tạp chí physics for tháng 4 năm 2015

edit
Vol. XXIII

No. 4

April 2015

t was a very highly inspiring advice given to the teachers and students by the
Hon’ble prime Minister, Shri Narendra Modi ji himself in a recent radio-talk.

plot 99, Sector 44 institutional area,
Gurgaon -122 003 (Hr). Tel : 0124-4951200
e-mail : website : www.mtg.in

Many books have been written on education by very learned authors. These are
addressed mainly to higher level students and professors. But the difficulties

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406, Taj apartment, Near Safdarjung Hospital,
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:
:

how to Excel in any competitive Exams, at any Level

i

Corporate Office :

Managing Editor
Editor

of the students are not spelt out. The prime Minister’s advice reaches directly
the students of all levels.

Mahabir Singh
anil ahlawat (BE, MBa)

The first is the fear of exams which all of us have felt. To combat this, the
advice is to hold a week-long examination festival, two times a year, with

contents
Physics Musing (Problem Set-21)

8

satirical poems on exams, cartoon contents and lectures on the psychological
effects of exams with debates to pepper the lectures.
Comparisons with others is wrong. Competition should be with yourself.

Thought Provoking Problems

11

Compete with yourself,

BITSAT Full Length

14

Compete for speed,

Practice Paper 2015
JEE Advanced

Compete to do more,

31

Brain Map

46

AIPMT

48

Practice Paper 2015
57

Practice Paper 2015
CBSE Board Class XII

Compete to achieve newer heights,
Focus on doing better every-time.

Practice Paper 2015

AIIMS

rial

66

The former Ukrainian pole vault champion Sergey Bubka had broken his own
record 35 times! perhaps you might have seen the film ‘Bhaag Milkha Bhaag’.
Milkha Singh, the famous champion runner was breaking his own record every
time. Even after winning, he runs round the track at the same speed waving
to the crowds! The final advice given by the prime Minister is valid not only to
the students but also to everyone working in any field. “live in the present,
struggle with the present. Victory will walk alongside”, Go ahead, students!
Victory is yours.

Solved Paper 2015

Anil Ahlawat

Core Concept

80

Physics Musing (Solution Set-20)

83

You Ask We Answer

84

Crossword

85

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physics for you | april ‘15

7


PHYSICS

P

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 solutions. 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.
By : Akhil Tewari

21
1. A uniform rod of mass m and length l starts rotating
with constant angular acceleration a in a horizontal
plane about a fixed vertical axis passing through
one end. The horizontal component of the net force
exerted on the rod by the axis when it has rotated by
an angle p/2, is
l

l
(a) ma
(b) ma 1 + p2
2
2
m
l
pa
(c)
(d) none of these
2
2. A battery is connected between two points A and
B on the circumference of a uniform conducting
ring of radius r and resistance R. One of the arcs AB
of the ring substends an angle q at the centre. The
value of the magnetic induction at the centre due to
the current in the ring is
(a) proportional to 2(180° – q)
(b) inversely proportional to r
(c) zero, only if q = 180°
(d) zero, for all values of q.
3. A particle is projected with a speed u in air at angle
q with the horizontal. The particle explodes at the
highest point of its path into two equal fragments,
one of the fragments moving up straight with a
speed u. The difference in time in which the two
particles fall on the ground is (Assume it is at a
height H at the time of explosion.)
u 2
(a) 2u

(b)
u − 2 gH
g
g
8

(c)

1 2
u + 2 gH
g

(d)

2 2
u + 2 gH
g

4. Consider the cube shaped carriage ABCDEFGH
of side l and a mass M and it can slide over two
frictionless rails PQ and RS. A shot of mass m is
thrown from corner A such that it lands at corner
F. The angle of projection as seen from the carriage
is 45°. While the shot is in the air, the velocity of
carriage as seen from the ground is
H
D

A

G

l

C

E
R

F
B S
Q

P

(a)

m gl 2
2(M + m)

(b)

m 2 gl
2(M + m)

(c)

m gl
(M + m)

(d)

m 2 gl
(M + m)

5. A particle of mass m kept at the origin is subjected

^
to a force F = ( pt − qx) i where t is the time elapsed
and x is the x co-ordinate of the position of the
particle. Particle starts its motion at t = 0 with zero
initial velocity. If p and q are positive constants,
then

physics for you | april ‘15

Page 8



(a) the acceleration of the particle will continuously
keep on increasing with time
(b) particle will execute simple harmonic motion
(c) the force on the particle will continuously keep
on decreasing with time
(d) the acceleration of particle will vary sinusoidally
with time.
6. Two rods of equal lengths and equal cross-sectional
areas are made of materials whose Young’s
modulii are in the ratio of 2:3. They are suspended

and loaded with the same mass. When stretched and
released, they will oscillate with time periods in the
ratio of
(a) 3 : 2
(b) 3 : 2
(c) 3 3 : 2 2
(d) 9 : 4

µ0 I
(3p + 4) 
8pR
µ I
(c) 0 (3p − 4) 
8pR

µ0 I
(3p + 4) ⊗
8pR
µ I
(d) 0 (3p − 4) ⊗
8pR

(a)

(b)

10. Two blocks of masses 2 kg and 4 kg are connected
through a massless inextensible string. The
co-efficient of friction between 2 kg block and
ground is 0.4 and the coefficient of friction between

4 kg block and ground is 0.6. Two forces F1 = 10 N
and F2 = 20 N are applied on the blocks as shown
in the figure. Calculate the frictional force between
4 kg block and ground. (Assume initially the tension
in the string was just zero before forces F1 and F2
were applied.)

7. A thin uniform annular disc (see figure) of mass M
has outer radius 4R and inner radius 3R. The work
required to take a unit mass from point P on its axis
to infinity is

(a) 24 N
(c) 18 N

(b) 8 N
(d) 10 N


solution of march
2015
crossword
1

(a)

2GM
(4 2 − 5)
7R

GM
(c)
4R

(b) −
(d)

2GM
(4 2 − 5)
7R

2GM
( 2 − 1)
5R

8. An ideal gas is expanded so that amount of heat
given is equal to the decrease in internal energy.
Find the adiabatic exponent if the gas undergoes
the process TV1/5 = constant.
(a) 7/5
(b) 6/5
(c) 8/5
(d) None of these
9. The magnetic field at the point P is given by

F L U I D I T Y
4
G E
E
7

R
S
8
Y
M
A
9
10
U L T
I C E L I N E
C
T
11
12
E
HA L L E F F E
N
X
I
15
U
C Y C L O N E
16
S C A
I
E
T
T
19
H Y P O T H E S I

O
N
O
24
23
M O M E N
B
26
R I P P L E
27
A
E
29
C
P
D
31
I
K
O
H
32
VO L C A
P
B
L
C
O
H
34

B L I N D S P O T
35
U
R
Y
G
I N
R
C
F
36
L
M E A N F R
F
P
E
E
38
R
H
E
Y

2
T
O D E S I C
T
A
R A S O N I C

C T

L A R
S
T

N

C
E
F

14

F O L I A T
17

18

F A H R

L
20
L U
A
22
T
W
O F I N E R T
I

T
U
G
30
R E D S H I F
E
T
L
O
33
E L E C T
S
A N D E S C E
N
37
E
E T I M E
S
F L U X S P E

3
C
B I O M 6A S S
H
M
E
A
S
L
I

G
13
O
A
L
M
I O N
N
E N H E I T
A
21
R I C A N T
W
O
25
M
A
A
V
P
E
28
R
E
A
P
S
O
L
S

D
A
Y
S
O M A G N E T
I
A
M
C
C E
I
I
A S T I C
T
S
Y
D
5

B
I

T

R
N
L
E

Winner (March 2015)

neha Gupta
Solution Senders
Sandeep Kumar rana
Atriz roy

10

physics for you | april ‘15

Page 10


By : Prof. Rajinder Singh Randhawa*

I

II

III
–Q

2. Two small identical balls lying on a horizontal
plane are connected by a weightless spring.
One ball (ball 2) is fixed at O and the other
(ball 1) is free. The balls are charged identically as
a resultant of which the spring length increases
h = 2 times. Determine the change in frequency?
O

2

+

+
+
+

+

4. A non-conducting ring y
of mass m and radius
R, the charge per unit
length l is shown in
–
figure. It is then placed
–
–
on
a rough non– –
x
conducting horizontal O

^
plane. At time t = 0, a uniform electric field E = E0 i
is switched on and the ring starts rolling without
sliding. Find the frictional force acting on the
ring.
5. A non-conducting solid cylindrical rod of length L
and radius R has uniformly distributed charge Q.
Find the electric field at point P, a distance L from
the centre of the rod.

R

O

P

L

1
+

Q2
3. A point charge Q1 = –125 mC is
fixed at the centre of an insulated
h
disc of mass 1 kg. The disc rests
Q1
on a rough horizontal plane.
Another charge Q2 = 125 m C is fixed vertically
above the centre of the disc at a height h = 1 m.
After the disc is displaced slightly in the horizontal
direction, find the time period of oscillation of
disc.

++

1. An infinitely long conducting wire of charge
density +l and a point charge –Q are at a
distance from each other. In which of the

three regions (I, II or III) are there points that
(a) lie on the line passing through point charge
perpendicular to the conductor and (b) at which
the field is zero?

