JEE Mains 10 April 2019 Question Paper Shift 2 | 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
Evening
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
Evening
JEE (MAIN)-2019 (Online) Phase-2
PART–A : PHYSICS
1. One mole of an ideal gas passes through a
process where pressure and volume obey the
relation 
??
??
????
??
?? ??
??
2
0
0
V 1
PP 1
2V
. Here P
0
 and V
0
 are
constants. Calculate the change in the
temperature of the gas if its volume changes
from V
0
 to 2V
0
.
(1)
00
PV 1
4R
(2)
00
PV 5
4R
(3)
00
PV 1
2R
(4)
00
PV 3
4R
Answer (2)
Sol. If V
1
 = V
0 
0
10
P 1
PP1
22
??
? ?? ?
??
??
If V
2
 = 2V
0
 ? P
2
 = 
0
0
7P 11
P1
24 8
???? ??
??
?? ?? ??
?? ??
??
11 22
PV PV PV
TT
nR nR nR
??
? ??? ?
??
??
? ?
00 00
11 2 2
PV 7PV 11
TPVPV
nR nR 2 4
?? ?? ??
?? ? ? ?
?? ????
?? ????
00
5P V
4nR
?
2. The correct figure that shows, schematically,
the wave pattern produced by superposition of
two waves of frequencies 9 Hz and 11 Hz, is :
(1) 0
12
t(s)
y
(2)
0
12
t(s)
y
(3)
0
12
t(s)
y
(4) 0
12
t(s)
y
Answer (3)
Sol. Beat frequency = |f
1
 – f
2
| = 11 – 9 = 2 Hz
3. A spaceship orbits around a planet at a height
of 20 km from its surface. Assuming that only
gravitational field of the planet acts on the
spaceship, what will be the number of
complete revolutions made by the spaceship in
24 hours around the planet?
[Given: Mass of planet = 8 × 10
22
 kg,
Radius of planet = 2 × 10
6
 m,
Gravitational constant G = 6.67 × 10
–11
 Nm
2
/kg
2
]
(1) 17 (2) 13
(3) 11 (4) 9
Answer (3)
Sol.
?
??
2r GM
T,v
vr
?? ??
3
rr
T2r 2
GM GM
?
?
??
???
312
11 22
(202) 10
T2
6.67 10 8 10
 sec
T = 7812.2 s
?  T 2.17hr 11revolutions
4. A plane is inclined at an angle ? = 30° with
respect to the horizontal. A particle is
projected with a speed u = 2 ms
–1
, from the
base of the plane, making an angle ? = 15° with
respect to the plane as shown in the figure. The
distance from the base, at which the particle
hits the plane is close to :
(Take g = 10 ms
–2
)
? = 30°
? = 15°
u
(1) 18 cm (2) 20 cm
(3) 14 cm (4) 26 cm
Answer (2)
Sol. Time of flight (T) 
?
?
?
2usin
gcos
??
(2)(2sin15) 4sin15
T
gcos30 10cos30
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
Evening
JEE (MAIN)-2019 (Online) Phase-2
PART–A : PHYSICS
1. One mole of an ideal gas passes through a
process where pressure and volume obey the
relation 
??
??
????
??
?? ??
??
2
0
0
V 1
PP 1
2V
. Here P
0
 and V
0
 are
constants. Calculate the change in the
