JEE Mains 9 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)
09/04/2019
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)
09/04/2019
Evening
PART–A : PHYSICS
1. A test particle is moving in a circular orbit in
the gravitational field produced by a mass
density 
??
2
K
(r)
r
. Identify the correct relation
between the radius R of the particle’s orbit and
its period T :
(1) T/R is a constant (2) TR is a constant
(3) T/R
2
 is a constant (4) T
2
/R
3
 is a constant
Answer (1)
Sol.
R
2
(in)
2
0
k
M4rdr
r
?? ?
?
(in)
M4KR ??
?
2
2
4KR V
Gv4GK
R R
?
?? ? ??
?
2R 2 R
T
v 4GK
???
??
?
?
T
constant
R
?
2. The specific heats, C
P
 and C
V
 of a gas of
diatomic molecules, A, are given (in units of
J mol
–1
 K
–1
) by 29 and 22, respectively. Another
gas of diatomic molecules, B, has the
corresponding values 30 and 21. If they are
treated as ideal gases, then :
(1) A is rigid but B has a vibrational mode.
(2) A has a vibrational mode but B has none.
(3) Both A and B have a vibrational mode each.
(4) A has one vibrational mode and B has two.
Answer (2)
Sol. For (A) C
P
 = 29, C
V
 = 22
For (B) C
P
 = 30 C
V
 = 21
?
A
A
P
A
V
C
29
1.31
C22
?? ? ?
When A has vibrational degree of freedom,
then ?
A
 = 9/7  1.29.
B
P
B
V
C
30
1.42
C21
?? ? ?
B
? B has no vibrational degree of freedom
3. The position vector of a particle changes with
time according to the relation
22
ˆˆ
r(t) 15t i (4–20t )j
?
??
. What is the
magnitude of the acceleration at t = 1?
(1) 50 (2) 100
(3) 40 (4) 25
Answer (1)
Sol.
22
ˆˆ ˆ
r15ti 4j 20t j ???

??

dr
ˆˆ
30ti 40tj
dt
??

2
2
dr
ˆˆ
30i 40j
dt
?
?

2
2
2
dr
50m/s
dt
4. 50 W/m
2
 energy density of sunlight is normally
incident on the surface of a solar panel. Some
part of incident energy (25%) is reflected from
the surface and the rest is absorbed. The force
exerted on 1 m
2
 surface area will be close to
(c = 3 × 10
8
 m/s) :
(1) 20 × 10
–8
 N
(2) 35 × 10
–8
 N
(3) 15 × 10
–8
 N
(4) 10 × 10
–8
 N
Answer (1)
Sol. ?
W
P
c
? and pressure = I/c
?
hh 5h
P
44
? ? ??
?? ?
 for one photon
?
5Nh
pressure
4tA
??
??
But 
2
Nhc
50W/m
tA
??
?
??
?? ? ??
?
550
pressure
4c
??
? F
n
 = 
2
8
550 1m
20 10 N
4c
?
??
??
?
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)
09/04/2019
Evening
PART–A : PHYSICS
1. A test particle is moving in a circular orbit in
the gravitational field produced by a mass
density 
??
2
K
(r)
r
. Identify the correct relation
between the radius R of the particle’s orbit and
its period T :
(1) T/R is a constant (2) TR is a constant
(3) T/R
2
 is a constant (4) T
2
/R
3
 is a constant
Answer (1)
Sol.
R
2
(in)
2
0
k
M4rdr
r
?? ?
?
(in)
M4KR ??
?
2
2
4KR V
Gv4GK
R R
?
?? ? ??
?
2R 2 R
T
v 4GK
???
??
?
?
T
constant
R
?
2. The specific heats, C
P
 and C
V
 of a gas of
diatomic molecules, A, are given (in units of
J mol
–1
 K
–1
) by 29 and 22, respectively. Another
gas of diatomic molecules, B, has the
corresponding values 30 and 21. If they are
treated as ideal gases, then :
(1) A is rigid but B has a vibrational mode.
(2) A has a vibrational mode but B has none.
(3) Both A and B have a vibrational mode each.
(4) A has one vibrational mode and B has two.
Answer (2)
Sol. For (A) C
P
 = 29, C
V
 = 22
For (B) C
P
 = 30 C
V
 = 21
?
A
A
P
A
V
C
29
1.31
C22
?? ? ?
When A has vibrational degree of freedom,
then ?
A
 = 9/7  1.29.
B
P
B
V
C
30
1.42
C21
?? ? ?
B
? B has no vibrational degree of freedom
3. The position vector of a particle changes with
time according to the relation
22
ˆˆ
r(t) 15t i (4–20t )j
?
??
. What is the
magnitude of the acceleration at t = 1?
(1) 50 (2) 100
(3) 40 (4) 25
Answer (1)
Sol.
22
ˆˆ ˆ
r15ti 4j 20t j ???