Solution

1. In the region II, the electric field of wire and point
charge point in the same direction, +ve x-axis. So
no point can exist where the field is zero. Now, we
take a point to the right of the point charge at a
distance x from it. Resultant field at this point is

ER =

Q
λ
^
^
i+
(− i )
2
2 πε0 (x + a)
4 πε0 x

*Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh, Ph. 09814527699
physics for you | april ‘15

11

Page 11


Resultant field is zero if
λ
Q
=
( x + a) 2 x 2

Q
Qa
=0
or 2lx2 = Qx + Qa or x 2 −
x−
2λ
2λ
On solving the quadratic equation in x, we get
Q
Q 2 Qa
±
+
4λ
16 λ2 2 λ
Here, there is only one value of x (with +ve sign)
because –ve sign would mean that the point is to
the left of point charge.
Now we take a point to the left of wire at a distance
x from it.
The resultant field is


Q
λ
^
^
ER =
i
(− i ) +
2
2 πε0 x
4 πε0 (a + x )
The two fields point in the opposite directions, so
resultant field can be zero if,
Q
λ
=
2 πε0 x 4 πε0 (a + x )2
x=

Q

or x 2 −  − 2a  x + a2 = 0
 2λ

On solving the quadratic equation in x, we get
2

1 Q
1Q



x =  − 2a  ±
− 2a  − a2


4  2 λ
2  2λ
If the discriminant of the quadratic equation is
real, we have two points where the field is zero.
Discriminant is +ve for Q ≥ 8al.
2. When the balls are uncharged, the frequency of
oscillation is
1 k
2π m
where k is the force constant of the spring and
m = mass of the oscillating ball (ball 1).
When the balls are identically charged,
q2
1
...(i)
= k(ηl − l ) = kl(η − 1)
4 πε0 (ηl )2
where l is the natural length of spring and hl is the
new length of spring after its extension.
q2
... (ii)
or l 3 =
4 πε0 η2 (η − 1)k
υ0 =

When the ball 1 is displaced by a small distance

x from the equilibrium position to the right, the
unbalanced force to the right is given by

12

q2
1
− k ( ηl + x − l )
4 πε0 (ηl + x )2
Using Newton’s law,
Fr =

−2

x
1 q2 
m
1 +  − kl(η − 1) − kx
=

2
2
2
4 πε0 η l  ηl 
dt
Expanding binomially and using eqn. (i), we get

 1
2q2
d2 x

m
= −
⋅
+ k x
dt 2

 4 πε0 η3l 3
Using eqn. (ii), we get

 1
2q2
⋅
+ k

d2 x
= −  4 πε0 3
m
q2
x
2
η
dt


2
4 πε0 η (η − 1)k


 3η − 2 
 2(η − 1)


k + k x = − 
kx
= −
 η 
 η

 3η − 2  k
d2 x
...(iii)
∴
= −
x
2
 η  m
dt
d2 x

By definition of simple harmonic motion,
d2 x
2

= −ω2 x

...(iv)

dt
From eqns. (iii) and (iv), we get
 3η − 2  k
1  3η − 2  k

ω2 = 
⇒ υ=

 η m
2 π  η  m
∴

υ
3η − 2
=
υ0
η

3η − 2
times.
η
Here, h = 2 so frequency increases 2 times.
Thus frequency is increased

3. Let the radius of the
Q2
disc be R. If the disc
F
is displaced x, then
h 
q = x/R. The restoring
Fsin
torque t about the point
x
of contact of the disc

with ground,
tP = (Fsinq)R = Ia

 MR2
=
+ MR2  α

 2
Q2
where, F =
4πε0 (h2 + x 2 )
and sin θ =

Fcos
Q1

P

x
2

h + x2

physics for you | april ‘15

Page 12


Hence,

or α =

R

2

Q x
2

6 πε0 MR(h + x )
Q x

For x < < h, α ≈ −

2

Q θR

=−

6 πε0 MRh
6 πε0 MRh3
Negative sign is being introduced because angular
acceleration and angular displacement are opposite
to each other.
Thus, α = −

3

Q2θ

6 πε0 Mh3

Hence, ω =

or T = 2 π

Q2
6 πε0 Mh3
6 πε0 Mh3
Q

h
6 πε0 Mh
Q

N
+

dF

–

–
–

+ L /2

 + L /2
(L − r )dr

dr − ∫

∫
2
2
2 1/2 
2 πR Lε0  − L /2
− L /2 [(L − r ) + R ] 
The second integral can be evaluated by substituting
(L – r)2 + R2 = t.
Differentiating both sides, we get,
–2(L – r)dr = dt
 + L /2 1 dt 
Q
∴ E=
[r ]− L /2 + 2 ∫ 1/2 
2
t 
2 πR Lε0 

E=

+

+
+
+

A force of same magnitude but in opposite direction
acts on a corresponding element in the region of

negative charge.
\ Equation of motion for pure rolling is
π /2

2
∫ 2 λRdθE0 R sin θ − fR = (mR )α

and f = ma
and a = Ra
Solving eqns. (i), (ii) and (iii), we get
f = lRE0 along +ve x-axis.

Q

=

dF = lRdqE0

0


(Q / L)dr 
(L − r )
1−


πR2 (2ε0 )  [(L − r )2 + R2 ]1/2 

E=

f

mg

or 2 λR2 E0 − fR = mR2 α

L


(Q / L)dr 
(L − r )
1−

2
2
2 1/2 
− L /2 πR (2ε0 )  [(L − r ) + R ] 

4. Consider a differential element subtending an angle
dq at the centre and at angle q as shown in figure.

– –

+L/2

+ L /2

T = 0.6 s



r dr

E = ∫ dE = ∫

On substituting the given values, we get

d

P

Consider a disc of radius R of thickness dr at a
distance r from the centre O of the cylinder.
Charge on the disc,
Q
Q
dq =
× πR2dr = dr
2
L
πR L
∵ Electric field due to disc along its axis

x
σ 
Ex =
1 − 2

2ε0  (x + R2 )1/2 
Hence,

dE =

= 2π

2

O

–L/2

2 3/2

2

L

5.


 MR2
=
+ MR2  α
4 πε0 (h2 + x 2 )3/2  2

Q 2 xR

Q
2 πR2 Lε0

[L + t ]

+ L /2 

L + [(L − r )2 + R2 ]1/2  − L /2 




2 πR2 Lε0 

Q

1/2
1/2
  2

 9L2
 
L
2
2

E=
L+ +R  −
+R  
2

4
4
2 πR Lε0  



 




Q

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

physics for you | april ‘15

13

Page 13


EXAM from
th

th

14 to 29 MAY

BITSAT
FULL LENGTH PRACTICE PAPER
physics

1.

2.

3.

4.

5.

14

Suppose speed of light (c), force (F) and kinetic
energy (K) are taken as the fundamental units, then
the dimensional formula for mass will be
(a) [Kc–2]
(b) [KF–2]
–2
(c) [cK ]
(d) [Fc–2]
A sand bag of mass M is suspended by a rope. A
bullet of mass m is fired at it with speed v and gets
embeded in it. The loss of kinetic energy of the
system is
M v2
M mv 2
(a)
(b)
2(M + m)
2(M + m)

2 2
m v
1
(c)
(d) (M + m) v 2
2 (M + m)
2
A steel wire with cross-section 3 cm2 has elastic
limit 2.4 × 108 N m–2. The maximum upward
acceleration that can be given to a 1200 kg elevator
supported by this cable wire if the stress is not to
exceed one-third of the elastic limit is (Take g = 10 m s–2)
(a) 12 m s–2
(b) 10 m s–2
–2
(c) 8 m s
(d) 7 m s–2
A body of density r at rest is dropped from a height
h into a lake of density s where s > r. Neglecting
all dissipative forces, find the maximum depth to
which the body sinks before returning to float on
the surface.
hρ
h
(a)
(b)
σ
σ−ρ
hρ
hσ

(c)
(d)
σ−ρ
σ−ρ
Consider two containers A and B containing
identical gases at the same pressure, volume and
temperature. The gas in container A is compressed
to half of its original volume isothermally while
the gas in container B is compressed to half of its
original volume adiabatically. The ratio of final
pressure of gas in B to that of gas in A is

(a) 2

g–1

 1 
(c) 
 1 − γ 

2

1
(b)  
2

γ −1

 1 
(d) 

 γ − 1 

2

6.

Soap water drips from a capillary. When the drop
breaks away, the diameter of its neck is 1 mm.
The mass of the drop is 0.0129 g. Find the surface
tension of soapy water. (Take g = 9.8 m s–2)
(a) 12.9 × 10–3 N m–1 (b) 31.2 × 10–3 N m–1
(c) 40.3 × 10–3 N m–1 (d) 58.6 × 10–3 N m–1

7.

An artificial satellite is moving in a circular orbit
around the Earth with a speed equal to half the
magnitude of escape velocity from the Earth. The
height of the satellite above the Earth’s surface is
(Take radius of Earth = 6400 km)
(a) 6000 km
(b) 5800 km
(c) 7500 km
(d) 6400 km
A needle placed 45 cm from a lens forms an image
on a screen placed 90 cm on the other side of the
lens. Its focal length and the size of image if the size
of the needle is 5 cm are respectively
(a) – 30 cm, 10 cm (b) + 30 cm, – 10 cm
(c) – 20 cm, 15 cm (d) + 20 cm, – 15 cm

8.

9.