temperature of the gas if its volume changes
from V
0
 to 2V
0
.
(1)
00
PV 1
4R
(2)
00
PV 5
4R
(3)
00
PV 1
2R
(4)
00
PV 3
4R
Answer (2)
Sol. If V
1
 = V
0 
0
10
P 1
PP1
22
??
? ?? ?
??
??
If V
2
 = 2V
0
 ? P
2
 = 
0
0
7P 11
P1
24 8
???? ??
??
?? ?? ??
?? ??
??
11 22
PV PV PV
TT
nR nR nR
??
? ??? ?
??
??
? ?
00 00
11 2 2
PV 7PV 11
TPVPV
nR nR 2 4
?? ?? ??
?? ? ? ?
?? ????
?? ????
00
5P V
4nR
?
2. The correct figure that shows, schematically,
the wave pattern produced by superposition of
two waves of frequencies 9 Hz and 11 Hz, is :
(1) 0
12
t(s)
y
(2)
0
12
t(s)
y
(3)
0
12
t(s)
y
(4) 0
12
t(s)
y
Answer (3)
Sol. Beat frequency = |f
1
 – f
2
| = 11 – 9 = 2 Hz
3. A spaceship orbits around a planet at a height
of 20 km from its surface. Assuming that only
gravitational field of the planet acts on the
spaceship, what will be the number of
complete revolutions made by the spaceship in
24 hours around the planet?
[Given: Mass of planet = 8 × 10
22
 kg,
Radius of planet = 2 × 10
6
 m,
Gravitational constant G = 6.67 × 10
–11
 Nm
2
/kg
2
]
(1) 17 (2) 13
(3) 11 (4) 9
Answer (3)
Sol.
?
??
2r GM
T,v
vr
?? ??
3
rr
T2r 2
GM GM
?
?
??
???
312
11 22
(202) 10
T2
6.67 10 8 10
 sec
T = 7812.2 s
?  T 2.17hr 11revolutions
4. A plane is inclined at an angle ? = 30° with
respect to the horizontal. A particle is
projected with a speed u = 2 ms
–1
, from the
base of the plane, making an angle ? = 15° with
respect to the plane as shown in the figure. The
distance from the base, at which the particle
hits the plane is close to :
(Take g = 10 ms
–2
)
? = 30°
? = 15°
u
(1) 18 cm (2) 20 cm
(3) 14 cm (4) 26 cm
Answer (2)
Sol. Time of flight (T) 
?
?
?
2usin
gcos
??
(2)(2sin15) 4sin15
T
gcos30 10cos30
JEE (MAIN)-2019 (Online) Phase-2
? = 30°
? = 15°
y
gcos30
g
x
2 m/s
gsin30
??
2
1
Range(R)(2cos15)T gsin30(T)
2
??
???
??
??
2
2
4 sin15 1 16 sin 15
(2cos15) 10sin30
10cos30 3 100cos 30
?
? 
16 3 16
0.1952m 20cm
60
5. A bullet of mass 20 g has an initial speed of
1 ms
–1
, just before it starts penetrating a mud
wall of thickness 20 cm. If the wall offers a
mean resistance of 2.5 × 10
–2
 N, the speed of
the bullet after emerging from the other side of
the wall is close to :
(1) 0.4 ms
–1
(2) 0.7 ms
–1
(3) 0.3 ms
–1
(4) 0.1 ms
–1
Answer (2)
Sol. v
2
 = u
2
 – 2aS
?
?
??
?
??
??
???
??
2
22
3
2.5 10 20
v(1) (2)
100 20 10
??
2
1
v1
2
?
??
1
v m/s 0.7m/s
2
6. The time dependence of the position of a
particle of mass m = 2 is given by
??