??

dr
ˆˆ
30ti 40tj
dt
??

2
2
dr
ˆˆ
30i 40j
dt
?
?

2
2
2
dr
50m/s
dt
4. 50 W/m
2
 energy density of sunlight is normally
incident on the surface of a solar panel. Some
part of incident energy (25%) is reflected from
the surface and the rest is absorbed. The force
exerted on 1 m
2
 surface area will be close to
(c = 3 × 10
8
 m/s) :
(1) 20 × 10
–8
 N
(2) 35 × 10
–8
 N
(3) 15 × 10
–8
 N
(4) 10 × 10
–8
 N
Answer (1)
Sol. ?
W
P
c
? and pressure = I/c
?
hh 5h
P
44
? ? ??
?? ?
 for one photon
?
5Nh
pressure
4tA
??
??
But 
2
Nhc
50W/m
tA
??
?
??
?? ? ??
?
550
pressure
4c
??
? F
n
 = 
2
8
550 1m
20 10 N
4c
?
??
??
?
5. A metal wire of resistance 3 ? is elongated to
make a uniform wire of double its previous
length. This new wire is now bent and the ends
joined to make a circle. If two points on this
circle make an angle 60° at the centre, the
equivalent resistance between these two points
will be :
(1)
?
5
 
3
(2)
?
5
 
2
(3)
?
7
 
2
(4)
?
12
 
5
Answer (1)
Sol.
A
B
R
1
R
2
0
l
R3
A
?? ? ?
Now if l = 2l
Then A = 
A
2
?
2l 2
R12
A
??
???
?? ?
1
12
R2
6
R
2
 = 10 ?
?
eq
11 1 6
R21010
?? ?
? R
eq
 = 
5
3
?
6. A thin smooth rod of length L and mass M is
rotating freely with angular speed ?
0
 about an
axis perpendicular to the rod and passing
through its center. Two beads of mass m and
negligible size are at the center of the rod
initially. The beads are free to slide along the
rod. The angular speed of the system, when
the beads reach the opposite ends of the rod,
will be :
(1)
?
?
0
M
M3m
(2)
?
?
0
M
Mm
(3)
?
?
0
M
M6m
(4)
?
?
0
M
M2m
Answer (3)
Sol. Initial angular momentum = Final Angular
Momentum
22 2
0
ML ML mL
2
12 12 4
??
?? ? ?
??
??
?
0
M
M6m
?
??
?
7. Moment of inertia of a body about a given axis
is 1.5 kg m
2
. Initially the body is at rest. In order
to produce a rotational kinetic energy of 1200 J,
the angular acceleration of 20 rad/s
2
 must be
applied about the axis for a duration of
(1) 2.5 s (2) 2 s
(3) 5 s (4) 3 s
Answer (2)
Sol. I = 150; ? = 20 rad/s
2
?? = ?t
2
1
E I 1200
2
?? ?
2
1
1.5 (20t) 1200J
2
?? ?
? t = 2 s
8. The resistance of a galvanometer is 50 ohm
and the maximum current which can be passed
through it is 0.002 A. What resistance must be
connected to it in order to convert it into an
ammeter of range 0 – 0.5 A?
(1) 0.2 ohm
(2) 0.002 ohm
(3) 0.5 ohm
(4) 0.02 ohm
Answer (1)
Sol.
0.5 A
R = 50 
g
?
I = 0.002 A
g
G
I
g
(0.5 – I ) S
g
We have
I
g
 R
g
 = (0.5 – I
g
) S
