JEE Mains 10 April 2019 Question Paper Shift 1 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


Time : 3 hrs. M.M. : 360
Answers & Solutions
for for for for for
JEE (MAIN)-2019 (Online) Phase-2
Important Instructions :
1. The test is of 3 hours duration.
2. The Test Booklet consists of 90 questions. The maximum marks are 360.
3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics
having 30 questions in each part of equal weightage.
4. Each question is allotted 4 (four) marks for each correct response. ¼ (one-fourth) marks will be deducted
for indicating incorrect response of each question. No deduction from the total score will be made if no
response is indicated for an item in the answer sheet.
5. There is only one correct response for each question.
(Physics, Chemistry and Mathematics)
10/04/2019
Morning
Page 2


Time : 3 hrs. M.M. : 360
Answers & Solutions
for for for for for
JEE (MAIN)-2019 (Online) Phase-2
Important Instructions :
1. The test is of 3 hours duration.
2. The Test Booklet consists of 90 questions. The maximum marks are 360.
3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics
having 30 questions in each part of equal weightage.
4. Each question is allotted 4 (four) marks for each correct response. ¼ (one-fourth) marks will be deducted
for indicating incorrect response of each question. No deduction from the total score will be made if no
response is indicated for an item in the answer sheet.
5. There is only one correct response for each question.
(Physics, Chemistry and Mathematics)
10/04/2019
Morning
PART–A : PHYSICS
1. Two coaxial discs, having moments of inertia I
1
and 
1
I
2
, are rotating with respective angular
velocities ?
1
 and 
1
2
?
 about their common axis.
They are brought in contact with each other
and thereafter they rotate with a common
angular velocity. If E
f
 and E
i
 are the final and
initial total energies, then (E
f
 – E
i
) is :
(1)
2
11
I
12
?
? (2)
2
11
3
I
8
?
(3)
2
11
I
6
?
(4)
2
11
I
24
?
?
Answer (4)
Sol. By applying conservation of angular momentum
(I
1
 + I
2
)??
common
 = I
1
?
1
 + I
2
?
2
?
??
??
?? ? ? ?
??
??
?
11
11
common 1
1
1
I
I
52
4
I
43
I
2
1
c
5
6
?
??
? Loss in KE = 
2 22
12 c 11 2 2
1 11
(II ) II
2 22
??
?? ? ?? ?
??
??
??KE = 
2
11
I
24
?
?
2. A ray of light AO in vacuum is incident on a
glass slab at angle 60° and refracted at angle
30° along OB as shown in the figure. The
optical path length of light ray from A to B is :
A
a
b
B
Vacuum
glass
60°
30°
O
(1)
23
2b
a
? (2)
2b
2a
3
?
(3)
2b
2a
3
? (4) 2a + 2b
Answer (4)
Sol. From the given figure
a
A
B
b
60°
O
30°
a
cos60
AO
??
AO = 2a
b
cos30
BO
??
2b
BO
3
?
? Length of optical path = AO + BO × 3
 = 2a + 2b
3. A ball is thrown upward with an initial velocity
V
0
 from the surface of the earth. The motion of
the ball is affected by a drag force equal to
m ?v
2
 (where m is mass of the ball, v is its
instantaneous velocity and ? is a constant).
Time taken by the ball to rise to its zenith is :
(1)
1
0
1
sin V
g g
?
??
?
??
?
??
(2)
0
1
ln1V
g g
??
?
?
??
?
??
(3)
1
0
1
tan V
g g
?
??
?
??
?
??
(4)
1
0
1 2
tan V
g 2g
?
??
?
??
?
??
Answer (3)
Sol. Retardation of the particle
a = – (g + ?v
2
)
0
0t
2
v0
dv
dt
gv
?
?
??
??
[for H
max
 v = 0]
1 0
v 1
tan t
gg
?
??
?
? ??
??
?
??
Page 3


