
(a) 4 s
(b) 2 s
(c) 8 s
(d) 10 s
19. A particle performs uniform circular motion with an
angular momentum L. If the frequency of particle’s
motion is halved and its kinetic energy doubled, the
angular momentum becomes
(a) 2 L
(b) 4 L
(c) L/2 (d) L/4
20. The moment of inertia of two spheres of equal masses
about their diameters are equal. If one of them is
solid and other is hollow, the ratio of their radii is

(a) 3 : 5
(b) 3 : 5
(c) 5 : 3
(d) 5 : 3
21. The curve between log L and log p is (L is angular
momentum and p is linear momentum)
log L

Physics for you | January ‘16

log L

(b)

(a)
log p

log p

log L
t

(d)

–1

P

t

P

V(m s )
15. Velocity-time graph of a particle
20
of mass 4 kg moving in a
straight line is as shown in
figure. Work done by all
t(s)
2
forces on the particle is
(a) 400 J (b) – 800 J (c) – 400 J
(d) 200 J
16. An ideal spring with spring constant k is hung from
the ceiling and a block of mass M is attached to
its lower end. The mass is released with the spring
initially unstretched. Then the maximum extension
in the spring is
4 Mg
2 Mg
Mg (d) Mg
(a)
(b)
(c)
2k
k
k
k

t

log L

(c)

(d)
log p

log p


22. A uniform rod of mass m and length L
is suspended by means of two light
inextensible strings as shown in figure.
B
Tension in one string immediately A
after the other string is cut is
mg
mg
(a)
(b) mg
(c) 2 mg (d)
4
2
23. A circular platform is free to rotate in a horizontal plane
about a vertical axis passing through its centre. A
tortoise is sitting at the edge of the platform. Now
the platform is given an angular velocity w0. When the
tortoise moves along a chord of the platform with a
constant velocity with respect to the platform, the angular

velocity of the platform will vary with the time t as
(t)

(a)

(b)

0

(t)

t

t

(t)

(c)

(d)
0

0

t

t

24. A wheel has angular acceleration of 3.0 rad s–2 and
an initial angular speed of 2.00 rad s–1. In a time of

2 s it has rotated through an angle (in radian) of
(a) 6
(b) 10
(c) 12
(d) 4
25. The moment of inertia of a rod about an axis through
1
its centre and perpendicular to it is
ML2 where,
12
M is the mass and L the length of the rod. The rod
is bent in the middle so that the two halves make an
angle of 60°. The moment of inertia of the bent rod
about the same axis would be
ML2
ML2
ML2
ML2
(a)
(b)
(c)
(d)
48
24
8 3
12
26. A body of mass 15 kg is suspended by the strings
making angles 60° and 30° with the horizontal as
shown in figure. Then (take g = 10 m s–2)
T2

60°

T1
15 kg

29. For shown atwood machine m1 = 8 kg,
m2 = 2 kg. The string and the pulley are
assumed to be smooth and massless.
Take g = 10 m s–2. The acceleration of
center of mass of the system is.
(a) 3.6 m s–2
(b) 6 m s–2
–2
(c) 5 m s
(d) 0

m2
m1

0

(t)

28. A 5000 kg rocket is set for vertical firing. The
exhaust speed is 800 m s–1. To give an initial upward
acceleration of 20 m s–2, the amount of gas ejected
per second to supply the needed thrust will be
(g = 10 m s–2)
(a) 127.5 kg s–1
(b) 187.5 kg s–1

–1
(c) 185.5 kg s
(d) 137.5 kg s–1

30°

(a) T2 = 75 N
(b) T2 = 75 3 N
(c) T1 = 25 3 N
(d) T1 = 12 N
27. A car of mass 1000 kg negotiates a banked curve of
radius 40 m on a frictionless road. If the banking
angle is 45°, the speed of the car is
(a) 20 m s–1
(b) 30 m s–1
–1
(c) 5 m s
(d) 10 m s–1