In Young’s double slit experiment distance between
two sources is 0.1 mm. The distance of screen from
the source is 20 cm. Wavelength of light used is
5460 Å. Then, angular position of the first dark
fringe is
(a) 0.08° (b) 0.16° (c) 0.20° (d) 0.32°

10. When light of wavelength 400 nm is incident on

the cathode of a photocell, the stopping potential
recorded is 6 V. If the wave of the incident light
is increased to 600 nm, then the new stopping
potential is
(a) 1.03 V
(b) 2.42 V
(c) 4.97 V
(d) 3.58 V

physics for you | april ‘15

Page 14


11. Two particles A and B describe S.H.M. of same

amplitude a and frequency u along the same
straight line. The maximum distance between
two particles is 3 a. The initial phase difference
between the particles is
(a) 2p/3 (b) p/6 (c) p/2
(d) p/2

12. A racing car moving towards a cliff sounds its

horn. The driver observes that the sound reflected
from the cliff has a pitch one octave higher than
the actual sound of the horn. If v be the velocity of
sound, the velocity of the car is
(a) v/ 2 (b) v/2
(c) v/3
(d) v/4

kinetic energy equals the original kinetic energy of
the lighter particle. What is the original speed of
the heavier particle ?
(a) (2 − 2 ) m s–1
(b) 2(1 + 2 ) m s–1
(c) (2 + 3 2 ) m s–1

(d) 4(1 − 2 ) m s–1

18. An equilateral triangle of side length l is formed from

a piece of wire of uniform resistance. The current I
is fed as shown in the figure. The magnitude of the

magnetic field at its centre O is
Q

13. The rate of cooling at 600 K, if surrounding

temperature is 300 K is R. Assume that the Stefan’s
law holds. The rate of cooling at 900 K is
16
2
(a)
(c) 3R
(d) R
R (b) 2R
3
3
14. The ratio of specific heat of gas at constant pressure
to that at constant volume is g. The change in
internal energy of one mole of gas when volume
changes from V to 2V at constant pressure P is
R
(a)
(b) PV
( γ −1)
γ PV
PV
(c)
(d)
.
γ −1
( γ −1)

O
P

(a)
(c)

16. A thin uniform rod AB of mass M and length L

is hinged at one end A to the level floor. Initially
it stands vertically and is allowed to fall freely to
the floor in the vertical plane. The angular velocity
of the rod, when its end B strikes the floor is (g is
acceleration due to gravity)
 Mg 
(a) 
 L 

 Mg 
(b) 
 3L 

g
(c)  
L

(d)  3 g 
 L 

1/2

1/2

17. A particle of mass m has half the kinetic energy of

another particle of mass m/2. If the speed of the
heavier particle is increased by 2 m s–1, its new

I

3 µ0 I

(b)

2 πl
µ0 I

3 3 µ0 I
2 πl

(d) zero

2 πl

19. Two inductors L1 and L2 are connected in parallel

and a time varying current flows as shown in figure.
The ratio of current I1/I2 at any time t is

15. A boy throws a ball upwards with velocity

u = 15 m s–1. The wind imparts a horizontal
acceleration of 3 m s–2 to the left. The angle q with
vertical at which the ball must be thrown so that
the ball returns to the boy’s hand is
(Take g = 10 m s–2)
(a) tan–1 (0.4)
(b) tan–1 (0.2)
(c) tan–1 (0.3)
(d) tan–1 (0.15)

R

I

I1

L1

I

I
I2

(a)
(c)

L2
L1

L2

(b)
L22

(d)

L1
L2
L21

(L1 + L2 )2
(L1 + L2 )2
20. An a.c. source is connected across an LCR series
circuit with L = 100 mH, C = 0.1 mF and R = 50 W.
The frequency of a.c. to make the power factor of
the circuit, unity is
10 4
(a) 2π Hz

103
(b) 2π Hz

10 −4
(c) 2π Hz

10 −3
(d) 2π Hz

21. The electric field (in N C–1) in an electromagnetic

wave is given by E = 50 sin w(t – x/c). The energy
stored in a cylinder of cross-section 10 cm2 and
physics for you | april ‘15

15

Page 15


length 100 cm along the x-axis will be
(a) 5.5 × 10–12 J
(b) 1.1 × 10–11 J
–11
(c) 2.2 × 10 J
(d) 3.3 × 10–11 J
22. In an interference experiment using waves of same

amplitude, path difference between the waves at a
point on the screen is l/4. The ratio of intensity at
this point with that at the central bright fringe is
(a) 1
(b) 0.5
(c) 1.5
(d) 2.0

23. A plane mirror is placed along the x-axis facing

negative y-axis. The mirror is fixed. A point object
^

^

is moving with 3 i + 4 j in front of the plane
mirror. The relative velocity of image with respect
to its object is
y

(c) Ex = 33 cos p × 1011 (t –

x
)
c

x
Bx = 11 × 10–7 cos p × 1011 (t – )
c
x
11
(d) Ey = 66 cos 2p × 10 (t – )
c
x
–7
Bz = 2.2 × 10 cos 2p × 1011 (t – )
c
27. The three stable isotopes of neon

10Ne

20

,

21
10Ne

and 10Ne22 have respective abundances of 90.51%,
0.27% and 9.22%. The atomic masses of the
three isotopes are 19.99 u, 20.99 u and 21.99 u
respectively. The average atomic mass of neon is
(a) 11.18 u
(b) 15.18 u
(c) 20.18 u
(d) 10.18 u
28. Light rays of wavelength 6000 Å and of photon

x

^

^

(a) − 8 j
^

(b) 8 j
^

(c) 3 i − 4 j

^

(d) − 6 i

24. A surface irradiated with light of wavelength 480 nm

gives out electrons with maximum velocity v m s–1,
the cut off wavelength being 600 nm. The same
surface would release electrons with maximum
velocity 2v m s–1 if it is irradiated by light of
wavelength
(a) 325 nm
(b) 360 nm
(c) 384 nm
(d) 300 nm

25. The ratio of the de Broglie wavelengths of proton

and alpha particle which have been accelerated
through same potential difference is
(a) 2 3 (b) 3 2 (c) 2 2 (d) 3 3

26. A plane electromagnetic wave travelling along the

x-direction has a wavelength of 3 mm. The variation
in the electric field occurs in the y-direction with an
amplitude 66 V m–1. The equations for the electric
and magnetic fields as a function of x and t are
x
(a) Ey = 33 cos p × 1011 (t – )

c
x
Bz = 1.1 × 10–7 cos p × 1011 (t – )
c
x
(b) Ey = 11 cos 2p × 1011 (t – )
c
x
By = 11 × 10–7 cos 2p × 1011 (t – )
c

16

intensity 39.6 W m–2 is incident on a metal
surface. If only 1% of photons incident on surface
emit photoelectrons, then the number of electrons
emitted per second per unit area from the surface
will be (Take h = 6.64 × 10–34 J s, c = 3 × 108 m s–1)
(a) 12 × 1018
(b) 10 × 1018
(c) 12 × 1017
(d) 12 × 1016

29. A sample contains 10–2 kg each of the two

substances A and B with half-lives 4 s and 8 s
respectively. Their atomic weights are in the ratio
of 1 : 2. Find the ratio of the amounts of A and B
after an interval of 16 seconds.
(a) 1 : 4 (b) 4 : 1 (c) 1 : 2 (d) 2 : 1

30. A diode having potential difference 0.5 V across

its junction which does not depend on current, is
connected in series with resistance of 20 W across
source. If 0.1 A current passes through resistance,
then what is the voltage of the source?
(a) 1.5 V (b) 2.0 V (c) 2.5 V (d) 5 V

31. If ground state ionisation energy of H-atom is

13.6 eV, the energy required to ionize a H-atom
from second excited state is
(a) 1.51 eV
(b) 3.4 eV
(c) 13.6 eV
(d) 12.1 eV

32. Two satellites S1 and S2 revolve around a planet in

coplanar circular orbits in the same sense. Their
periods of revolution are 1 h and 8 h respectively.
The radius of orbit of S1 is 104 km. When S2 is
closest to S1, the speed of S2 relative to S1 is
(a) p × 104 km h–1
(b) 2p × 104 km h–1
4
–1
(c) 3p × 10 km h
(d) 4p × 104 km h–1

physics for you | april ‘15

Page 16


fragments A, B and C. The momentum of A is P i
and that of B is 3 P j where P is positive number.
The momentum of C is
(a) (1 + 3 ) P in a direction making 120° with A
(b) 2 P in a direction making 150° with A
(c) 2 P in a direction making 150° with B
(d) (1 + 3 ) P in a direction making 150° with B.