Time : 3 hrs. M.M. : 360
Answers & Solutions
for for for for for
JEE (MAIN)-2019 (Online) Phase-2
Important Instructions :
1. The test is of 3 hours duration.
2. The Test Booklet consists of 90 questions. The maximum marks are 360.
3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics
having 30 questions in each part of equal weightage.
4. Each question is allotted 4 (four) marks for each correct response. ¼ (one-fourth) marks will be deducted
for indicating incorrect response of each question. No deduction from the total score will be made if no
response is indicated for an item in the answer sheet.
5. There is only one correct response for each question.
(Physics, Chemistry and Mathematics)
10/04/2019
Morning
PART–A : PHYSICS
1. Two coaxial discs, having moments of inertia I
1
and 
1
I
2
, are rotating with respective angular
velocities ?
1
 and 
1
2
?
 about their common axis.
They are brought in contact with each other
and thereafter they rotate with a common
angular velocity. If E
f
 and E
i
 are the final and
initial total energies, then (E
f
 – E
i
) is :
(1)
2
11
I
12
?
? (2)
2
11
3
I
8
?
(3)
2
11
I
6
?
(4)
2
11
I
24
?
?
Answer (4)
Sol. By applying conservation of angular momentum
(I
1
 + I
2
)??
common
 = I
1
?
1
 + I
2
?
2
?
??
??
?? ? ? ?
??
??
?
11
11
common 1
1
1
I
I
52
4
I
43
I
2
1
c
5
6
?
??
? Loss in KE = 
2 22
12 c 11 2 2
1 11
(II ) II
2 22
??
?? ? ?? ?
??
??
??KE = 
2
11
I
24
?
?
2. A ray of light AO in vacuum is incident on a
glass slab at angle 60° and refracted at angle
30° along OB as shown in the figure. The
optical path length of light ray from A to B is :
A
a
b
B
Vacuum
glass
60°
30°
O
(1)
23
2b
a
? (2)
2b
2a
3
?
(3)
2b
2a
3
? (4) 2a + 2b
Answer (4)
Sol. From the given figure
a
A
B
b
60°
O
30°
a
cos60
AO
??
AO = 2a
b
cos30
BO
??
2b
BO
3
?
? Length of optical path = AO + BO × 3
 = 2a + 2b
3. A ball is thrown upward with an initial velocity
V
0
 from the surface of the earth. The motion of
the ball is affected by a drag force equal to
m ?v
2
 (where m is mass of the ball, v is its
instantaneous velocity and ? is a constant).
Time taken by the ball to rise to its zenith is :
(1)
1
0
1
sin V
g g
?
??
?
??
?
??
(2)
0
1
ln1V
g g
??
?
?
??
?
??
(3)
1
0
1
tan V
g g
?
??
?
??
?
??
(4)
1
0
1 2
tan V
g 2g
?
??
?
??
?
??
Answer (3)
Sol. Retardation of the particle
a = – (g + ?v
2
)
0
0t
2
v0
dv
dt
gv
?
?
??
??
[for H
max
 v = 0]
1 0
v 1
tan t
gg
?
??
?
? ??
??
?
??
4. A particle of mass m is moving along a
trajectory given by
x = x
0
 + a cos ?
1
t
y = y
0
 + b sin ?
2
t
The torque, acting on the particle about the
origin, at t = 0 is :
(1)
22
02 0 1
ˆ
m(xb y a )k ??? ?
(2)
2
00 1
ˆ
m( x b y a) k ?? ?
(3)
2
01
ˆ
myak ??
(4) Zero
Answer (3)
Sol. x = x
0
 + a cos ?
1
t
y = y
0
 + b sin ?
2
t
? v
x
 = –a ?
1
 sin ( ?
1
t), v
y
 = b ?
2
 cos ( ?
2
t)
2
x1 1
aa cos(t) ?? ? ? ,
2
y 22
ab sin( t) ?? ? ?
At t = 0, x = x
0
 + a, y = y
0
a
x
 = 
2
1
a ??,a
y
 = 0
?
2
10
ˆ
m( a ) y ( k) ?? ? ? ? ?

2
01
ˆ
my a k ?? ? .
5. In a photoelectric effect experiment the
threshold wavelength of light is 380 nm. If the
wavelength of incident light is 260 nm, the
maximum kinetic energy of emitted electrons
will be :
Given E (in eV) = 
1237
(innm) ?
(1) 4.5 eV (2) 15.1 eV
(3) 3.0 eV (4) 1.5 eV
Answer (4)
Sol. Wavelength of incident wave ( ?) = 260 nm
Cut off (threshold) wavelength ( ?
0
) = 380 nm
Then
max
0
hc hc
KE ??
??
 