30. A wide hose pipe is held horizontally by a
fireman. It delivers water through a nozzle at two
litre per second. On increasing the pressure, this
increases to four litres per second. The fireman
has now to
(a) push forward twice as hard
(b) push forward four times as hard
(c) push backward four times as hard
(d) push backward twice as hard.
31. A block of mass m is placed on a smooth wedge
of inclination q. The whole system is accelerated

horizontally so that the block does not slip on the
wedge. The force exerted by the wedge on the block
has a magnitude
(a) mg tan q
(b) mg cos q
(c) mg sec q
(d) mg
32. A system consists of 3 particles each of same mass
and located at points (1, 2), (2, 4) and (3, 6). The
co-ordinates of the center of mass are
(a) (1, 2) (b) (2, 4) (c) (4, 2) (d) (3, 6)
33. Out of the following bodies of same mass, which
one will have maximum moment of inertia about
an axis passing through its center of gravity and
perpendicular to its plane ?
(a) ring of radius r
(b) disc of radius r
(c) square frame of sides 2r
(d) square lamina of sides 2r
34. Consider a two particle system with particles having
masses m1 and m2. If the first particle is pushed
towards the center of mass through a distance d,
by what distance should the second particle be
moved so as to keep the center of mass at the same
position ?
Physics for you | January ‘16

27

m1
d
m2
m2
(c) m d
1

(b) d

(a)

(d)

m1
d
m1 + m2

35. Two particles of mass m1 and m2 (m1 > m2) attract
each other with a force inversely proportional to the
square of the distance between them. The particles are
initially held at rest and then released. Then the CM
(a) moves towards m1 (b) moves towards m2
(c) remains at rest
(d) moves at right to the line joining m1 and m2
36. Two skaters A and B of masses 50 kg and 70 kg
respectively stand facing each other 6 m apart. Then
they pull on a rope stretched between them. How
far has each moved when they meet?
(a) both have moved 3 m
(b) A moves 2.5 m and B 2.5 m

(c) A moves 3.5 m and B 2.5 m
(d) A moves 2 m and B 4 m
37. A thin uniform circular disc of mass M and radius
R is rotating in a horizontal plane about an axis
passing through its center and perpendicular to its
plane with an angular velocity w. Another disc of
1
same dimensions but of mass M is placed gently
4
on the first disc co-axially. The angular velocity of
the system is
3
4
1
(a) 2 w (b) w (c) w (d) w
4
5
3
38. A wheel is rolling uniformly along a level road (see
figure). The speed of translational motion of the
wheel axis is v. What are the speeds of the points
A and B on the wheel rim relative to the road at the
instant shown in the figure?
B
(a) vA = v ; vB = 0
(b) vA = 0 ; vB = v
v
(c) vA = 0 ; vB = 0
(d) vA = 0 ; vB = 2v
A

^

^

^

39. If a force 10 i + 15 j − 25 k acts on a system and gives
^

^

^

an acceleration 2 i + 3 j − 5 k to the centre of mass of
the system, the mass of the system is
(a) 5 units
(b) 38 units
(c) 5 38 units
(d) None of these
u = 5 m s–1
40. Two particles A and B
are situated at a distance
v
d = 3 m apart. Particle A
–1
has a velocity of 5 m s at
30°
60°
an angle of 60° and particle A
B has a velocity v at an angle of 30° as shown in

28

Physics for you | January ‘16

figure. The distance d between A and B is constant.
The angular velocity of B with respect to A is
5
(a) 5 3 rad s −1
(b) rad s −1
3
5
10
−1
rad s −1
rad s
(c)
(d)
3
3
41. The ratio of the dimensions of Planck’s constant and
that of the moment of inertia is the dimension of
(a) Frequency
(b) Velocity
(c) Angular momentum (d) Time
42. A ball is dropped to the ground from a height of
8 m. The coefficient of restitution is 0.5. To what
height will the ball rebound?
(a) 2 m (b) 1.42 m (c) 4 m (d) 0.5 m
43. A mass m moving horizontally with velocity vo

strikes a pendulum of mass 2m. If the two masses
stick together after the collision, then the maximum
height reached by the pendulum is
v2
v02
v2
v2
(c) 0
(d) 0
(a) 0
(b)
18 g
2g
6g
12 g
44. A ball is projected in vacuum as shown. Average
power delivered by gravitational force
B
(a) for A to C is positive
(b) for B to C is zero.
u
(c) for A to B is negative.
C

(d) for A to B is zero.
A
45. 1. During any collision, velocity along common
tangent doesn’t change.
2. In an elastic collision with equal masses, the
velocity along common normal is interchanged.