2
ˆˆ
r(t) 2ti 3t j . Its angular momentum, with
respect to the origin, at time t = 2 is :
(1)
??
??
ˆ ˆ
34 k i (2)
??
ˆˆ
48 i j ?
(3)
ˆ
36k (4) ?
ˆ
48k
Answer (4)
Sol. ??

2
ˆˆ
r2ti 3t j
?? ?
 dr
ˆˆ
v2i6tj
dt
???
 dv
ˆ
a6j
dt
???


ˆ
Fma 12j
?? ?

ˆˆ
r(at t 2) 4i 12j
?? ? ? ? ? ??

 
ˆ ˆˆ ˆˆ
Lm(rv) 2(4i 12j) (2i 12j) 48k
7. The figure represents a voltage regulator
circuit using a Zener diode. The breakdown
voltage of the Zener diode is 6 V and the load
resistance is, R
L
 = 4 k ?. The series resistance
of the circuit is R
i
 = 1 k ?. If the battery voltage
V
B
 varies from 8 V to 16 V, what are the
minimum and maximum values of the current
through Zener diode?
R
i
R
L V
B
(1) 0.5 mA; 6 mA (2) 0.5 mA; 8.5 mA
(3) 1.5 mA; 8.5 mA (4) 1 mA; 8.5 mA
Answer (2)
Sol. 4 k ?
V
I
1
1 k ?
1
3
I
2
??
? ??
??
6
mA
4
1
31
I8–6– 0.5mA
22
??
???
??
??
??
??
??
??
2
3
I 16–6– 8.5mA
2
8. A metal coin of mass 5 g and radius 1 cm is fixed
to a thin stick AB of negligible mass as shown in
the figure. The system is initially at rest. The
constant torque, that will make the system rotate
about AB at 25 rotations per second in 5 s, is
closed to :
A
B
1 cm
(1) 4.0 × 10
–6
 Nm (2) 7.9 × 10
–6
 Nm
(3) 2.0 × 10
–5
 Nm (4) 1.6 × 10
–5
 Nm
Answer (3)
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
Evening
JEE (MAIN)-2019 (Online) Phase-2
PART–A : PHYSICS
1. One mole of an ideal gas passes through a
process where pressure and volume obey the
relation 
??
??
????
??
?? ??
??
2
0
0
V 1
PP 1
2V
. Here P
0
 and V
0
 are
constants. Calculate the change in the
temperature of the gas if its volume changes
from V
0
 to 2V
0
.
(1)
00
PV 1
4R
(2)
00
PV 5
4R
(3)
00
PV 1
2R
(4)
00
PV 3
4R
Answer (2)
Sol. If V
1
 = V
0 
0
10
P 1
PP1
22
??
? ?? ?
??
??
If V
2
 = 2V
0
 ? P
2
 = 
0
0
7P 11
P1
24 8
???? ??
??
?? ?? ??
?? ??
??
11 22
PV PV PV
TT
nR nR nR
??
? ??? ?
??
??
? ?
00 00
11 2 2
PV 7PV 11
TPVPV
nR nR 2 4
?? ?? ??
?? ? ? ?
?? ????
?? ????
00
5P V
4nR
?
2. The correct figure that shows, schematically,
the wave pattern produced by superposition of
two waves of frequencies 9 Hz and 11 Hz, is :
(1) 0
12
t(s)
y
(2)
0
12
t(s)
y
(3)
0
12
t(s)
y
(4) 0
12
t(s)
y
Answer (3)
Sol. Beat frequency = |f
1
 – f
2
| = 11 – 9 = 2 Hz
3. A spaceship orbits around a planet at a height
of 20 km from its surface. Assuming that only
gravitational field of the planet acts on the
spaceship, what will be the number of
complete revolutions made by the spaceship in
24 hours around the planet?
[Given: Mass of planet = 8 × 10
22
 kg,
Radius of planet = 2 × 10
6
 m,
Gravitational constant G = 6.67 × 10
–11
 Nm
2
/kg
2
]
(1) 17 (2) 13
(3) 11 (4) 9
Answer (3)
Sol.
?
??
2r GM
T,v
vr
?? ??
3
rr
T2r 2
GM GM
?
?
??
???
312
11 22
(202) 10
T2
6.67 10 8 10
 sec
T = 7812.2 s
?  T 2.17hr 11revolutions
4. A plane is inclined at an angle ? = 30° with
respect to the horizontal. A particle is
projected with a speed u = 2 ms
–1
, from the
base of the plane, making an angle ? = 15° with
respect to the plane as shown in the figure. The
distance from the base, at which the particle
hits the plane is close to :
(Take g = 10 ms
–2
)
? = 30°
? = 15°
u
(1) 18 cm (2) 20 cm
(3) 14 cm (4) 26 cm
Answer (2)
Sol. Time of flight (T) 
?
?
?
2usin
gcos
??
(2)(2sin15) 4sin15
T
gcos30 10cos30
JEE (MAIN)-2019 (Online) Phase-2
? = 30°
? = 15°
y
gcos30
g
x
2 m/s
gsin30
??
2
1
Range(R)(2cos15)T gsin30(T)
2
??
???
??
??
2
2
4 sin15 1 16 sin 15
(2cos15) 10sin30
10cos30 3 100cos 30
?
? 
16 3 16
0.1952m 20cm
60
5. A bullet of mass 20 g has an initial speed of
1 ms
–1
, just before it starts penetrating a mud
wall of thickness 20 cm. If the wall offers a
mean resistance of 2.5 × 10
–2
 N, the speed of
the bullet after emerging from the other side of
the wall is close to :
(1) 0.4 ms
–1
(2) 0.7 ms
–1
(3) 0.3 ms
–1
(4) 0.1 ms
–1
Answer (2)
Sol. v
2
 = u
2
 – 2aS
?
?
??
?
??
??
???
??
2
22
3
2.5 10 20
v(1) (2)
100 20 10
??
2
1
v1
2
?
??
1
v m/s 0.7m/s
2
6. The time dependence of the position of a
particle of mass m = 2 is given by
??