? (0.002) (50) = (0.5 – I
g
) S
?
0.002 50
S0.2
0.5
?
?? 
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)
09/04/2019
Evening
PART–A : PHYSICS
1. A test particle is moving in a circular orbit in
the gravitational field produced by a mass
density 
??
2
K
(r)
r
. Identify the correct relation
between the radius R of the particle’s orbit and
its period T :
(1) T/R is a constant (2) TR is a constant
(3) T/R
2
 is a constant (4) T
2
/R
3
 is a constant
Answer (1)
Sol.
R
2
(in)
2
0
k
M4rdr
r
?? ?
?
(in)
M4KR ??
?
2
2
4KR V
Gv4GK
R R
?
?? ? ??
?
2R 2 R
T
v 4GK
???
??
?
?
T
constant
R
?
2. The specific heats, C
P
 and C
V
 of a gas of
diatomic molecules, A, are given (in units of
J mol
–1
 K
–1
) by 29 and 22, respectively. Another
gas of diatomic molecules, B, has the
corresponding values 30 and 21. If they are
treated as ideal gases, then :
(1) A is rigid but B has a vibrational mode.
(2) A has a vibrational mode but B has none.
(3) Both A and B have a vibrational mode each.
(4) A has one vibrational mode and B has two.
Answer (2)
Sol. For (A) C
P
 = 29, C
V
 = 22
For (B) C
P
 = 30 C
V
 = 21
?
A
A
P
A
V
C
29
1.31
C22
?? ? ?
When A has vibrational degree of freedom,
then ?
A
 = 9/7  1.29.
B
P
B
V
C
30
1.42
C21
?? ? ?
B
? B has no vibrational degree of freedom
3. The position vector of a particle changes with
time according to the relation
22
ˆˆ
r(t) 15t i (4–20t )j
?
??
. What is the
magnitude of the acceleration at t = 1?
(1) 50 (2) 100
(3) 40 (4) 25
Answer (1)
Sol.
22
ˆˆ ˆ
r15ti 4j 20t j ???

??

dr
ˆˆ
30ti 40tj
dt
??

2
2
dr
ˆˆ
30i 40j
dt
?
?

2
2
2
dr
50m/s
dt
4. 50 W/m
2
 energy density of sunlight is normally
incident on the surface of a solar panel. Some
part of incident energy (25%) is reflected from
the surface and the rest is absorbed. The force
exerted on 1 m
2
 surface area will be close to
(c = 3 × 10
8
 m/s) :
(1) 20 × 10
–8
 N
(2) 35 × 10
–8
 N
(3) 15 × 10
–8
 N
(4) 10 × 10
–8
 N
Answer (1)
Sol. ?
W
P
c
? and pressure = I/c
?
hh 5h
P
44
? ? ??
?? ?
 for one photon
?
5Nh
pressure
4tA
??
??
But 
2
Nhc
50W/m
tA
??
?
??
?? ? ??
?
550
pressure
4c
??
? F
n
 = 
2
8
550 1m
20 10 N
4c
?
??
??
?
5. A metal wire of resistance 3 ? is elongated to
make a uniform wire of double its previous
length. This new wire is now bent and the ends
joined to make a circle. If two points on this
circle make an angle 60° at the centre, the
equivalent resistance between these two points
will be :
(1)
?
5
 