11
1237
260 380
? ?
? ?
? ?
? ?
 
1237 120
1.5eV
380 260
?
??
?
6. In an experiment, the resistance of a material
is plotted as a function of temperature (in some
range). As shown in the figure, it is a straight
line.
One may conclude that :
(1) ?
0
2
R
R(T)
T
(2)
?
?
22
0
T/T
0
R(T) R e
(3)
?
?
22
0
T/T
0
R(T) R e
(4) ?
22
0
T/T
0
R(T) R e
Answer (2)
Sol.
?? ?
2
a1
lnR(T) a
b T
a, b are constant
?
?
2
0
2
T
T
0
R(T) R e
7. An npn transistor operates as a common
emitter amplifier, with a power gain of 60 dB.
The input circuit resistance is 100 ? and the
output load resistance is 10 k ?. The common
emitter current gain ? is :
(1) 10
4
(2) 6 × 10
2
(3) 10
2
(4) 60
Answer (3)
Sol.
??
?? ??
??
??
2 out
gain gain
in
R
P&I
R
?
??
??
??
??
62
10000
10
100
? = 100
Page 4


Time : 3 hrs. M.M. : 360
Answers & Solutions
for for for for for
JEE (MAIN)-2019 (Online) Phase-2
Important Instructions :
1. The test is of 3 hours duration.
2. The Test Booklet consists of 90 questions. The maximum marks are 360.
3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics
having 30 questions in each part of equal weightage.
4. Each question is allotted 4 (four) marks for each correct response. ¼ (one-fourth) marks will be deducted
for indicating incorrect response of each question. No deduction from the total score will be made if no
response is indicated for an item in the answer sheet.
5. There is only one correct response for each question.
(Physics, Chemistry and Mathematics)
10/04/2019
Morning
PART–A : PHYSICS
1. Two coaxial discs, having moments of inertia I
1
and 
1
I
2
, are rotating with respective angular
velocities ?
1
 and 
1
2
?
 about their common axis.
They are brought in contact with each other
and thereafter they rotate with a common
angular velocity. If E
f
 and E
i
 are the final and
initial total energies, then (E
f
 – E
i
) is :
(1)
2
11
I
12
?
? (2)
2
11
3
I
8
?
(3)
2
11
I
6
?
(4)
2
11
I
24
?
?
Answer (4)
Sol. By applying conservation of angular momentum
(I
1
 + I
2
)??
common
 = I
1
?
1
 + I
2
?
2
?
??
??
?? ? ? ?
??
??
?
11
11
common 1
1
1
I
I
52
4
I
43
I
2
1
c
5
6
?
??
? Loss in KE = 
2 22
12 c 11 2 2
1 11
(II ) II
2 22
??
?? ? ?? ?
??
??
??KE = 
2
11
I
24
?
?
2. A ray of light AO in vacuum is incident on a
glass slab at angle 60° and refracted at angle
30° along OB as shown in the figure. The
optical path length of light ray from A to B is :
A
a
b
B
Vacuum
glass
60°
30°
O
(1)
23
2b
a
? (2)
2b
2a
3
?
(3)
2b
2a
3
? (4) 2a + 2b
Answer (4)
Sol. From the given figure
a
A
B
b
60°
O
30°
a
cos60
AO
??
AO = 2a
b
cos30
BO
??
2b
BO
3
?
? Length of optical path = AO + BO × 3
 = 2a + 2b
3. A ball is thrown upward with an initial velocity
V
0
 from the surface of the earth. The motion of
the ball is affected by a drag force equal to
m ?v
2
 (where m is mass of the ball, v is its
instantaneous velocity and ? is a constant).
Time taken by the ball to rise to its zenith is :
(1)
1
0
1
sin V
g g
?
??
?
??
?
??
(2)
0
1
ln1V
g g
??
?
?
??
?
??
(3)
1
0
1
tan V
g g
?
??
?
??
?
??
(4)
1
0
1 2
tan V
g 2g
?
??
?
??
?
??
Answer (3)
Sol. Retardation of the particle
a = – (g + ?v
2
)
0
0t
2
v0
dv
dt
gv
?
?
??
??
[for H
max
 v = 0]
1 0
v 1
tan t
gg
?
??
?
? ??
??
?
??
4. A particle of mass m is moving along a
trajectory given by
x = x
0
 + a cos ?
1
t
y = y
0
 + b sin ?
2
t
The torque, acting on the particle about the
origin, at t = 0 is :
(1)
22
02 0 1
ˆ
m(xb y a )k ??? ?
(2)
2
00 1
ˆ
m( x b y a) k ?? ?
(3)
2
01
ˆ
myak ??
(4) Zero
Answer (3)
Sol. x = x
0
 + a cos ?
1
t
y = y
0
 + b sin ?
2
t
? v
x
 = –a ?
1
 sin ( ?
1
t), v
y
 = b ?
2
 cos ( ?
2
t)
2
x1 1
aa cos(t) ?? ? ? ,
2
y 22
ab sin( t) ?? ? ?
At t = 0, x = x
0
 + a, y = y
0
a
x
 = 
2
1
a ??,a
y
 = 0
?
2
10
ˆ
m( a ) y ( k) ?? ? ? ? ?