2
ˆˆ
r(t) 2ti 3t j . Its angular momentum, with
respect to the origin, at time t = 2 is :
(1)
??
??
ˆ ˆ
34 k i (2)
??
ˆˆ
48 i j ?
(3)
ˆ
36k (4) ?
ˆ
48k
Answer (4)
Sol. ??

2
ˆˆ
r2ti 3t j
?? ?
 dr
ˆˆ
v2i6tj
dt
???
 dv
ˆ
a6j
dt
???


ˆ
Fma 12j
?? ?

ˆˆ
r(at t 2) 4i 12j
?? ? ? ? ? ??

 
ˆ ˆˆ ˆˆ
Lm(rv) 2(4i 12j) (2i 12j) 48k
7. The figure represents a voltage regulator
circuit using a Zener diode. The breakdown
voltage of the Zener diode is 6 V and the load
resistance is, R
L
 = 4 k ?. The series resistance
of the circuit is R
i
 = 1 k ?. If the battery voltage
V
B
 varies from 8 V to 16 V, what are the
minimum and maximum values of the current
through Zener diode?
R
i
R
L V
B
(1) 0.5 mA; 6 mA (2) 0.5 mA; 8.5 mA
(3) 1.5 mA; 8.5 mA (4) 1 mA; 8.5 mA
Answer (2)
Sol. 4 k ?
V
I
1
1 k ?
1
3
I
2
??
? ??
??
6
mA
4
1
31
I8–6– 0.5mA
22
??
???
??
??
??
??
??
??
2
3
I 16–6– 8.5mA
2
8. A metal coin of mass 5 g and radius 1 cm is fixed
to a thin stick AB of negligible mass as shown in
the figure. The system is initially at rest. The
constant torque, that will make the system rotate
about AB at 25 rotations per second in 5 s, is
closed to :
A
B
1 cm
(1) 4.0 × 10
–6
 Nm (2) 7.9 × 10
–6
 Nm
(3) 2.0 × 10
–5
 Nm (4) 1.6 × 10
–5
 Nm
Answer (3)
Sol. ? = I ?
??= ?
0 
+ ?t
? 25 × 2 ? = ( ?)5
??= 10 ?
?
2
5
mR
4
??
?? ?
??
??
??? ?
–3 –4
5
510 10 10
4
??
?? ?
??
??
= 2.0 × 10
–5 
Nm
9. In a Young’s double slit experiment, the ratio of
the slit’s width is 4 : 1. The ratio of the intensity
of maxima to minima, close to the central fringe
on the screen, will be :
(1)
??
4
31 :16 ? (2) 4 : 1
(3) 25 : 9 (4) 9 : 1
Answer (4)
Sol. I
1
 = 4 I
0
I
2
 = I
0
??
??
?
??
??
??
??
2
12
max
2
min
12
II
I
9
I1
I– I
10. Two radioactive substances A and B have
decay constants 5 ? and ? respectively. At
t = 0, a sample has the same number of the two
nuclei. The time taken for the ratio of the
number of nuclei to become 
2
1
e
??
??
??
 will be :
(1)
1
2 ?
(2)
1
4 ?
(3)
1
?
(4)
2
?
Answer (1)
Sol. N
x
(at t) = N
0
e
–5 ?t
N
y
(at t) = N
0
e
– ?t
–4t x
2
y
N
1
e
N e
?
??
? 4 ?t = 2
?
21
t
42
??
??
??
?? ??
11. A cubical block of side 0.5 m floats on water
with 30% of its volume under water. What is the
maximum weight that can be put on the block
without fully submerging it under water?
[Take, density of water = 10
3
 kg/m
3
]
(1) 30.1 kg (2) 46.3 kg
(3) 87.5 kg (4) 65.4 kg
Answer (3)
Sol. Given 
?? ??
3
cube
30
50 1 g M g
100
?? ?? …(i)
Let m mass should be placed
Hence 
?? ??
??
3
cube
50 1 g M m g ?? ? ? …(ii)
equation (ii) – equation (i)
? mg = (50)
3
 × g(1 – 0.3) = 125 × 0.7 × 10
3 
g
? m = 87.5 kg
12. The graph shows how the magnification m
produced by a thin lens varies with image
distance v. What is the focal length of the lens
used?
C
a b
v
m
(1)
2
b
ac
(2)
2
bc
a
(3)
a
c
(4)
b
c
Answer (4)
Sol. As the graph between magnification (m) and
image distance (v) varies linearly, then