3
(2)
?
5
 
2
(3)
?
7
 
2
(4)
?
12
 
5
Answer (1)
Sol.
A
B
R
1
R
2
0
l
R3
A
?? ? ?
Now if l = 2l
Then A = 
A
2
?
2l 2
R12
A
??
???
?? ?
1
12
R2
6
R
2
 = 10 ?
?
eq
11 1 6
R21010
?? ?
? R
eq
 = 
5
3
?
6. A thin smooth rod of length L and mass M is
rotating freely with angular speed ?
0
 about an
axis perpendicular to the rod and passing
through its center. Two beads of mass m and
negligible size are at the center of the rod
initially. The beads are free to slide along the
rod. The angular speed of the system, when
the beads reach the opposite ends of the rod,
will be :
(1)
?
?
0
M
M3m
(2)
?
?
0
M
Mm
(3)
?
?
0
M
M6m
(4)
?
?
0
M
M2m
Answer (3)
Sol. Initial angular momentum = Final Angular
Momentum
22 2
0
ML ML mL
2
12 12 4
??
?? ? ?
??
??
?
0
M
M6m
?
??
?
7. Moment of inertia of a body about a given axis
is 1.5 kg m
2
. Initially the body is at rest. In order
to produce a rotational kinetic energy of 1200 J,
the angular acceleration of 20 rad/s
2
 must be
applied about the axis for a duration of
(1) 2.5 s (2) 2 s
(3) 5 s (4) 3 s
Answer (2)
Sol. I = 150; ? = 20 rad/s
2
?? = ?t
2
1
E I 1200
2
?? ?
2
1
1.5 (20t) 1200J
2
?? ?
? t = 2 s
8. The resistance of a galvanometer is 50 ohm
and the maximum current which can be passed
through it is 0.002 A. What resistance must be
connected to it in order to convert it into an
ammeter of range 0 – 0.5 A?
(1) 0.2 ohm
(2) 0.002 ohm
(3) 0.5 ohm
(4) 0.02 ohm
Answer (1)
Sol.
0.5 A
R = 50 
g
?
I = 0.002 A
g
G
I
g
(0.5 – I ) S
g
We have
I
g
 R
g
 = (0.5 – I
g
) S
? (0.002) (50) = (0.5 – I
g
) S
?
0.002 50
S0.2
0.5
?
?? 
9. A He
+
 ion is in its first excited state. Its
ionization energy is :
(1) 13.60 eV (2) 6.04 eV
(3) 48.36 eV (4) 54.40 eV
Answer (1)
Sol. E
n
: 
2
00
0
2
Ez E 4
E
4 n
???
???
To ionise it E
0
 energy must be supplied.
? E
0
 = 13.6 eV.
10. Two cars A and B are moving away from each
other in opposite directions. Both the cars are
moving with a speed of 20 ms
–1
 with respect to
the ground. If an observer in car A detects a
frequency 2000 Hz of the sound coming from
car B, what is the natural frequency of the
sound source in car B?
(speed of sound in air = 340 ms
–1
)
(1) 2150 Hz (2) 2300 Hz
(3) 2060 Hz (4) 2250 Hz
Answer (4)
Sol.
S
20 m/s
O
20 m/s
0
00
s
(v u ) (v20)
ff f
(vu)(v20)
? ?
??? ?
??
? ? ? ? ?
0
320
2000 f
360
??
? ? ? ? ?
0
2000 9
f 2250Hz
8
?
?? .
11. A convex lens of focal length 20 cm produces
images of the same magnification 2 when an
object is kept at two distances x
1
 and x
2
(x
1
 > x
2
) from the lens. The ratio of x
1
 and x
2
 is :
(1) 3 : 1 (2) 2 : 1
(3) 4 : 3 (4) 5 : 3
Answer (1)
Sol. ?
11 1
vu f
??
? ? ? ? ? v = (± 2u)
? ? ? ? ?
11 1 3 1
2uu20 2u20
?? ? ?
? ? ? ? ? u
1
 = 30 cm
And
11 1
u2u20
??
? ? ? ? ?
2
11
u10
2u 20
?? ?
? ? ? ? ?
30
3
10
? .
12. The position of a particle as a function of
time t, is given by
x(t) = at + bt
2
 – ct
3
where a, b and c are constants. When the
particle attains zero acceleration, then its
velocity will be :
(1)
2
b
a
4c
?
(2)
2
b
a
3c
?
(3)
2
b
a
2c
?
(4)
2
b
a
c
?
Answer (2)
Sol. x = at + bt
2
 – ct
3
x

 = a + 2bt – 3ct
2
x

 = 2b – 6ct
For x0 ?

b
t
3c
??
?
2
b b
vx a 2b 3c
3c 3c 3c
??
? ??
?? ? ?
?? ??
? ??
??