3. When a ball makes an oblique inelastic collision
with a fixed target the reflection angle is less than
incidence.
4. In a one dimensional elastic collision the fraction
of kinetic energy transferred by a projectile to a
4 m1 m2
stationary target is
.
(m1 + m2 )2
(a) Only 3 is wrong
(c) Only 4 is wrong

(b) 2 and 3 are wrong
(d) 3 and 4 are wrong

solutions

1. (a) : 7m . 0 = 4mv1 + 3mv2 = p1 + p2
⇒ p2 = – p1 = p (given)
K=
=

p12
p2
+ 2
2m1 2m2

p2  1 1  7 p2
p2
p2

=
+
 + =
2(4m) 2(3m) 2m  4 3  24m


2. (d) : acp =
⇒v =

2

v2 4
=
r r2

2m
.
. . p = mv =
r

r
3. (c) : For m2 : m2 a = mg – T
For m1 : T = m1a
m2 g
On adding, (m1 + m2) a = mg or a =
m1 + m2
Fnet
3
−2
4. (c) : a =

=
=1m s
mtotal 2 + 1
Case 1: for m1 : 3 – N = 2 (1) ⇒ N = 1 N
Case 2: for m2 : 3 – N = (1) (1) ⇒ N = 2 N
5. (c) : Torque of a force about a point is zero only
when the line of action of the force passes through
that point.
6. (a) : Check: When r
0, it becomes disc and
when r
R, it becomes ring.
7. (c) : P = M1 L2 T –3 = constant
⇒ x2 t–3 = constant or x2 ∝ t3 or x ∝ t3/2
2t
2
t2
⇒v =
v(t = 0) = (0) = 0
3 ,
3
3
2
4
v(t = 2) = (2) =
3
3
2

 1 4 4 8

1
4

W = DK = .3.   − 0  = .3. . = = 2.6 J
 2 3 3 3
2  3 



8. (d) : s =

9. (d) : p = 5a + 7bt2
dp
F=
= 0 + 7 × 2bt ⇒ F ∝ t
dt
10. (c) : F ∝ x2 , W = ∫F dx ∝ x3
dU
b
b
= −a − 2
11. (a) :U = −ax + ⇒ −F =
dx
x
x
2b
b d 2U
or F = a +
=0+ 3
x 2 dx 2

x
At stable equilibrium, F = 0,

d 2U

dx 2
2b
b
−b
a + 2 = 0 ⇒ a = 2 Also,
>0
x
x
x3

12. (b) : For rope:
F = T2
For pulley B: 2 T2 = T1
For pulley A: 2 T1 = W = 100
Using above eqns.
100
F=
= 25 N
4

>0

13. (c)
14. (a) : P = Fv = (constant) (u + at)
⇒ P ∝ t, so linear variation.

15. (b) : Speed becomes 0 from 20 m s–1.
1
... W = DK = .4. 0 − (20)2  = −800 J

2 
16. (b) : Let the maximum extension in the spring be x.
By work energy theorem, W = DK
2 Mg
1
Mgx − kx 2 = 0 − 0 ⇒ x =
k
2
17. (c) W = DK

1
20 × 4 × cos q = 40 ⇒ cos q = or q = 60°
2
1
18. (b) : K = I (w0 + at )2
2
1
1500 = ×1.2 × (0 + 25t )2 ⇒ t = 2 s
2
1
K ′ L′ w′
19. (b) : K = Lw ⇒
=
2
K L w
⇒ 2=

L′  1 
  ⇒ L′ = 4 L
L 2

5
2
2
2 ⇒ RS
20. (c): MRS2 = MRH
=
RH
3
5
3
21. (b): L = pr ⇒ log L = log (pr)
log L = log p + log r; y = mx + C
So, straight line with positive intercept.
22. (a): Before cutting, TA + TB = mg
For rotational equilibrium about center
L
L
TA = TB ⇒ TA = TB
2
2
mg
So, 2TA = mg
or TA =
2
If string B is cut, just after cutting tension in

mg
A remains same i.e.,
.
2
23. (c): Moment of inertia first decreases and then
increases, thus by law of conservation of angular
momentum,
L = Iw = constant
w first increases and then decreases.
1
1
2
24. (b) : q = w0t + at 2 = 2(2) + (3)(2) = 10 radian
2
2
2