34. One end of a uniform

A

rod of length l and
B
mass m is hinged at
A. It is released from
rest from horizontal
C
position AB as shown
in figure. The force
exerted by the rod B′
on the hinge when it
becomes vertical is
3

5
(a) mg (b) 3mg (c) 5mg (d) mg
2
2
35. Two bulbs 40 W and 60 W and rated voltage
240 V are connected in series across a potential
difference of 420 V. Which bulb will work at above
its rated voltage?
(a) 40 W
(b) 60 W
(c) Both 40 W and 60 W (d) None of the bulbs
36. The three resistances A, B and C have values 3R, 6R

and R respectively. When some potential difference
is applied across the network, the thermal powers
dissipated by A, B and C are in the ratio
3R
A
6R
B

R
C

(a) 2 : 3 : 4
(b) 2 : 4 : 3
(c) 4 : 2 : 3
(d) 3 : 2 : 4
37. The masses of three wires of copper are in the ratio
1 : 3 : 5 and lengths are in the ratio 5 : 3 : 1. Then the

ratio of their electrical resistances are
(a) 1 : 3 : 5
(b) 5 : 3 : 1
(c) 1 : 15 : 25
(d) 125 : 15 : 1
38. A pendulum bob of mass m carrying a charge q is
at rest with its string making an angle q with the
vertical in a uniform horizontal electric field E. The
tension in the string is
mg
qE
mg
qE
(a) sin θ and cos θ (b) cos θ and sin θ
qE
mg
(c) mg
(d) qE

39. A modulating signal is a square wave as shown in

figure.
The carrier wave is given by c(t) = 2sin(8pt) volt.
What is the modulation index?
1
O

m(t) V

33. An object initially at rest explodes into three

0.5

(a) 0.2
(c) 0.4

1

1.5

2.0

t(s)

(b) 0.3
(d) 0.5

40. A Zener diode when used as a voltage regulator is

connected
(i) in forward bias (ii) in reverse bias
(iii) in parallel with load
(iv) in series with load
(a) (i) and (ii) are correct
(b) (ii) and (iii) are correct
(c) (i) only correct
(d) (iv) only correct
chemistry

41. The correct order of first ionisation potential

among the following elements Be, B, C, N, O is
(a) B < Be < C < O < N
(b) B < Be < C < N < O
(c) Be < B < C < N < O
(d) Be < B < C < O < N

42. Amongst TiF62–, CoF63–, Cu2Cl2 and NiCl42– (At.

nos. of Ti = 22, Co = 27, Cu = 29, Ni = 28). The
colourless species are
(a) CoF63– and NiCl42– (b) TiF62– and CoF63–
(c) Cu2Cl2 and NiCl42– (d) TiF62– and Cu2Cl2

43. Identify ‘Z’ in the sequence :

(a) C6H5CN
(c) C6H5COOH

(b) C6H5CONH2
(d) C6H5CH2NH2

44. The oxidation number of S in Caro’s acid (H2SO5) is

(a) + 5
(c) + 2

(b) + 3
(d) + 6

45. The solubility of sulphates in water down the IIA

group follows the order Be > Mg > Ca > Sr > Ba.
This is due to
(a) increase in melting point
(b) increasing molecular mass
(c) decreasing lattice energy
(d) high heat of solvation of smaller ions.
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17

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46. A hydrocarbon P of the formula C7H12 on

ozonolysis gives a compound Q which undergoes
aldol condensation giving 1-acetylcyclopentene.
The compound P is
(a)

(b)

(c)

(d)

47. Identify the pair of species in which the central

atom has the same type of hybridisation.
(a) BF3 and NCl3
(b) H2S and BeCl2
(c) NCl3 and H2S
(d) SF4 and BeCl2
compound is

CH3

(b) CH3 CH2 CH CH2
(c) CH3 C CH2

(c) 0 , 1, 1

(d) 0 , 1 , 1

n

n

55. The equilibrium constant of the reaction,

1
O
2 2(g)
is 0.15 at 900 K. The equilibrium constant for
2SO2(g) + O2(g)
2SO3(g) is
SO2(g) +

(b) 49.72 mol–1 L
(d) 44.44 mol–1 L

56. In the Hoope’s process for refining of aluminium,

CH3

(d) CH3 CH CH2 CH2
CH3

49. The cubic unit cell of aluminium (molar mass

27.0 g mol–1) has an edge length of 405 pm and
density 2.70 g cm–3. What type of unit cell is it?
(a) Face-centred
(b) Body-centred
(c) Simple cubic
(d) None of these.

50. Al2O3 on heating with carbon in an atmosphere of

Cl2 at high temperature produces
(a) Al + CO2
(b) Al + CO2 + NO
(c) Al4C3 + CO2
(d) AlCl3 + CO
following

does

not

show

(b) CH3CHO
(d) C6H5COC(CH3)3

52. Which of the following has least covalent P H

(b) P2H6
+
(d) PH6

53. What products are expected from the dispropor-

tionation reaction of hypochlorous acid?
(a) HClO3 and Cl2O (b) HClO2 and HClO4
(c) HCl and Cl2O
(d) HCl and HClO3

18

(b) 1 , 1, 0

n

(a) 52.52 mol–1 L
(c) 63.34 mol–1 L

CH3

bond?
(a) PH3
(c) P2H5

(a) 1, 1 , 0

SO3(g)

(a) CH3 CH2 CH2 CH2 CH2

of the
tautomerism?
(a) C6H5COCH3
(c) CH3COCH3

x, y and z are respectively

n

48. The structure of neo pentyl group in an organic

51. Which

54. In the given Freundlich adsorption isotherm plot,

the fused materials form three different layers and
they remain separated during electrolysis also. This
is because
(a) there is a special arrangement in the cell to keep

the layers separate
(b) the three layers have different densities
(c) the three layers are maintained at different
temperature
(d) the upper layer is kept attracted by the cathode
and the lower layer is kept attracted by the
anode.
57. Which of the following complexes will give white
precipitate with BaCl2(aq)?
(a) [Co(NH3)4SO4]NO2
(b) [Cr(NH3)4SO4]Cl
(c) [Cr(NH3)5Cl]SO4
(d) Both (b) and (c).
58. At 25°C, the molar conductivity of 0.001 M
hydrofluoric acid is 184.5 W–1 cm2 mol–1. If its L°m
is 502.4 W–1 cm2 mol–1, then equilibrium constant
at the given concentration is
(a) 3.607 × 10–4 M
(b) 5.404 × 10–4 M
(c) 2.127 × 10–4 M
(d) 6.032 × 10–4 M
59. Zinc on reacting with cold, dil. HNO3, gives

(a) ZnNO3
(c) NO2

(b) NH4NO3
(d) NO

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60. Which of the following is aromatic?

(a)

+

(c)

(b)

–

(d)

61. Which of the following statements is not correct?

(a) Some antiseptics can be added to soaps.
(b) Dilute solutions of some disinfectants can be
used as an antiseptic.
(c) Disinfectants are antimicrobial drugs.
(d) Antiseptic medicines can be ingested.

62. Which molecule/ion out of the following does not

contain unpaired electrons?
(a) N2+ (b) O2

(c) O22–

(d) B2

63. Energy of an electron in hydrogen atom is given by
E=

13.6
n2

eV. Which one of the following statements

is true if n is changed from 1 to 3?
Energy will
(a) decrease three times
(b) increase three times
(c) increase nine times
(d) decrease nine times.
64. Structure of the compound whose IUPAC name is

3-Ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid is
(a)

(b)

(b)
(c)
(d)

+ Zn/Hg and conc. HCl

+ CrO2Cl2 in CS2 followed by H3O+
+ H2 in presence of Pd-BaSO4

68. A reaction having equal energies of activation for

forward and reverse reactions has
(a) DH = 0
(b) DH = DG = DS = 0
(c) DS = 0
(d) DG = 0
69. Amongst the acids,
(i) CH CCOOH
(ii) CH2 CHCOOH and
(iii)CH3CH2COOH,
the acid strength follows the sequence
(a) (i) < (ii) > (iii)
(b) (i) > (ii) > (iii)
(c) (i) = (ii) = (iii)
(d) (i) = (ii) > (iii)

70. Ionization potential of hydrogen atom is 13.6 eV.

Hydrogen atom in ground state is excited by
monochromatic light of energy 12.1 eV. The
spectral lines emitted by hydrogen according to
Bohr’s theory will be
(a) one (b) two (c) three (d) four.

71. Which of the following reactions will yield

2-propanol?
H+
→
I. CH2 CH CH3 + H2O 
(i) CH MgI

3
→
II. CH3 CHO 
(ii) H O

(i) C H MgI

2

2 5
→
III. CH2O 
(ii) H O
2

Neutral KMnO

(c)

(d)

65. The value of Planck’s constant is 6.63 × 10–34 J s.

The speed of light is 3 × 1017 nm s–1. Which value

is closest to the wavelength in nanometer of a
quantum of light with frequency of 6 × 1015 s–1?
(a) 50
(b) 75
(c) 10
(d) 25

66. Identify the correct order of solubility in aqueous

medium.
(a) Na2S > CuS > ZnS (b) Na2S > ZnS > CuS
(c) CuS > ZnS > Na2S (d) ZnS > Na2S > CuS

67. Reaction by which benzaldehyde cannot be

prepared
(a)

+ CO + HCl in presence of anhydrous
AlCl3

4
→
IV. CH2 CH CH3 
(a) I and II
(b) II and III
(c) III and I
(d) II and IV
72. The heat liberated when 1.89 g of benzoic acid is
burnt in a bomb calorimeter at 25°C and it increases

the temperature of 18.94 kg of water by 0.632°C. If
the specific heat of water at 25°C is 0.998 cal/g-deg,
the value of the heat of combustion of benzoic acid is
(a) 881.1 kcal
(b) 771.12 kcal
(c) 981.1 kcal
(d) 871.2 kcal
73. Which of the following combinations illustrates
the law of reciprocal proportions?
(a) N2O3, N2O4, N2O5(b) NaCl, NaBr, NaI
(c) CS2, CO2, SO2
(d) PH3, P2O3, P2O5
74. Which of the following is formed, when
benzaldehyde reacts with alcoholic KCN?
(a) Benzoin
(b) Benzyl alcohol
(c) Benzoic acid
(d) Ethyl benzoate

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19

Page 19


75. The standard electrode potentials for Pb2+| Pb and

Zn2+ | Zn are – 0.126 V and – 0.763 V respectively.
The e.m.f. of the cell

Zn | Zn2+(0.1 M) || Pb2+(0.1 M)|Pb is
(a) 0.637 V
(b) < 0.637 V
(c) > 0.637 V
(d) 0.889 V

76. Which of the following will be most acidic?

(a)

(b)

(c)

(d)

KCN forms yellow, white and reddish-brown
precipitate. (X) gives insoluble complex with
excess of KCN and no ppt. upon passing H2S gas.
(Y) also gives insoluble complex with excess of
KCN but gives yellow ppt. on passing H2S gas. (Z)
gives yellow solution with excess of KCN. Then X,
Y and Z respectively are
(a) Cu2+, Cd2+, Fe3+ (b) Cu2+, Fe2+, Cd2+
(c) Pb2+, Cd2+, Cu2+ (d) Fe3+, Pb2+, Fe2+

78. Which of the following represents the correct order

of increasing electron gain enthalpy with negative
sign for the elements O, S, F and Cl?