?
22 2
b2b b
vaa
3c 3c 3c
?? ? ? ? ?
13. The parallel combination of two air filled
parallel plate capacitors of capacitance C and
nC is connected to a battery of voltage, V.
When the capacitors are fully charged, the
battery is removed and after that a dielectric
material of dielectric constant K is placed
between the two plates of the first capacitor.
The new potential difference of the combined
system is :
(1)
??
??
n1V
Kn
?
?
(2)
V
Kn ?
(3) V (4)
nV
Kn ?
Answer (1)
Sol. Initially
Q = CV(1 + n)
? C
eq
 = (K + n)C
CV
nCV
?
??
??
??
??
CV 1 n V 1 n
V
KnC K n
??
??
??
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)
09/04/2019
Evening
PART–A : PHYSICS
1. A test particle is moving in a circular orbit in
the gravitational field produced by a mass
density 
??
2
K
(r)
r
. Identify the correct relation
between the radius R of the particle’s orbit and
its period T :
(1) T/R is a constant (2) TR is a constant
(3) T/R
2
 is a constant (4) T
2
/R
3
 is a constant
Answer (1)
Sol.
R
2
(in)
2
0
k
M4rdr
r
?? ?
?
(in)
M4KR ??
?
2
2
4KR V
Gv4GK
R R
?
?? ? ??
?
2R 2 R
T
v 4GK
???
??
?
?
T
constant
R
?
2. The specific heats, C
P
 and C
V
 of a gas of
diatomic molecules, A, are given (in units of
J mol
–1
 K
–1
) by 29 and 22, respectively. Another
gas of diatomic molecules, B, has the
corresponding values 30 and 21. If they are
treated as ideal gases, then :
(1) A is rigid but B has a vibrational mode.
(2) A has a vibrational mode but B has none.
(3) Both A and B have a vibrational mode each.
(4) A has one vibrational mode and B has two.
Answer (2)
Sol. For (A) C
P
 = 29, C
V
 = 22
For (B) C
P
 = 30 C
V
 = 21
?
A
A
P
A
V
C
29
1.31
C22
?? ? ?
When A has vibrational degree of freedom,
then ?
A
 = 9/7  1.29.
B
P
B
V
C
30
1.42
C21
?? ? ?
B
? B has no vibrational degree of freedom
3. The position vector of a particle changes with
time according to the relation
22
ˆˆ
r(t) 15t i (4–20t )j
?
??
. What is the
magnitude of the acceleration at t = 1?
(1) 50 (2) 100
(3) 40 (4) 25
Answer (1)
Sol.
22
ˆˆ ˆ
r15ti 4j 20t j ???

??

dr
ˆˆ
30ti 40tj
dt
??

2
2
dr
ˆˆ
30i 40j
dt
?
?

2
2
2
dr
50m/s
dt
4. 50 W/m
2
 energy density of sunlight is normally
incident on the surface of a solar panel. Some
part of incident energy (25%) is reflected from
the surface and the rest is absorbed. The force
exerted on 1 m
2
 surface area will be close to
(c = 3 × 10
8
 m/s) :
(1) 20 × 10
–8
 N
(2) 35 × 10
–8
 N
(3) 15 × 10
–8
 N
(4) 10 × 10
–8
 N
Answer (1)
Sol. ?
W
P
c
? and pressure = I/c
?
hh 5h
P
44
? ? ??
?? ?
 for one photon
?
5Nh
pressure
4tA
??
??
But 
2
Nhc
50W/m
tA
??
?
??
?? ? ??
?
550
pressure
4c
??
? F
n
 = 
2
8
550 1m
20 10 N
4c
?
??
??
?
5. A metal wire of resistance 3 ? is elongated to
make a uniform wire of double its previous
length. This new wire is now bent and the ends
joined to make a circle. If two points on this
circle make an angle 60° at the centre, the
equivalent resistance between these two points
will be :
(1)
?
5
 