1 M L
1 M L
25. (b) : I =     +    
3  2  2  3  2  2 

2

Physics for you | January ‘16

29


35. (c) : Internal forces cannot change velocity of CM

of a system. As CM was initially at rest, it will
remain at rest.
36. (c) : Two bodies under mutual internal forces
always meet at their CM. From theorem of moment
of masses, we can write
50 x = 70 (6 – x) ⇒ 120 x = 420 ⇒ x = 3.5 m
37. (b) : From law of conservation of angular
momentum,

2

2
1  M  L 
ML
=    ×2 =
3  2  2 
12

26. (b) : Horizontally: T2 cos 60° = T1 cos 30°
⇒ T2 = T1 3
Vertically : T2 sin 60° + T1 sin 30° = 150
Thus T1 = 75 N and T2 = 75 3 N
27. (a) : As tan q =

v2
gr

M

M 2

1 M 
4
R w + 0 =  R2 +   R2  w′ ⇒ w′ = w
2
2
2
4
5




5
As total moment of inertia increases to times,
4
4
then w becomes th.
5
38. (d) : For pure rolling, point of contact is at rest and
topmost point has double speed of that of CM.

v2
tan 45° =
⇒ v = 20 m s–1
10(40)
dm
28. (b) : vgr
= m( g + a)
dt
dm

dm
= 187.5 kg s −1
800
= 5000(10 + 20) ⇒
dt
dt
2

29. (a) : acm

→

2
 m − m2 
 8 − 2
= 1
10 = 3.6 m s −2
 g=
 8 + 2 
 m1 + m2 

^

dm dm
,
= ρAv
dt dt
dm
On doubling
, v also doubles, this increases

dt
force to 4 times.

30. (b) : F = v

N

31. (c) : N = m g 2 + a2 ...(i)
For no slip of block with
respect to wedge
ma cos q = mg sin q
or a = g tan q
...(ii)
Using (ii) in (i), we get

a

ma
mg



N = m g 2 + g 2 tan2 q ⇒ N = mg sec q
→

→

32. (b) : r =
CM
=

→

→

m1 r1 + m2 r2 + m3 r3
m1 + m2 + m3

m(1, 2) + m(2, 4) + m(3, 6) (6, 12)
=
= (2, 4)
m+m+m
3

33. (c) : Iring = mr2 , Idisc = 0.5 mr2
4 2
1 m
2 m
2
Isq.frame
12 4 (2 r ) + 4 (r )  = 3 mr


1
2
2
2
Isq. lamina = m (2r ) + (2r )  = mr 2



12
3
34. (a) : (m1 + m2) DxCM = m1 Dx1 + m2 Dx2
(m1+ m2) (0) = m1 (–d) + m2 Dx2
md
⇒ Dx2 = 1
m2
= 4×

30

Physics for you | January ‘16

→

39. (a) : | F | = m | a |
^

^

^

^

^

10 i + 15 j − 25 k = m 2 i + 3 j − 5 k ⇒ m = 5 units
40. (b) : v cos 30° = 5 cos 60°
⇒ v = 5 m s −1
3

v v AB 5 sin 600 − v sin 300
w= =
=
r AB
3
3 5 1
−
.
2
2
5
3
1
=
= rad s–
3
3
41. (a) : Planck’s constant and angular momentum have
same dimensions.
[ h] [ L ]
=
=w
[I ] [I ]
5

42. (a) : hr = e 2 hs = (0.5)2 (8) = 2 m
43. (a) : Using law of conservation of linear momentum:
mv0 + 0 = (m + 2m) vs
⇒
...

vs = v0 / 3
v 2 (v / 3)2 v02
h= s = 0
=
2g
2g
18 g

44. (c) : During upward motion gravity does negative
work and thus negative power is delivered.
45. (a) : When a ball makes an oblique inelastic collision
with a fixed target, the reflection angle is more than
tan i
incidence, as tan r =
.
e
nn