(a) O < S < F < Cl
(b) F < S < O < Cl
(c) S < O < Cl < F
(d) Cl < F < O < S

79. Potassium permanganate has intense purple colour

due to
(a) weak d-d transitions
(b) metal to ligand charge transfer
(c) ligand to metal charge transfer
(d) both metal and ligand transitions.

TsCl

H

(b)
N

CH NO

HO

O

HO
HO

N

(d)

H

20

Q

Sn/HCl, 

H

(c)

82. Let R be a relation in N defined by

R = {(x, y) : x + 2y = 8}, then range of R is
(a) {2, 4, 6}
(b) {1, 2, 3, 4, 6}
(c) {1, 2, 3}
(d) none of these.

N
H

set B = {1, 3, 5} i.e. (a, b) ∈ R ⇔ a < b, then RoR–1
equals
(a) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
(b) {(3, 3), (5, 3), (3, 5), (5, 5)}

(c) {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}
(d) {(4, 5), (3, 4), (3, 3)}.

84. The straight lines joining the origin to the points of

intersection of the straight line hx + ky = 2hk and
the curve (x – k)2 + (y – h)2 = c2 are at right angles, then
(a) h2 + k2 + c2 = 0
(b) h2 – k2 – c2 = 0
2
2
2
(c) h + k – c = 0
(d) none of these.

85. The exhaustive range of values of a such that the

angle between the pair of tangents drawn from
(a, a) to the circle x2 + y2 – 2x – 2y – 6 = 0 lies in the
range (p/3, p) is
(a) (1, ∞)
(b) (–5, –3) ∪ (3, 5)
(c) (−∞, − 2 2 ) ∪ (2 2 , ∞)
(d) (–3, –1) ∪ (3, 5)

86. If a = tan27q – tanq and

sin θ sin 3θ sin 9θ
+
+

, then
cos 3θ cos 9θ cos 27θ
(a) a = b
(b) a = 2b
(c) b = 2a
(d) none of these.

2
P NaOH,3 MeOH

R

(a)

(a) 2 sin x sin y sin z (b) 2 cos x cos y cos z
(c) 2 sin x cos y cos z (d) 2 sin x sin y cos z

β=

80. Identify ‘R’ in the following series of reactions.
OH Pyridine

81. If x + y = π + z, then sin2x + sin2y – sin2z is equal to

83. If R be a relation ‘<’ from A = {1, 2, 3, 4} to the

77. Three metal ions (X), (Y), (Z) on treatment with

O

mathematics

NH2

87. The graph of f(x) = cosx cos(x + 2) – cos2(x + 1) is

(a) a straight line through (p/2, –sin21) and parallel
to x-axis
(b) a parabola with vertex (1, –sin21)
(c) a straight line passing through origin
(d) none of these.
88. Seven coupons are selected at random one
at a time with replacement from 15 coupons
numbered 1 to 15. The probability that the largest
number appearing on a selected coupon is 9, is
9
(a)  
 16 
(c)  3 
5

7

6

8
(b)  
 15 

7

(d) none of these

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89. If the letters of the word MATHEMATICS are

arranged arbitrarily, the probability that C comes
before E, E before H,H before I and I before S is
1
1
1
1
(a)
(b)
(c)
(d)
120
720
24
75
90. Suppose A1, A2, A3, ..., A30 are thirty sets each with
five elements and B1, B2, B3, ..., Bn are n sets each
30

n

i =1

i =1

with three elements such that ∪ Ai = ∪ Bi = S .
If each element of S belongs to exactly ten of the
Ai’s and exactly 9 of the Bi’s, then the value of n is
(a) 15
(b) 135
(c) 45
(d) 35
91. The value of
 50   5   40   5   30   5   20   5  10 
 6  −  1   6  +  2   6  −  3   6  +  4   6 
n 
where   denotes nCr , is
r 
(a) 15625
(b) 0
(c) 1000000
(d) 2250000
∞

 3 
92. The value of ∫  2  dx {where [·] is greatest
0  x + 1
integer function} is
1
3
(a)

(b)
2
2
2
(c)
(d) none of these
3
93. The value of

3

∫

[x 2 − 2x + 2]dx , where [·] is greatest

0

integer function is
(a) 6 − 3 − 2
(c) 8 − 3 − 2

(b) 6 + 3 + 2
(d) 2 + 2

94. f (x) is a continuous function such that

f(x + 4) = f (x + 2) – f (x). The value of
12

(a)

(c)

∫

6

f (x)dx

(b)

0
8

95. The range of

(c)  π 
2

f (x)dx is

λ

f (x)dx

(d) none of these
1

sin xdt

∫ 1 − 2t cos x + t 2

−1

(a) {p}

∫

0

∫ f (x)dx
0

∫

λ+12

is x ∈ (0, 2p)

 π π
(b) − , 
 2 2
(d) {– p, p}

96. Given a function g continuous everywhere such that
x

1

1
2

g(1) = 5 and ∫ g (t )dt = 2 . If f (x) = ∫ ( x − t ) g (t )dt ,
2
0
0
then f ′′′(1) − f ′′(1) is
(a) 2
(b) 4
(c) 3
(d) 5
97. The least value of the function
x
 π 3π 
F(x) = ∫ (4 sin t + 3 cos t )dt in  , 
4 4 
π /6

(a)

4 3− 2
2

(b)

4 3 −3− 2
2

−4 3 + 3 + 2
3+ 2
(d)
2

2
2
98. For x – (a + 3)|x| + 4 = 0 to have real solutions, the
range of a is
(a) (– ∞, –7] ∪ [1, ∞) (b) (–3, ∞)
(c) (– ∞, –7]
(d) [1, ∞)
(c)

99. If the equation x2 + ax + b = 0 has distinct real roots

and x2 + a|x| + b = 0 has only one real root, then
which of the following is true?
(a) b = 0, a > 0
(b) a = 0, b > 0
(c) b > 0, a < 0
(d) b < 0, a > 0
2 − 1
1
2 , then det(Adj(Adj A)) =


1
 2 −1
 1

100. If A =  −1

(a) 144
(b) 143

2
(c) 14
(d) 14
101. The existence of the unique solution of the system
of equations x + y + z = l, 5x – y + mz = 10,
2x + 3y – z = 6 depends on
(a) m only
(b) l only
(c) both l and m
(d) neither l nor m
102. If A and B are square matrices such that

B = –A–1BA , then
(a) AB + BA = 0
(b) (A + B)0 = A2 + B2 + AB
(c) (A + B)2 = A2 + 2AB + B2
(d) (A + B)2 = A + B
1

1
103. If A = 2

2
1
2

3 a

(a) a = 2, b = 1
(c) a = 2, b = –1

2
− 2 is an orthogonal matrix, then

b
(b) a = –2, b = –1
(d) a = –2, b = 1
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21

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104. If A and B are square matrices of order 3 × 3 such

that A is an orthogonal matrix and B is a skew symmetric matrix, then which of the following statements is true?
(a) |AB| = 1
(b) |AB| = 0
(c) |AB| = –1
(d) none of these

105. The least value of the expression

2log10 x – logx(0.01) for x > 1 is
(a) 10
(b) 2
(c) – 0.01 (d) 4

y = 2x – 2x2 is

(a) 7/12 (b) 5/2

(c) 1/12 (d) 4

49
113. sin 51x (sin x) dx equals

∫

(a)

sin 50x (sin x)50
50
cos 50x (sin x)50

+C

x
x

2
107. If f (x) = (a − 3a + 2)  cos2 − sin2  + (a − 1)x + sin 1

4
4

+C
50
cos 50x (cos x)50
(c)

+C
50
sin 50x (sin x)51
(d)
+C
51
114. If f be a polynomial function satisfying
f (x2 + x + 3) + 2 f (x2 – 3x + 5) = 6x2 – 10x + 17
∀ x ∈R then
(a) f is a decreasing function
(b) f(x) = 0 has a root in (0, 2)
(c) f(x) is an odd function
(d) no such polynomial exist

108. If x ∈ (– ∞, –2) and y3 – 3y + x = 0, then

115. If |f ′′(x)| ≤ 1 ∀ x ∈ R and f(0) = 0 = f ′(0) then

106. The expression {x + (x 3 − 1)1/2}5 + {x − (x 3 − 1)1/2}5is

a polynomial of degree
(a) 5
(b) 6
(c) 7

(d) 8

possesses critical points, then the set of values of ‘a’
are
(a) (– ∞, 0] ∪ [4, ∞)

(b) (– ∞, 0] ∪ [4, ∞) ∪ {1}
(c) (– ∞, 0) ∪ (4, ∞) ∪ (4, ∞) ∪ {1}
(d) none of these
(a) y is not a function of x
(b) y is not a monotonic function
(c) y is an increasing function of x
(d) y is a decreasing function of x

1
 1

ln x, x ∈  , 3  ,
2
 3

then greatest value of f (x) is
π 1
π 1
(a) − ln 3
(b) + ln 3
6 4
6 2
1
π
π
(c) ln 3 −
(d) + ln 3
4
6
6 4