2
01
ˆ
my a k ?? ? .
5. In a photoelectric effect experiment the
threshold wavelength of light is 380 nm. If the
wavelength of incident light is 260 nm, the
maximum kinetic energy of emitted electrons
will be :
Given E (in eV) = 
1237
(innm) ?
(1) 4.5 eV (2) 15.1 eV
(3) 3.0 eV (4) 1.5 eV
Answer (4)
Sol. Wavelength of incident wave ( ?) = 260 nm
Cut off (threshold) wavelength ( ?
0
) = 380 nm
Then
max
0
hc hc
KE ??
??
 
11
1237
260 380
? ?
? ?
? ?
? ?
 
1237 120
1.5eV
380 260
?
??
?
6. In an experiment, the resistance of a material
is plotted as a function of temperature (in some
range). As shown in the figure, it is a straight
line.
One may conclude that :
(1) ?
0
2
R
R(T)
T
(2)
?
?
22
0
T/T
0
R(T) R e
(3)
?
?
22
0
T/T
0
R(T) R e
(4) ?
22
0
T/T
0
R(T) R e
Answer (2)
Sol.
?? ?
2
a1
lnR(T) a
b T
a, b are constant
?
?
2
0
2
T
T
0
R(T) R e
7. An npn transistor operates as a common
emitter amplifier, with a power gain of 60 dB.
The input circuit resistance is 100 ? and the
output load resistance is 10 k ?. The common
emitter current gain ? is :
(1) 10
4
(2) 6 × 10
2
(3) 10
2
(4) 60
Answer (3)
Sol.
??
?? ??
??
??
2 out
gain gain
in
R
P&I
R
?
??
??
??
??
62
10000
10
100
? = 100
8. Two wires A and B are carrying currents
I
1
 and I
2
 as shown in the figure. The separation
between them is d. A third wire C carrying a
current I is to be kept parallel to them at a
distance x from A such that the net force acting
on it is zero. The possible values of x are :
(1) ??
?
1
12
Id
x
(II )
(2)
?? ? ?
??
?? ? ?
??
?? ? ?
1 2
12 1 2
I I
xdandx d
II I I
(3)
?? ? ?
??
?? ? ?
??
?? ? ?
22
12 1 2
andx d
II I I
(4)
?? ? ?
??
?? ? ?
??
?? ? ?
1 2
12 1 2
I I
xdandx d
II I I
Answer (1)
Sol. ??