3
(2)
?
5
 
2
(3)
?
7
 
2
(4)
?
12
 
5
Answer (1)
Sol.
A
B
R
1
R
2
0
l
R3
A
?? ? ?
Now if l = 2l
Then A = 
A
2
?
2l 2
R12
A
??
???
?? ?
1
12
R2
6
R
2
 = 10 ?
?
eq
11 1 6
R21010
?? ?
? R
eq
 = 
5
3
?
6. A thin smooth rod of length L and mass M is
rotating freely with angular speed ?
0
 about an
axis perpendicular to the rod and passing
through its center. Two beads of mass m and
negligible size are at the center of the rod
initially. The beads are free to slide along the
rod. The angular speed of the system, when
the beads reach the opposite ends of the rod,
will be :
(1)
?
?
0
M
M3m
(2)
?
?
0
M
Mm
(3)
?
?
0
M
M6m
(4)
?
?
0
M
M2m
Answer (3)
Sol. Initial angular momentum = Final Angular
Momentum
22 2
0
ML ML mL
2
12 12 4
??
?? ? ?
??
??
?
0
M
M6m
?
??
?
7. Moment of inertia of a body about a given axis
is 1.5 kg m
2
. Initially the body is at rest. In order
to produce a rotational kinetic energy of 1200 J,
the angular acceleration of 20 rad/s
2
 must be
applied about the axis for a duration of
(1) 2.5 s (2) 2 s
(3) 5 s (4) 3 s
Answer (2)
Sol. I = 150; ? = 20 rad/s
2
?? = ?t
2
1
E I 1200
2
?? ?
2
1
1.5 (20t) 1200J
2
?? ?
? t = 2 s
8. The resistance of a galvanometer is 50 ohm
and the maximum current which can be passed
through it is 0.002 A. What resistance must be
connected to it in order to convert it into an
ammeter of range 0 – 0.5 A?
(1) 0.2 ohm
(2) 0.002 ohm
(3) 0.5 ohm
(4) 0.02 ohm
Answer (1)
Sol.
0.5 A
R = 50 
g
?
I = 0.002 A
g
G
I
g
(0.5 – I ) S
g
We have
I
g
 R
g
 = (0.5 – I
g
) S
? (0.002) (50) = (0.5 – I
g
) S
?
0.002 50
S0.2
0.5
?
?? 
9. A He
+
 ion is in its first excited state. Its
ionization energy is :
(1) 13.60 eV (2) 6.04 eV
(3) 48.36 eV (4) 54.40 eV
Answer (1)
Sol. E
n
: 
2
00
0
2
Ez E 4
E
4 n
???
???
To ionise it E
0
 energy must be supplied.
? E
0
 = 13.6 eV.
10. Two cars A and B are moving away from each
other in opposite directions. Both the cars are
moving with a speed of 20 ms
–1
 with respect to
the ground. If an observer in car A detects a
frequency 2000 Hz of the sound coming from
car B, what is the natural frequency of the
sound source in car B?
(speed of sound in air = 340 ms
–1
)
(1) 2150 Hz (2) 2300 Hz
(3) 2060 Hz (4) 2250 Hz
Answer (4)
Sol.
S
20 m/s
O
20 m/s
0
00
s
(v u ) (v20)
ff f
(vu)(v20)
? ?
??? ?
??
? ? ? ? ?
0
320
2000 f
360
??
? ? ? ? ?
0
2000 9
f 2250Hz
8
?
?? .
11. A convex lens of focal length 20 cm produces
images of the same magnification 2 when an
object is kept at two distances x
1
 and x
2
(x
1
 > x
2
) from the lens. The ratio of x
1
 and x
2
 is :
(1) 3 : 1 (2) 2 : 1
(3) 4 : 3 (4) 5 : 3
Answer (1)
Sol. ?
11 1
vu f
??
? ? ? ? ? v = (± 2u)
? ? ? ? ?
11 1 3 1
2uu20 2u20
?? ? ?
? ? ? ? ? u
1
 = 30 cm
And
11 1
u2u20
??
? ? ? ? ?
2
11
u10
2u 20
?? ?
? ? ? ? ?
30
3
10
? .
12. The position of a particle as a function of
time t, is given by
x(t) = at + bt
2
 – ct
3
where a, b and c are constants. When the
particle attains zero acceleration, then its
velocity will be :
(1)
2
b
a
4c
?
(2)
2
b
a
3c
?
(3)
2
b
a
2c
?
(4)
2
b
a
c
?
Answer (2)
Sol. x = at + bt
2
 – ct
3
x

 = a + 2bt – 3ct
2
x

 = 2b – 6ct
For x0 ?

b
t
3c
??
?
2
b b
vx a 2b 3c
3c 3c 3c
??
? ??
?? ? ?
?? ??
? ??
??