m = k
1
v + k
2
?
12
v
kv k
u
??
?
2
1
k
1
k
uv
??
?
2
1
k
1
– k
vu
?
Clearly, 
12
1
kandk1
f
?? here
?
1b
f
slopeofm vgraphc
??
?
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
Evening
JEE (MAIN)-2019 (Online) Phase-2
PART–A : PHYSICS
1. One mole of an ideal gas passes through a
process where pressure and volume obey the
relation 
??
??
????
??
?? ??
??
2
0
0
V 1
PP 1
2V
. Here P
0
 and V
0
 are
constants. Calculate the change in the
temperature of the gas if its volume changes
from V
0
 to 2V
0
.
(1)
00
PV 1
4R
(2)
00
PV 5
4R
(3)
00
PV 1
2R
(4)
00
PV 3
4R
Answer (2)
Sol. If V
1
 = V
0 
0
10
P 1
PP1
22
??
? ?? ?
??
??
If V
2
 = 2V
0
 ? P
2
 = 
0
0
7P 11
P1
24 8
???? ??
??
?? ?? ??
?? ??
??
11 22
PV PV PV
TT
nR nR nR
??
? ??? ?
??
??
? ?
00 00
11 2 2
PV 7PV 11
TPVPV
nR nR 2 4
?? ?? ??
?? ? ? ?
?? ????
?? ????
00
5P V
4nR
?
2. The correct figure that shows, schematically,
the wave pattern produced by superposition of
two waves of frequencies 9 Hz and 11 Hz, is :
(1) 0
12
t(s)
y
(2)
0
12
t(s)
y
(3)
0
12
t(s)
y
(4) 0
12
t(s)
y
Answer (3)
Sol. Beat frequency = |f
1
 – f
2
| = 11 – 9 = 2 Hz
3. A spaceship orbits around a planet at a height
of 20 km from its surface. Assuming that only
gravitational field of the planet acts on the
spaceship, what will be the number of
complete revolutions made by the spaceship in
24 hours around the planet?
[Given: Mass of planet = 8 × 10
22
 kg,
Radius of planet = 2 × 10
6
 m,
Gravitational constant G = 6.67 × 10
–11
 Nm
2
/kg
2
]
(1) 17 (2) 13
(3) 11 (4) 9
Answer (3)
Sol.
?
??
2r GM
T,v
vr
?? ??
3
rr
T2r 2
GM GM
?
?
??
???
312
11 22
(202) 10
T2
6.67 10 8 10
 sec
T = 7812.2 s
?  T 2.17hr 11revolutions
4. A plane is inclined at an angle ? = 30° with
respect to the horizontal. A particle is
projected with a speed u = 2 ms
–1
, from the
base of the plane, making an angle ? = 15° with
respect to the plane as shown in the figure. The
distance from the base, at which the particle
hits the plane is close to :
(Take g = 10 ms
–2
)
? = 30°
? = 15°
u
(1) 18 cm (2) 20 cm
(3) 14 cm (4) 26 cm
Answer (2)
Sol. Time of flight (T) 
?
?
?
2usin
gcos
??
(2)(2sin15) 4sin15
T
gcos30 10cos30
JEE (MAIN)-2019 (Online) Phase-2
? = 30°
? = 15°
y
gcos30
g
x
2 m/s
gsin30
??
2
1
Range(R)(2cos15)T gsin30(T)
2
??
???
??
??
2
2
4 sin15 1 16 sin 15
(2cos15) 10sin30
10cos30 3 100cos 30
?
? 
16 3 16
0.1952m 20cm
60
5. A bullet of mass 20 g has an initial speed of
1 ms
–1
, just before it starts penetrating a mud
wall of thickness 20 cm. If the wall offers a
mean resistance of 2.5 × 10
–2
 N, the speed of
the bullet after emerging from the other side of
the wall is close to :
(1) 0.4 ms
–1
(2) 0.7 ms
–1
(3) 0.3 ms
–1
(4) 0.1 ms
–1
Answer (2)
Sol. v
2
 = u
2
 – 2aS
?
?
??
?
??
??
???
??
2
22
3
2.5 10 20
v(1) (2)
100 20 10
??
2
1
v1
2
?
??
1
v m/s 0.7m/s
2
6. The time dependence of the position of a
particle of mass m = 2 is given by
??