−1
109. If f (x) = cot x +

1
−1
[x] + [−x] + 3− | x | + 2
110. The range of f (x) = tan
x

is (where [.] denotes G.I.F)
1 
1
(a)  , ∞ 
(b)   ∪ 2, ∞ )
4 
4
1
5


5

(c)  , , 1 + 2 
(d)  , 1 + 2 
9 4

4



111. The solution of differential equation
x2dy – y2dx + xy2(x – y)dy = 0 is
(a) ln

x + y y2
xy
y2
+
=C
+
= C (b) ln
x−y 2
x−y 2

2
2
(c) ln x − y + y = C (d) ln x − y + y = C
2
xy
x+y 2

22

112. The area of the region bounded by y = x ln x,

(b)

which of the following cannot be true?
1 1
 1 1

(a) f   =
(b) f  −  =
3 5
 3  12
1
(c) f (3) = 4
(d) f (−3) =
3
 1 
f
(
x
)
=
116. Let
 sin{x}  , where [.] and {.} denote


the greatest integer and fractional part function
respectively. The range of f is
(a) the set of integers = I
(b) the set of natural numbers = N
(c) the set of whole numbers = W
(d) {2, 3, 4, ....}
117. The chord of contact of tangents from any point

of circle x2 + y2 = a2 with respect to the circle
x2 + y2 = b2 touches the circle x2 + y2 = c2 where
(a, b, c > 0) then
a+c

(a) b <
2
1
1
1
(b)
,
,
are in A.P.
1 + log a 1 + log b 1 + log c
(c) a, b, c are in A.P.
(d) b > ac

118. The locus of mid-point of the chord of the circle with

diameter as minor axis of the ellipse

x2

y2
+
=1
a 2 b2

physics for you | april ‘15

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(a > b) which subtend right angle at centre of ellipse is

2

a +b
2
2
b
2
2
(c) 2(x2 + y2) = b2
(d) x + y =
4
119. Let a, b, g are the roots of the equation
(b) x 2 + y 2 =

(a) x2 + y2 = 2b2

2

 x2 + x + 1

x →∞

 29 23 
(c) circumcentre of the triangle is  , − 
 8 16 
(d) centroid of the triangle is  1 , − 1

 4
sec2 x

∫ (sec x + tan x)9/2 dx

equals (for

some arbitrary constant K)
1
1 1
(a) −
− (sec x + tan x)2 + K
11/2 11 7
(sec x + tan x)
(b)

1

(sec x + tan x)

(c) −
(d)

11/2

1

}
}
}
}

1 1

− (sec x + tan x)2 + K
11 7

11/2

(sec x + tan x)

1

{

{

{

{

colours can be distributed among 3 persons so that
each person gets at least one ball is
(a) 75
(b) 150
(c) 210 (d) 243

124. If lim 


1 
 1  1 
x3 + 3x2 – 6x – 8 = 0. If  , α  ,  , β  and  , γ 
α  β 

γ 
are the vertices of the triangle, then
 3

(a) centroid of the triangle is  − , − 1
 4

 1

(b) orthocentre of the triangle is  − , − 8 
 8


120. The integral

123. The total number of ways in which 5 balls of different

1 1
+ (sec x + tan x)2 + K
11 7

1 1
+ (sec x + tan x)2 + K
11/2 11 7
(sec x + tan x)

121. Let z be a complex number such that the imaginary

part of z is nonzero and a = z2 + z + 1 is real. Then
a cannot take the value

1
3
1
(a) –1
(b) 3
(c)
(d) 4
2


π
 x 2 cos , x ≠ 0
, x ∈ R, then f is
122. Let f (x) = 
x

x =0
0,
(a) differentiable both at x = 0 and at x = 2
(b) differentiable at x = 0 but not differentiable at
x=2
(c) not differentiable at x = 0 but differentiable at
x=2
(d) differentiable at neither at x = 0 nor at x = 2

x +1

(a) a = 1, b = 4
(c) a = 2, b = –3


− ax − b  = 4, then

(b) a = 1, b = – 4
(d) a = 2, b = 3

125. The function f : [0, 3] → [1, 29], defined by

f(x) = 2x3 –15x2 + 36x + 1, is
(a) one-one and onto
(b) onto but not one-one
(c) one-one but not onto
(d) neither one-one nor onto

english and logical reasoning

Directions (Questions 126 to 128) : Read the passage
and answer the following questions.
Books are, by far, the most lasting product of human
effort. Temples crumble into ruin, pictures and statues
decay, but books survive. Time does not destroy the
great thoughts which are as fresh today as when they
first passed through their author’s mind. These thoughts
speak to us through the printed page. The only effect
of time has been to throw out of currency the bad
products. Nothing in literature which is not good can
live for long. Good books have always helped man in
various spheres of life. No wonder that the world keeps
its books with great care.
126. Of the product of human effort, books are the

most
(a) Permanent
(c) Enjoyable

(b) Important
(d) Useful.

127. Time does not destroy books because they contain

(a) Useful material
(b) Subject matter for education
(c) High ideals
(d) Great ideas.

128. “To throw out of currency” means

(a) Destroy
(c) Extinguish

(b) Put out of use
(d) Forget.

Directions (Questions 129 to 130) : Pick out the
correct synonyms for each of the following words.
129. Eradicate

(a) Dedicate
(c) Complicate

(b) Eliminate
(d) Indicate
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23

Page 23


130. Myopic

(a) Astigmatic
(c) Blind

(b) Cross-eyed
(d) Short-sighted

Directions (Questions 131 to 132) : In each of the
following questions, an idiomatic expression/a proverb
has been given, followed by some alternatives. Choose the
one which best expresses the meaning of the given idiom
or proverb.
131. To take the wind out of another’s sails

(a) To defeat the motives of another
(b) To anticipate another and to gain advantage
over him
(c) To manouevre to mislead another on the high
seas
(d) To cause harm to another

132. To keep the ball rolling

(a) To keep the conversation going
(b) To make the best use of
(c) To earn more and more
(d) To work constantly

Directions (Questions 133 to 137) : Rearrange
the given five sentences A, B, C, D and E in the proper
sequence so as to form a meaningful paragraph and then
answer the questions given below them.
A. The alternative was a blitz by the health workers
to popularise preventive measures.
B. This information was considered inadequate.
C. People have been reading about AIDS in the
mass media.
D. Nobody is sure as to how effective this would
be.
E. People were also not being influenced enough
to take preventive measures.
133. Which sentence should come first in the

paragraph?
(a) C
(b) D

(c) B

(d) A

134. Which sentence should come second in the

paragraph?
(a) C
(b) D

(c) B

137. Which sentence should come last in the

paragraph?
(a) A
(b) D

(c) B

(d) E

Directions (Questions 138 to 140) : In each of the
following questions, a word has been written in four
different ways out of which only one is correctly spelt.
Find the correctly spelt word.
138. (a) Garuntee

(b) Guaruntee
(d) Guarantee

139. (a) Benefited

(b) Benifitted
(d) Benifited

140. (a) Efflorescance

(b) Eflorescence
(d) Efflorescence

(c) Gaurantee
(c) Benefeted
(c) Efflorascence

141. There is a certain relationship between two

given words on one side of : : establish a similar
relationship on another side of : : by selecting a
word from the given options.
Doctor : Patient : : Politician : ?
(a) Masses
(b) Voter
(c) Power
(d) Chair

142. In the following question, four words have been

given, out of which three are alike in some manner, while
the fourth one is different. Choose out the odd one.
(a) Seminar
(b) Semicolon
(c) Semifinal

(d) Semicircle
143. Find the missing character in the following :
?

235

4
7

117
59

(a) 327

(b) 386

29

(c) 438

15

(d) 469

144. Select a figure from amongst the Answer Figures

which will continue the same series as established
by the five Problem Figures.
Problem Figures

(d) E

135. Which sentence should come third in the

paragraph?
(a) C
(b) D

(c) B

(d) E

136. Which sentence should come fourth in the

paragraph?
(a) C
(b) D

24

(c) B

(d) A

(a)

(b)

(c)

(d)

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145. In the following question, find out which of the

figures (a), (b), (c) and (d) can be formed from the
pieces given in fig. (X).

150. In the following number series, one term is wrong.

Find out the wrong term.
1, 3, 10, 21, 64, 129, 356, 777
(a) 21
(b) 129
(c) 10

(d) 356

solution
x

(a)

(b)

(c)

(d)

146. Select the missing term.

A, D, H, M, ?, Z
(a) T
(b) G

(c) N

(d) S

147. In the following question, find out which of the

answer figures (a), (b), (c) and (d) completes the
figure matrix ?

2. (a) : According to law of conservation of linear
momentum, mv = (M + m) V
or
V = m v / (M + m)
1
Initial KE of the system = mv 2
2
1
Final KE of the system = (M + m) V2
2

?

(a)

(b)

(c)

(d)

148. In the following question, a set of figures carrying

certain characters, is given. Assuming that the
characters in each set follow a similar pattern, find
the missing character.
15
5

(a) 35

80

2

9

6

4

(b) 48

65

7

13

6

11

(c) 72

?