F0
? ?
??
???
01 02
I I
since(xd)
2x 2 (x d)
I
1
x – I
1
d = I
2
x
?
?
1
12
Id
x
II
9. A stationary source emits sound waves of
frequency 500 Hz. Two observers moving along
a line passing through the source detect sound
to be of frequencies 480 Hz and 530 Hz. Their
respective speeds are, in ms
–1
,
(Given speed of sound = 300 m/s)
(1) 12, 16 (2) 16, 14
(3) 8, 18 (4) 12, 18
Answer (4)
Sol. Frequency of sound source (f
0
) = 500 Hz
When observer is moving away from the source
Apparent frequency 
? ?? ? ??
??
??
?
??
0
10
f 480 f ...(i)
And when observer is moving towards the
source 
?? ?? ? ??
??
??
?
??
0
20
f530 f ...(ii)
From equation (i)
? ?? ??
?
??
??
0
300
480 500
300
??
0
 = 12 m/s
From equation (ii)
?? ? ??
??
??
?
??
0
530 500 1
????
?
 = 18 m/s
10. A transformer consisting of 300 turns in the
primary and 150 turns in the secondary gives
output power of 2.2 kW. If the current in the
secondary coil is 10 A, then the input voltage
and current in the primary coil are :
(1) 440 V and 5 A (2) 220 V and  20 A
(3) 220 V and 10 A (4) 440 V and 20 A
Answer (1)
Sol. Power output (V
2
I
2
) = 2.2 kW
?
??
2
2.2kW
V 220 volts
(10A)
? Input voltage for step-down transformer
??
11
22
VN
2
VN
V
input
 = 2 × V
output
 = 2 × 220
= 440 V
Also ?
1 2
21
I N
IN
? ?? ?
1
1
I105A
2
11. A uniformly charged ring of radius 3a and total
charge q is placed in xy-plane centred at
origin. A point charge q is moving towards the
ring along the z-axis and has speed v at
z = 4a. The minimum value of v such that it
crosses the origin is :
(1)
??
??
??
??
1/2
2
0
21 q
m154 a
(2)
??
??
??
??
1/2
2
0
21 q
m54 a
(3)
??
??
??
??
1/2
2
0
24 q
m154 a
(4)
??
??
??
??
1/2
2
0
22 q
m154 a
II
xd
Page 5


Time : 3 hrs. M.M. : 360
Answers & Solutions
for for for for for
JEE (MAIN)-2019 (Online) Phase-2
Important Instructions :
1. The test is of 3 hours duration.
2. The Test Booklet consists of 90 questions. The maximum marks are 360.
3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics
having 30 questions in each part of equal weightage.
4. Each question is allotted 4 (four) marks for each correct response. ¼ (one-fourth) marks will be deducted
for indicating incorrect response of each question. No deduction from the total score will be made if no
response is indicated for an item in the answer sheet.
5. There is only one correct response for each question.
(Physics, Chemistry and Mathematics)
10/04/2019
Morning
PART–A : PHYSICS
1. Two coaxial discs, having moments of inertia I
1
and 
1
I
2
, are rotating with respective angular
velocities ?
1
 and 
1
2
?
 about their common axis.
They are brought in contact with each other
and thereafter they rotate with a common
angular velocity. If E
f
 and E
i
 are the final and
initial total energies, then (E
f
 – E
i
) is :
(1)
2
11
I
12
?
? (2)
2
11
3
I
8
?
(3)
2
11
I
6
?
(4)
2
11
I
24
?
?
Answer (4)
Sol. By applying conservation of angular momentum
(I
1
 + I
2
)??
common
 = I
1
?
1
 + I
2
?
2
?
??
??
?? ? ? ?
??
??
?
11
11
common 1
1
1
I
I
52
4
I
43
I
2
1
c
5
6
?
??
? Loss in KE = 
2 22
12 c 11 2 2
1 11
(II ) II
2 22
??
?? ? ?? ?
??
??
??KE = 
2
11
I
24
?
?
2. A ray of light AO in vacuum is incident on a
glass slab at angle 60° and refracted at angle
30° along OB as shown in the figure. The
optical path length of light ray from A to B is :
A
a
b
B
Vacuum
glass
60°
30°
O
(1)
23
2b
a
? (2)
2b
2a
3
?
(3)
2b
2a
3
? (4) 2a + 2b
Answer (4)
Sol. From the given figure
a
A
B
b
60°
O
30°
a
cos60
AO
??
AO = 2a
b
cos30
BO
??
2b
BO
3
?
? Length of optical path = AO + BO × 3
 = 2a + 2b
3. A ball is thrown upward with an initial velocity
V
0
 from the surface of the earth. The motion of
the ball is affected by a drag force equal to
m ?v
2
 (where m is mass of the ball, v is its
instantaneous velocity and ? is a constant).
Time taken by the ball to rise to its zenith is :
(1)
1
0
1
sin V
g g
?
??
?
??
?
??
(2)
0
1
ln1V
g g
??
?
?
??
?
??
(3)
1
0
1
tan V
g g
?
??
?
??
?
??
(4)
1
0
1 2
tan V
g 2g
?
??
?
??
?
??
Answer (3)
Sol. Retardation of the particle
a = – (g + ?v
2
)
0
0t
2
v0
dv
dt
gv
?
?
??
??
[for H
max
 v = 0]
1 0
v 1
tan t
gg
?
??
?
? ??
??
?
??
4. A particle of mass m is moving along a
trajectory given by
x = x
0
 + a cos ?
1
t
y = y
0
 + b sin ?
2
t
The torque, acting on the particle about the
origin, at t = 0 is :
(1)
22
02 0 1
ˆ
m(xb y a )k ??? ?
(2)
2
00 1
ˆ
m( x b y a) k ?? ?
(3)
2
01
ˆ
myak ??
(4) Zero
Answer (3)
Sol. x = x
0
 + a cos ?
1
t
y = y
0
 + b sin ?
2
t
? v
x
 = –a ?
1
 sin ( ?
1
t), v
y
 = b ?
2
 cos ( ?
2
t)
2
x1 1
aa cos(t) ?? ? ? ,
2
y 22
ab sin( t) ?? ? ?
At t = 0, x = x
0
 + a, y = y
0
a
x
 = 
2
1
a ??,a
y
 = 0
?
2
10
ˆ
m( a ) y ( k) ?? ? ? ? ?