2
ˆˆ
r(t) 2ti 3t j . Its angular momentum, with
respect to the origin, at time t = 2 is :
(1)
??
??
ˆ ˆ
34 k i (2)
??
ˆˆ
48 i j ?
(3)
ˆ
36k (4) ?
ˆ
48k
Answer (4)
Sol. ??

2
ˆˆ
r2ti 3t j
?? ?
 dr
ˆˆ
v2i6tj
dt
???
 dv
ˆ
a6j
dt
???


ˆ
Fma 12j
?? ?

ˆˆ
r(at t 2) 4i 12j
?? ? ? ? ? ??

 
ˆ ˆˆ ˆˆ
Lm(rv) 2(4i 12j) (2i 12j) 48k
7. The figure represents a voltage regulator
circuit using a Zener diode. The breakdown
voltage of the Zener diode is 6 V and the load
resistance is, R
L
 = 4 k ?. The series resistance
of the circuit is R
i
 = 1 k ?. If the battery voltage
V
B
 varies from 8 V to 16 V, what are the
minimum and maximum values of the current
through Zener diode?
R
i
R
L V
B
(1) 0.5 mA; 6 mA (2) 0.5 mA; 8.5 mA
(3) 1.5 mA; 8.5 mA (4) 1 mA; 8.5 mA
Answer (2)
Sol. 4 k ?
V
I
1
1 k ?
1
3
I
2
??
? ??
??
6
mA
4
1
31
I8–6– 0.5mA
22
??
???
??
??
??
??
??
??
2
3
I 16–6– 8.5mA
2
8. A metal coin of mass 5 g and radius 1 cm is fixed
to a thin stick AB of negligible mass as shown in
the figure. The system is initially at rest. The
constant torque, that will make the system rotate
about AB at 25 rotations per second in 5 s, is
closed to :
A
B
1 cm
(1) 4.0 × 10
–6
 Nm (2) 7.9 × 10
–6
 Nm
(3) 2.0 × 10
–5
 Nm (4) 1.6 × 10
–5
 Nm
Answer (3)
Sol. ? = I ?
??= ?
0 
+ ?t
? 25 × 2 ? = ( ?)5
??= 10 ?
?
2
5
mR
4
??
?? ?
??
??
??? ?
–3 –4
5
510 10 10
4
??
?? ?
??
??
= 2.0 × 10
–5 
Nm
9. In a Young’s double slit experiment, the ratio of
the slit’s width is 4 : 1. The ratio of the intensity
of maxima to minima, close to the central fringe
on the screen, will be :
(1)
??
4
31 :16 ? (2) 4 : 1
(3) 25 : 9 (4) 9 : 1
Answer (4)
Sol. I
1
 = 4 I
0
I
2
 = I
0
??
??
?
??
??
??
??
2
12
max
2
min
12
II
I
9
I1
I– I
10. Two radioactive substances A and B have
decay constants 5 ? and ? respectively. At
t = 0, a sample has the same number of the two
nuclei. The time taken for the ratio of the
number of nuclei to become 
2
1
e
??
??
??
 will be :
(1)
1
2 ?
(2)
1
4 ?
(3)
1
?
(4)
2
?
Answer (1)
Sol. N
x
(at t) = N
0
e
–5 ?t
N
y
(at t) = N
0
e
– ?t
–4t x
2
y
N
1
e
N e
?
??
? 4 ?t = 2
?
21
t
42
??
??
??
?? ??
11. A cubical block of side 0.5 m floats on water
with 30% of its volume under water. What is the
maximum weight that can be put on the block
without fully submerging it under water?
[Take, density of water = 10
3
 kg/m
3
]
(1) 30.1 kg (2) 46.3 kg
(3) 87.5 kg (4) 65.4 kg
Answer (3)
Sol. Given 
?? ??
3
cube
30
50 1 g M g
100
?? ?? …(i)
Let m mass should be placed
Hence 
?? ??
??
3
cube
50 1 g M m g ?? ? ? …(ii)
equation (ii) – equation (i)
? mg = (50)
3
 × g(1 – 0.3) = 125 × 0.7 × 10
3 
g
? m = 87.5 kg
12. The graph shows how the magnification m
produced by a thin lens varies with image
distance v. What is the focal length of the lens
used?
C
a b
v
m
(1)
2
b
ac
(2)
2
bc
a
(3)
a
c
(4)
b
c
Answer (4)
Sol. As the graph between magnification (m) and
image distance (v) varies linearly, then
m = k
1
v + k
2
?
12
v
kv k
u
??
?
2
1
k
1
k
uv
??
?
2
1
k
1
– k
vu
?
Clearly, 
12
1
kandk1
f
?? here
?
1b
f
slopeofm vgraphc
??
?
13. In Li
++
, electron in first Bohr orbit is excited to
a level by a radiation of wavelength ?. When the
ion gets deexcited to the ground state in all
possible ways (including intermediate
emissions), a total of six spectral lines are
observed. What is the value of ??
(Given: h = 6.63 × 10
–34 
Js; c = 3 × 10
8
 ms
–1
)
(1) 11.4 nm
(2) 12.3 nm
(3) 9.4 nm
(4) 10.8 nm
Answer (4)
Sol.
hc
E ??
?
2
1
(13.4)(3)1eV
16
??
??
??
??
?
1242 16
nm 10.8nm
(13.4) (9)(15)
?
??
?