16
8

(d) 120

149. In the following question, choose the set of figures

which follows the given rule.
Rule : Closed figures gradually become open and
open figures gradually become closed.
(a)
(b)
(c)
(d)

1. (a) : Let M = kc F y K z
where k is a dimensionless constant.
\ [M1 L0 T0] = [LT–1]x [MLT–2]y [ML2 T–2]z
= M y+z Lx+y+2z T –x–2y–2z
Applying principle of homogeneity of dimensions,
y+z=1
...(i)
x + y + 2z = 0
...(ii)
– x – 2y – 2z = 0
...(iii)
Adding (ii) and (iii), we get, y = 0
Now, from (i) z = 1 – y = 1
from (ii) x = – y – 2z = 0 – 2
\
[M] = [c–2 F 0 K1] = [Kc–2]

2
2
1
1 (m v )
 mv 
= (M + m) 
=
 M + m 
2
2 (M + m)
2
1
1 (m v )

Loss of KE = m v 2 −
2
2 (M + m)

mv 2  M + m − m  m M v 2
=
=
2  M + m  2 (M + m)
3. (b) : Maximum tension an elevator can tolerate is
1
T = stress × area of cross-section
3
1
= × (2.4 × 108) × (3 × 10–4) = 2.4 × 104 N
3
If a is the maximum upward acceleration of
elevator then T = m (g + a)
or 2.4 × 104 = 1200 (10 + a)
On solving, a = 10 m s–2.
4. (c) : The speed of the body just before entering the
liquid is u = 2 gh . The buoyant force FB of the lake
(i.e., upward thrust of liquid on the body) is greater
than the weight of the body W, since s > r. If V is
the volume of the body and a is the acceleration of
the body inside the liquid, then FB – W = ma
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25

Page 25

sVg–rVg=rVa

(σ − ρ) g
or (s – r) g = r a or a =
ρ
Using the relation, v2 = u2 + 2as, we have
hρ
2
(σ − ρ)
0 = 2 gh − 2 g
s or s =
ρ
σ−ρ

(

)

5. (a) : For isothermal compression of gas A,
PV
P f = i i = 2Pi
Vf

1
(∵ Vf = Vi )
2

For adiabatic compression of gas B,

9. (b) : Angular position of first dark fringe,
λ
λ
θ1 = (2 × 1 − 1) =
2d 2d
5460 × 10 −10
=
= 2730 × 10 −6 rad
−3
2 × 0.1 × 10
180°
= 2730 × 10 −6 ×
= 0.16°.
π
10. (c) : As Kmax = hn – f0
hc
hc φ
or eV0 = − φ 0 or V0 = − 0
λ
eλ e
\ DV0 = (V0)1 – (V0)2
 hc φ 0   hc φ 0  hc  1
1 
− =  − 
=
− −
 eλ1 e   eλ 2 e  e  λ1 λ 2 

γ

V 
P f′ = Pi  i  = 2 γ Pi
 Vf 
P f′

2 γ Pi
Thus,
=
= 2 γ −1
2Pi
Pf
6. (c) : When the drop breaks away from the capillary,
weight of the drop = force of surface tension acting
on it due to capillary, i.e.,
mg
mg = (πD) × T or T =
...(i)
πD
Here, m = 0.0129 g = 1.29 × 10–5 kg, g = 9.8 m s–2,
D = 1 mm = 10–3 m
From eqn. (i),
T=

(1.29 × 10 −5 kg )(9.8 m s −2 )
3.14 × 10 −3 m

= 40.3 × 10 −3 N m −1

v

As v 0 = e ,
2

⇒

2 gR
gR 2
gR 2 gR
=
or
=
,
2
R+h
R+h 2
h = R = 6400 km

8. (b) : Here, u = – 45 cm, v = + 90 cm
Using thin lens formula,
1 1 1 1 1 1+ 2
= − = + =
f v u 90 45 90
\ f = + 30 cm
h2 v
h
90
or 2 =
=
Magnification, m =
5 − 45

h1 u
\
26

Size of image, h2 = – 10 cm.

\

11. (a) : y1 = a sin wt and y2 = a sin (wt + q)
y2 – y1 = a 3 = a sin (wt + q) – a sin wt
(ωt + θ) + ωt
(ωt + θ) − ωt
sin
2
2
= 2a cos (wt + q / 2) sin q / 2
For maximum value, cos (wt + q / 2) = 1
\
3 a = 2 a sin q / 2
or

or

3 a = 2 a cos

sin θ / 2 =

3
π
θ π

2π
= sin or =
or θ = .
2
3
2 3
3

12. (c)
13. (a) : Rate of cooling is proportional to (T 4 – T 40),
as per Stefan’s law.
R′ 900 4 − 300 4
\
=
R 600 4 − 300 4

gR 2
7. (d) : v 0 =
and v e = 2 gR
R+h

\

6.6 × 10 −34 × 3 × 108  1
1 
−


−19
−7

 4 × 10
1.6 × 10
6 × 10 −7 
= 1.03 V
(V0)2 = (V0)1 – 1.03 = 6 – 1.03 = 4.97 V
=

[∵ h1 = 5 cm]

=

9 4 − 34
4

4

=

34 (34 − 1)
4

4

=

80 16
=
15 3

6 −3

3 (2 − 1)
16
or R′ = R.
3
CP
C − CV
14. (c) : As
\ P
=γ
= γ −1
CV
CV
C − CV
R
or CV = P
=
γ −1
γ −1
RdT P dV
∆U = nCV dT = n
=
( γ − 1) γ − 1
P (2V − V ) PV
=
=
γ −1
γ −1

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15. (c) : Here, uy = u cos q = 15 cos q
ux = u sin q = 15 sin q
y

u = 15 m s–1

θ

x

Time of flight of the ball is
T=

2u y

=

2 × 15 cos θ

= 3 cos θ

...(i)
g
10
The boy will catch the ball if in time T, displacement
of the ball in horizontal direction should also be
1

zero. So 0 = uxT − a xT 2
2
2ux 2(15 sin θ)
or T =
...(ii)
=
= 10 sin θ
ax
3

From (i) and (ii), 3 cos q = 10 sin q
3
or tan θ = = 0.3 or θ = tan −1 (0.3).
10
16. (d) : As the rod is hinged at one end, its moment of
ML2
inertia about this end is I =
.
3
Total energy in upright position = total energy on
striking the floor
0+

MgL 1 2
1 ML2 2
= Iω + 0 =
ω
2
2
2 3

g=

3g
Lω 2
or ω =
3
L

17. (b)
18. (d) :

P
I

O
r
2I/3

I/3

e1 = e2
 dI 
 dI 
L1  1  = L2  2 
 dt 
 dt 
Integrating both sides, we get
I
L

L1I1 = L2 I 2 ∴ 1 = 2
I 2 L1
20. (a)
21. (b) : Energy contained in a cylinder
U = average energy density × volume
1
= ∈0 E02 × Al
2
1
= × (8.85 × 10 −12 ) × (50)2 × (10 × 10 −4 ) × 1
2
= 1.1 × 10–11 J
22. (b) : As IR = I1 + I2 + 2 I1I 2 cos φ
2π λ π
Here, φ =
× = , I =I =I
λ 4 2 1 2
π
\ I R = I + I + 2I cos = 2 I
2
At the central bright fringe, I′ = 4I
\ I R = 2 I = 0. 5
I ′ 4I
^
^

23. (a) : Velocity of object, v ob = 3 i + 4 j

^
^

^
^
^



v rel = v image − v ob = (3 i − 4 j) − (3 i + 4 j) = − 8 j

R
I

The magnetic field induction at O due to current
through PR is
B1 =

19. (a) : As the inductors are in parallel, therefore,
induced e.m.f. across the two inductors is the same i.e.

^
^

Velocity of image v image = 3 i − 4 j
Relative velocity of image with respect to its object

Q
I/3

µ 0 (I / 3)
µ 2I
[sin 30° + sin 30°] = 0 .

4π r
4 π 3r
It is directed inside the paper.
\ Resultant magnetic field induction at O is
B1 – B2 = 0.
B2 = 2 ×

µ0 2I / 3
µ 2I
[sin 30° + sin 30°] = 0
.
4π r
4π 3 r

It is directed outside the paper.
The magnetic field induction at O due to current
through PQR is

24. (d) 25. (c)

26. (d) : E0 = 66 V m–1
E
66
= 2.2 × 10 −7 T
B0 = 0 =
c 3 × 108
Since electromagnetic wave is of transverse nature,
hence if electric field is along y-axis the magnetic
field must be along z-axis, since the propagation of
wave is along x-axis. Thus the equations given in

option (d) are correct.
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27

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27. (c) : The masses of three isotopes are 19.99 u,
20.99 u, 21.99 u.
Their relative abundances are 90.51%, 0.27% and 9.22%.
\ Average atomic mass of neon is
90.51 × 19.99 + 0.27 × 20.99 + 9.22 × 21.99
m=
(90.51 + 0.27 + 9.22)
1809.29 + 5.67 + 202.75 2017.7
=
=
= 20.18 u
100
100
28. (c) : Useful intensity for the emission of electron is
1
× 39.6 = 0.396 W m −2
I ′ = 1% of I =
100
hc
Energy of each photon =
λ
=

(6.64 × 10

−34

8

) × (3 × 10 )

−19

= 3.32 × 10
J
6000 × 10 −10
Number of photoelectrons emitted per second per
unit area
0.396
=
≈ 12 × 1017
3.32 × 10 −19
29. (a) : Let N0 be the initial amount of a radioactive
substance. Then the amount left after n half-lives
n
1
will be N = N 0  
2
16 s
t
=
=4

For A : nA =
T1/2 4 s
4
1
N A = 10 −2 kg   = 6.25 × 10 −4 kg
2
16 s
For B : nB =
=2
8s
2
1
\
N B = 10 −2 kg   = 2.5 × 10 −3 kg
2
NA 1
=
NB 4
30. (c) : The circuit diagram is as shown below :