2
01
ˆ
my a k ?? ? .
5. In a photoelectric effect experiment the
threshold wavelength of light is 380 nm. If the
wavelength of incident light is 260 nm, the
maximum kinetic energy of emitted electrons
will be :
Given E (in eV) = 
1237
(innm) ?
(1) 4.5 eV (2) 15.1 eV
(3) 3.0 eV (4) 1.5 eV
Answer (4)
Sol. Wavelength of incident wave ( ?) = 260 nm
Cut off (threshold) wavelength ( ?
0
) = 380 nm
Then
max
0
hc hc
KE ??
??
 
11
1237
260 380
? ?
? ?
? ?
? ?
 
1237 120
1.5eV
380 260
?
??
?
6. In an experiment, the resistance of a material
is plotted as a function of temperature (in some
range). As shown in the figure, it is a straight
line.
One may conclude that :
(1) ?
0
2
R
R(T)
T
(2)
?
?
22
0
T/T
0
R(T) R e
(3)
?
?
22
0
T/T
0
R(T) R e
(4) ?
22
0
T/T
0
R(T) R e
Answer (2)
Sol.
?? ?
2
a1
lnR(T) a
b T
a, b are constant
?
?
2
0
2
T
T
0
R(T) R e
7. An npn transistor operates as a common
emitter amplifier, with a power gain of 60 dB.
The input circuit resistance is 100 ? and the
output load resistance is 10 k ?. The common
emitter current gain ? is :
(1) 10
4
(2) 6 × 10
2
(3) 10
2
(4) 60
Answer (3)
Sol.
??
?? ??
??
??
2 out
gain gain
in
R
P&I
R
?
??
??
??
??
62
10000
10
100
? = 100
8. Two wires A and B are carrying currents
I
1
 and I
2
 as shown in the figure. The separation
between them is d. A third wire C carrying a
current I is to be kept parallel to them at a
distance x from A such that the net force acting
on it is zero. The possible values of x are :
(1) ??
?
1
12
Id
x
(II )
(2)
?? ? ?
??
?? ? ?
??
?? ? ?
1 2
12 1 2
I I
xdandx d
II I I
(3)
?? ? ?
??
?? ? ?
??
?? ? ?
22
12 1 2
andx d
II I I
(4)
?? ? ?
??
?? ? ?
??
?? ? ?
1 2
12 1 2
I I
xdandx d
II I I
Answer (1)
Sol. ??