?
22 2
b2b b
vaa
3c 3c 3c
?? ? ? ? ?
13. The parallel combination of two air filled
parallel plate capacitors of capacitance C and
nC is connected to a battery of voltage, V.
When the capacitors are fully charged, the
battery is removed and after that a dielectric
material of dielectric constant K is placed
between the two plates of the first capacitor.
The new potential difference of the combined
system is :
(1)
??
??
n1V
Kn
?
?
(2)
V
Kn ?
(3) V (4)
nV
Kn ?
Answer (1)
Sol. Initially
Q = CV(1 + n)
? C
eq
 = (K + n)C
CV
nCV
?
??
??
??
??
CV 1 n V 1 n
V
KnC K n
??
??
??
14. The physical sizes of the transmitter and
receiver antenna in a communication system
are:
(1) Inversely proportional to modulation
frequency
(2) Inversely proportional to carrier frequency
(3) Proportional to carrier frequency
(4) Independent of both carrier and modulation
frequency
Answer (2)
Sol. Size of antenna depends on wavelength of
carrier wave.
15. A wedge of mass M = 4m lies on a frictionless
plane. A particle of mass m approaches the
wedge with speed v. There is no friction
between the particle and the plane or between
the particle and the wedge. The maximum
height climbed by the particle on the wedge is
given by :
(1)
2
v
g
(2)
2
2v
5g
(3)
2
2v
7g
(4)
2
v
2g
Answer (2)
Sol. mv = (4m + m)v ?
? Common speed 
v
v
5
??
mgh + 
2
2
1v 1
5m mv
2252
??
? mgh = 
22
111 4
mv 1– mv
252 5
??
??
??
??
?
22
2mv 2v
h
5mg 5g
??
?
16. A wooden block floating in a bucket of water
has 
4
5
 of its volume submerged. When certain
amount of an oil is poured into the bucket, it is
found that the block is just under the oil surface
with half of its volume under water and half in
oil. The density of oil relative to that of water is
(1) 0.5 (2) 0.8
(3) 0.7 (4) 0.6
Answer (4)
Sol.
4
Vg v g
5
?
?? ? …(1)
0
vv
Vg g g
22
?
?? ? ? ?
?
oil
4
22 5
?
?
?? ??
???
??
??
?
oil
41 3
–
25210
??
?
??
?? ? ?
??
??
?
oil
3
0.6
5
??
?? ? ? ?
17. Two materials having coefficients of thermal
conductivity ‘3K’ and ‘K’ and thickness ‘d’ and
‘3d’, respectively, are joined to form a slab as
shown in the figure. The temperatures of the
outer surfaces are ‘ ?
2
’ and ‘ ?
1
’ respectively,
( ?
2
 > ?
1
). The temperature at the interface is:
?
2
3d d
3K K
?
1
(1)
1 2
5
66
? ?
? (2)
1 2
9
10 10
? ?
?
(3)
1 2
2
33
? ?
?
(4)
21
2
???
Answer (2)
Sol. ?? ??
21
3KA KA
H– –
d3d
???? ??
?
1 2
9
10 10
? ?
?? ?
18. The area of a square is 5.29 cm
2
. The area of
7 such squares taking into account the
significant figures is :
(1) 37.03 cm
2
(2) 37.0 cm
2
(3) 37.030 cm
2
(4) 37 cm
2
Answer (2)
Sol. 5.29 × 7 = 37.0 cm
2
Answer should be in 3 significant digits.
19. Diameter of the objective lens of a telescope is
250 cm. For light of wavelength 600 nm.
coming from a distant object, the limit of
resolution of the telescope is close to :
(1) 1.5 × 10
–7 
rad
(2) 3.0 × 10
–7
 rad
(3) 2.0 × 10
–7 
rad
(4) 4.5 × 10
–7
 rad