GM
mv 2 GMm
=
⇒ v2 =
2
R
R
R
2πR
4

π2R 2 GM
Also,
v=
⇒ v2 =
=
2
T
R
T
4π 2R 3
2
\
T =
GM
If T1 and T2 are the time periods for satellite S1 and
S2 respectively

32. (a) :

2

 T1   R1 
 T  =  R 
2
2

T 
⇒ R2 =  2 
 T1 

2 /3

R1

Here, T1 = 1 h, T2 = 8 h, R1 = 104 km
\

8
R2 =  
1

2 /3

× 10 4 km = 4 × 10 4 km

2πR1 2π × 10 4
=
= 2π × 10 4 km h −1
T1
1
2πR2 2π × 4 × 10 4
v2 =
=
= π × 10 4 km h −1
T2
8
v1 =

Relative velocity of S2 with respect to S1 is
v = v2 – v1 = (p × 104 – 2p × 104) km h–1

|v| = p × 104 km h–1
33. (c) : Here Px = P and Py =

\

3P

Resultant momentum of A and B
P = Px2 + Py2 = P 2 + ( 3 P)2 = 2 P

It is along OC′.
As is clear from figure

Y

C

B


X

20 
V

3

O
C

0.1 A

A

X

Y

BC ′ OA
P
1
=
=
=
or q = 30°
OB OB
3P
3
As the object was initially at rest, the vector sum
of linear momenta of A, B and C must be zero.
Therefore, momentum of C = 2 P along OC opposite
to OC′. It makes an angle with B = ∠YOC
= ∠YOX′ + ∠X′OC = 90° + 60° = 150°
tan θ =

V = V′ + IR = 0.5 + 0.1 × 20 = 0.5 + 2.0 = 2.5 V
31. (a) : Here E∞ – E1 = 13.6 or E1 = – 13.6 eV
For second excited state,
E
13.6

E3 = 21 = −
= −1.51 eV
9
3
Energy required to ionise H-atom from second
excited state
= E∞ – E3 = 0 + 1.51 = 1.51 eV.
28

34. (d) : As the rod rotates about A, therefore, from

conservation of mechanical energy, decrease in
potential energy = increase in rotational kinetic
energy about A

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Page 28


1  ml  2
3g
l 1
mg   = IA ω 2 = 
 ω or ω 2 =
2
2
2 3
l
Centripetal force of centre of mass of the rod in this

l 3g 3 mg
position is = mrw2 = m
=
.
2 l
2
If F is the force exerted by the hinge on the rod
3 mg
(upwards), then F − mg =
2
3 mg
5
F=
+ mg = mg .
2
2
(240)2
Ω
35. (a) : Resistance of 40 W bulb, R1 =
40
(240)2
Resistance of 60 W bulb, R2 =
Ω
60
When bulbs are in series, the effective resistance
2

1  (240)2
1
R = R1 + R2 = (240)2  +  =

Ω
 40 60 
24
420 × 24 21
Current I =
=
A
120
(240)2

Potential difference across 40 W bulb
21 (240)2
=
×
= 252 V
120
40
Potential difference across 60 W bulb
21 (240)2
=
×
= 168 V
120
60
Since potential difference of 40 W is greater than
240 V, so it will work at above its rated voltage.

mg
qE
; Tsinq = qE or T =

cos θ
sin θ
39. (d) : Here, Am = 1 V, Ac = 2 V,
Am 1
Modulation index, µ = A = 2 = 0.5
c
40. (b) : The zener diode when used as a voltage
regulator is connected in reverse bias and a load
is connected in parallel to zener diode for output
voltage.
Tcosq = mg or T =

41. (a) : Due to the extra stability of half-filled p-orbitals
of N, its first ionisation potential is higher than
those of O and C. Further because of higher nuclear
charge, first ionisation potential of C is higher
than that of Be and B. Amongst Be and B, the first
ionisation potential of Be is higher than that of B
because in case of Be, an electron is to be removed
from 2s2 orbital while in case of B, an electron is to
removed from 2p1 orbital. Thus, the overall order is
B < Be < C < O < N.
42. (d) : Colour of salts is a property of partially filled
d-orbitals. Since TiF62– has completely empty and
Cu2Cl2 has completely filled d-subshells, hence
these are colourless salts.
43. (c) :

36. (c) :

44. (d) :
2

4
 2I 
Thermal power in A, PA =   3R = I 2R
3
3
2
2
I
Thermal power in B, PB =   6R = I 2R
3
3
2
Thermal power in C, PC = I R.
4
2
PA : PB : PC = I 2R : I 2R : I 2R = 4 : 2 : 3
3
3

37. (d)
38. (b) :

θ

T
Tsinθ

Tcosθ
θ

+q

mg

From figure at equilibrium,

CH3

46. (a) :

O
Ozonolysis

1-Methylcyclohexene
(P)

qE
E

Two oxygen atoms are in peroxo linkage i.e.,
oxidation number = –1
3 (–2) + 2(–1) + x + 2(+1) = 0 ⇒ – 8 + x + 2 = 0 ⇒ x = + 6
45. (d)
OH–

H2C—C—CH3 Aldol
O condenH2C C

sation
H
H2C—CH2
(Q)

O—
—C—CH3
C
H2C

CH

H2C —CH2
1-Acetylcyclopentene
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47. (c) : N atom in NCl3 and S atom in H2S are sp3
hybridised.
48. (c)

K=

Cα 2 0.001 M × (0.367 )2
=
= 2.127 × 10–4 M

1− α
0.633

59. (b) : 4Zn + 10HNO3

4Zn(NO3)2 + NH4NO3
+ 3H2O
60. (b) : Aromaticity can be predicted by the use of
Huckle’s rule which says that (4n + 2) p-electrons
are required in delocalisation system to give it
aromaticity.
61. (d)
62. (c)
(dilute)

49. (a) : ρ =
Z=

Z× M
N0 × a

3

or Z = ρ × N0 × a

3

M

23

2.7 × 6.023 × 10 × ( 405 × 10 −10 )3
=4
27.0

i.e., number of atoms per unit cell is 4. Hence, unit
cell is face-centred type.
50. (d) : Al2O3 + 3C + 3Cl2

1000°C

2AlCl3 + 3CO

51. (d) : C6H5COC(CH3)3 does not contain an
a-hydrogen and hence does not show tautomerism.
52. (d) : Due to the +ve charge on P, it attracts the
electrons of the P—H bond towards itself. As a
result, it has some ionic character. In other words,
the P—H bond in PH6+ is least covalent.
53. (d) : 3HClO(aq)
HClO3(aq) + 2HCl(aq)
It is a disproportionation reaction of hypochlorous
acid where the oxidation number of Cl changes
from +1 (in ClO–) to +5 (in ClO3–) and –1 (in Cl–).
54. (a)

Kc =
1

Kc =

2

On making square,

… (i)

…(ii)

[SO3 ]
1
=
K c [SO2 ][O2 ]1/2
1

2

 1 
[SO3 ]2
= Kc

 =
2
2
 K c1  [SO2 ] [O2 ]
\
56. (b)

[By Eq. (ii)]

2

 1 
Kc = 
= 44.44 mol −1 L
2
 0.15 

57. (c) : Sulphate ion is present outside the
coordination sphere so it can form white ppt. of
BaSO4 with BaCl2(aq).
58. (c) : α =
30

Λm
Λ°m

=

184.5
= 0.367
502.4

65. (a)

66. (b) : Sodium sulphide is soluble in water. The
solubility product (and hence solubility) of ZnS is
larger than that of CuS.
67. (b) : Reduction in presence of Zn-Hg and conc. HCl
is useful for aldehyde and ketone but carboxylic
acid group remains unaffected.

68. (a) : DH = (Ea)f – (Ea)b = 0
69. (b) : Since sp-hybridized carbon is more
electronegative than a sp2-hybridized carbon which
in turn is more electronegative than sp3-hybridized
sp2

[SO3 ]2

By reversing Eq. (i),

i.e., when n = 3; E decreases nine times.

acid than CH2 CH COOH which in turn, is a

2SO3(g)

[SO2 ]2[O2 ]

n2

sp

[SO2 ][O2 ]1/2
= 0.15
[SO3 ]

For 2SO2(g) + O2(g)

64. (d)

1

carbon, therefore, CH C COOH is a stronger

1
SO2(g) + O2(g)
2

55. (d) : For SO3(g)

63. (d) : E ∝

sp3

stronger acid than CH3 CH2 COOH . Thus, the
overall order of acid strength is (i) > (ii) > (iii).
70. (c) : Since ionization potential of hydrogen atom is
13.6 eV.
\ E1 = – 13.6 eV
Now, En − E1 =
−13.6
n2

−13.6
n2

− ( −13.6) = 12.1

+ 13.6 = 12.1 \ n = 3

After absorbing 12.1 eV the electron in H atom is
excited to 3rd shell.
Thus, possible transitions are 3
i.e., 3 → 2, 2 → 1 and 3 → 1.
71. (a)
72. (b) : Given : Weight of benzoic acid = 1.89 g;
Temperature of bomb calorimeter = 25°C = 298 K;
Mass of water (m) = 18.94 kg = 18940 g;
Increase in temperature (t) = 0.632°C and
specific heat of water (s) = 0.998 cal/g-deg.
We know that heat gained by water or heat liberated
by benzoic acid (Q) = msDt
= 18940 × 0.998 × 0.632 = 11946.14 cal
…Contd. on Page no. 76

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