F0
? ?
??
???
01 02
I I
since(xd)
2x 2 (x d)
I
1
x – I
1
d = I
2
x
?
?
1
12
Id
x
II
9. A stationary source emits sound waves of
frequency 500 Hz. Two observers moving along
a line passing through the source detect sound
to be of frequencies 480 Hz and 530 Hz. Their
respective speeds are, in ms
–1
,
(Given speed of sound = 300 m/s)
(1) 12, 16 (2) 16, 14
(3) 8, 18 (4) 12, 18
Answer (4)
Sol. Frequency of sound source (f
0
) = 500 Hz
When observer is moving away from the source
Apparent frequency 
? ?? ? ??
??
??
?
??
0
10
f 480 f ...(i)
And when observer is moving towards the
source 
?? ?? ? ??
??
??
?
??
0
20
f530 f ...(ii)
From equation (i)
? ?? ??
?
??
??
0
300
480 500
300
??
0
 = 12 m/s
From equation (ii)
?? ? ??
??
??
?
??
0
530 500 1
????
?
 = 18 m/s
10. A transformer consisting of 300 turns in the
primary and 150 turns in the secondary gives
output power of 2.2 kW. If the current in the
secondary coil is 10 A, then the input voltage
and current in the primary coil are :
(1) 440 V and 5 A (2) 220 V and  20 A
(3) 220 V and 10 A (4) 440 V and 20 A
Answer (1)
Sol. Power output (V
2
I
2
) = 2.2 kW
?
??
2
2.2kW
V 220 volts
(10A)
? Input voltage for step-down transformer
??
11
22
VN
2
VN
V
input
 = 2 × V
output
 = 2 × 220
= 440 V
Also ?
1 2
21
I N
IN
? ?? ?
1
1
I105A
2
11. A uniformly charged ring of radius 3a and total
charge q is placed in xy-plane centred at
origin. A point charge q is moving towards the
ring along the z-axis and has speed v at
z = 4a. The minimum value of v such that it
crosses the origin is :
(1)
??
??
??
??
1/2
2
0
21 q
m154 a
(2)
??
??
??
??
1/2
2
0
21 q
m54 a
(3)
??
??
??
??
1/2
2
0
24 q
m154 a
(4)
??
??
??
??
1/2
2
0
22 q
m154 a
II
xd
Answer (4)
Sol.
+ , Q R
Q
Potential at any point of the charged ring
P
22
KQ
V
Rx
?
?
The minimum velocity (v
0
) should just
sufficient to reach the point charge at the
center, therefore
2
0CP
1
mv Q[V V ]
2
??
KQ KQ
Q
3a 5a
??
??
??
??
22
2
0
0
4KQ 4 1 q
v
15ma 154 ma
??
??
?
1
2
2
0
0
22q
v
m15 4 a
??
?
??
???
??
12. n moles of an ideal gas with constant volume
heat capacity C
V
 undergo an isobaric
expansion by certain volume. The ratio of the
work done in the process, to the heat
supplied is :
(1)
?
V
4nR
CnR
(2)
?
V
nR
CnR
(3)
?
V
4nR
CnR
(4)
?
V
nR
CnR
Answer (Bonus)
Sol. For Isobaric process
Work done (W) = nR ?T
and Heat given (Q) = nC
P
?T
?
PV
WR R
QC C R
??
?
13. A 25 × 10
–3
 m
3
 volume cylinder is filled with
1 mol of O
2
 gas at room temperature (300 K).
The molecular diameter of O
2
, and its root
mean square speed, are found to be 0.3 nm
and 200 m/s, respectively. What is the average
collision rate (per second) for an O
2
molecule ?
(1) ~10
12
(2) ~10
10
(3) ~10
11
(4) ~10
13
Answer (2)
Sol. V = 25 × 10
–3
 m
3
, N = 1 mole of O
2
T = 300 K
V
rms
 = 200 m/s
?
2
1
2N r
??
?
Average time 
2
1V
200 N r 2
??
?? ???
??
23
18
3
2 200 6.023 10
10 0.09
25 10
?
?
?? ?
?????
?
Average no. of collision 
10
10 ?
14. Two radioactive materials A and B have decay
constants 10 ? and ?, respectively. If initially
they have the same number of nuclei, then the
ratio of the number of nuclei of A to that of B
will be 1/e after a time :
(1)
?
1
10
(2)
?
11
10
(3)
?
1
9
(4)
?
1
11
Answer (3)
Sol. Number of nuclei present at any time t
N = N
0
e
– ?t
?
BA
()t A
B
N 1
e
Ne
???
??
( ?
A
 – ?
B
) ? t = 1
?
11
t
10 9
??
?? ? ? ?
15. Figure shows charge (q) versus voltage (V)
graph for series and parallel combination of
two given capacitors. The capacitances are :
10 V V (Volt)
A
B
q( C) ?
500
80
(1)?? 40 Fand10 F (2)?? 20 Fand30 F
(3)?? 60 Fand40 F (4)?? 50 Fand30 F