14. Two blocks A and B of masses m
A
 = 1 kg and
m
B
 = 3 kg are kept on the table as shown in
figure. The coefficient of friction between A
and B is 0.2 and between B and the surface of
the table is also 0.2. The maximum force F that
can be applied on B horizontally, so that the
block A does not slide over the block B is:
[Take g = 10 m/s
2
]
B
A
F
(1) 40 N
(2) 12 N
(3) 16 N
(4) 8 N
Answer (3)
Sol. c
Ff
a
Mm
? ??
?
??
?
??
F (0.2)410 F 8
a
44
?? ? ??
??
??
??
We have 
F8
(0.2)10
4
?
?
?
F8 8 ??
?
F16 ?
15. In free space, a particle A of charge 1 ?C is
held fixed at a point P . Another particle B of the
same charge and mass 4 ?g is kept at a
distance of 1 mm from P. If B is released, then
its velocity at a distance of 9 mm from P is :
922
0
1
Take 9 10 Nm C
4
?
??
??
??
??
??
(1) 2.0 × 10
3
 m/s (2) 3.0 × 10
4
 m/s
(3) 1.5 × 10
2
 m/s (4) 1.0 m/s
Answer (Bonus)
1 C ?
1 mm
BB ?
m= 4 gram 
B 
?
= 4×10 kg
–9
A
(Fixed) 12
i
1
kq q
U
r
?
12
f
2
kq q
U
r
?
Conservation of energy
2 12 12
12
kqq kqq 1
mv
rr 2
??
2 12
12
2kq q 11
v
mr r
??
??
??
??
= 
912
93
29 10 10 1
1
9 410 10
?
??
?? ? ??
?
??
??
??
= 4 × 10
9
44
v 40 10 m/s 6.32 10 m/s ?? ? ?
16. Water from a tap emerges vertically
downwards with an initial speed of 1.0 ms
–1
.
The cross-sectional area of the tap is 10
–4
 m
2
.
Assume that the pressure is constant
throughout the stream of water and that the
flow is streamlined. The cross-sectional area of
stream, 0.15 m below the tap would be:
(Take g = 10 ms
–2
)
(1) 1 × 10
–5
 m
2
(2) 5 × 10
–5
 m
2
(3) 5 × 10
–4
 m
2
(4) 2 × 10
–5
 m
2
Answer (2)
Sol. Using Bernoullie’s equation 
2
21
vv2gh ??
Equation of continuity
A
1
V
1
 = A
2
V
2
22
2
15
(1cm )(1m/s) (A ) (1) 2 10
100
??
????
??
??
??
?
??
2
2
1
Alncm
2
?
? A
2
 = 5 × 10
–5
 m
2