JEE Main 2022 Physics June 29 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


 
  
PHYSICS 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. A small toy starts moving from the position of rest 
under a constant acceleration. If it travels a 
distance of 10 m in t s, the distance travelled by the 
toy in the next t s will be : 
 (A) 10 m (B) 20 m 
 (C) 30 m (D) 40 m 
Answer (C) 
Sol. 
2
1
10 m
2
at = 
 
2
1
(2 ) 40 m
2
at = 
 ? Distance travelled in next t s 
  = 40 – 10 = 30 m 
2. At what temperature a gold ring of diameter 
6.230 cm be heated so that it can be fitted on a 
wooden bangle of diameter 6.241 cm? Both the 
diameters have been measured at room 
temperature (27°C). 
 (Given: coefficient of linear thermal expansion of 
gold aL = 1.4 × 10
–5
 K
–1
) 
 (A) 125.7°C (B) 91.7°C 
 (C) 425.7°C (D) 152.7°C 
Answer (D) 
Sol. ?D = Da?T 
 
5
0.011
6.230 1.4 10
T
-
?=
××
 
 = 126.11°C 
 ? T f = T + ?T 
  = (27 + 126.11)°C 
  = 153.11°C 
 ? Option (D) is correct 
3. Two point charges Q each are placed at a distance 
d apart. A third point charge q is placed at a 
distance x from mid-point on the perpendicular 
bisector. The value of x at which charge q will 
experience the maximum Coulombs force is : 
 (A) x = d (B) 
2
=
d
x 
 (C) 
2
=
d
x (D) 
22
=
d
x 
Answer (D) 
Sol. Force experienced by the charge q 
 
3
2
2
2
2
kQqx
F
d
x
=
? ?
??
+ ? ? ??
?? ? ?
 
 For maximum Coulomb’s force for x 
 0
dF
dx
= 
 On solving 
22
d
x = 
4. The speed of light in media ‘A’ and ‘B’ are 
2.0 × 10
10
 cm/s and 1.5 × 10
10
 cm/s respectively. A 
ray of light enters from the medium B to A at an 
incident angle ‘?’. If the ray suffers total internal 
reflection, then 
 (A) 
1
3
sin
4
-
??
?=
??
??
 (B) 
1
2
sin
3
-
? ?
?>
? ?
? ?
 
 (C) 
1
3
sin
4
-
??
?<
??
??
 (D) 
1
3
sin
4
-
??
?>
??
??
 
Answer (D) 
Sol. 
8
8
3 10
1.5
2 10
A
×
µ= =
×
 
 
8
8
3 10
2
1.5 10
B
×
µ= =
×
 
 For TIR 
 ? > i c 
 
1
1.5
sin
2
-
??
?>
??
??
 
 
1
3
sin
4
-
??
?>
??
??
 
Page 2


 
  
PHYSICS 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. A small toy starts moving from the position of rest 
under a constant acceleration. If it travels a 
distance of 10 m in t s, the distance travelled by the 
toy in the next t s will be : 
 (A) 10 m (B) 20 m 
 (C) 30 m (D) 40 m 
Answer (C) 
Sol. 
2
1
10 m
2
at = 
 
2
1
(2 ) 40 m
2
at = 
 ? Distance travelled in next t s 
  = 40 – 10 = 30 m 
2. At what temperature a gold ring of diameter 
6.230 cm be heated so that it can be fitted on a 
wooden bangle of diameter 6.241 cm? Both the 
diameters have been measured at room 
temperature (27°C). 
 (Given: coefficient of linear thermal expansion of 
gold aL = 1.4 × 10
–5
 K
–1
) 
 (A) 125.7°C (B) 91.7°C 
 (C) 425.7°C (D) 152.7°C 
Answer (D) 
Sol. ?D = Da?T 
 
5
0.011
6.230 1.4 10
T
-
?=
××
 
 = 126.11°C 
 ? T f = T + ?T 
  = (27 + 126.11)°C 
  = 153.11°C 
 ? Option (D) is correct 
3. Two point charges Q each are placed at a distance 
d apart. A third point charge q is placed at a 
distance x from mid-point on the perpendicular 
bisector. The value of x at which charge q will 
experience the maximum Coulombs force is : 
 (A) x = d (B) 
2
=
d
x 
 (C) 
2
=
d
x (D) 
22
=
d
x 
Answer (D) 
Sol. Force experienced by the charge q 
 
3
2
2
2
2
kQqx
F
d
x
=
? ?
??
+ ? ? ??
?? ? ?
 
 For maximum Coulomb’s force for x 
 0
dF
dx
= 
 On solving 
22
d
x = 
4. The speed of light in media ‘A’ and ‘B’ are 
2.0 × 10
10
 cm/s and 1.5 × 10
10
 cm/s respectively. A 
ray of light enters from the medium B to A at an 
incident angle ‘?’. If the ray suffers total internal 
reflection, then 
 (A) 
1
3
sin
4
-
??
?=
??
??
 (B) 
1
2
sin
3
-
? ?
?>
? ?
? ?
 
 (C) 
1
3
sin
4
-
??
?<
??
??
 (D) 
1
3
sin
4
-
??
?>
??
??
 
Answer (D) 
Sol. 
8
8
3 10
1.5
2 10
A
×
µ= =
×
 
 
8
8
3 10
2
1.5 10
B
×
µ= =
×
 
 For TIR 
 ? > i c 
 
1
1.5
sin
2
-
??
?>
??
??
 
 
1
3
sin
4
-
??
?>
??
??
 
 
  
5. In the following nuclear reaction, 
 
1 2 34
-
aß a ?
?? ? ?? ? ? ?? ? ?? ? DD D D D 
 Mass number of D is 182 and atomic number is 74. 
Mass number and atomic number of D4 respectively 
will be _____ 
 (A) 174 and 71 
 (B) 174 and 69 
 (C) 172 and 69 
 (D) 172 and 71 
Answer (A) 
Sol. Equivalent reaction can be written as 
 D ?? D4 + 2a + ß
–
 + ? 
 ? Mass number of D4 = Mass number of D – 2 × 4 
   = 182 – 8 = 174 
 ? Atomic number of D4 
  = Atomic number of D – 2 × 2 + 1 
  = 74 – 4 + 1 = 71 
6. The electric field at a point associated with a light 
wave is given by 
 E = 200[sin(6 × 10
15
)t + sin(9 × 10
15
)t] Vm
–1 
 Given : h = 4.14 × 10
–15
 eVs 
 If this light falls on a metal surface having a work 
function of 2.50 eV, the maximum kinetic energy of 
the photoelectrons will be 
 (A) 1.90 eV 
 (B) 3.27 eV 
 (C) 3.60 eV 
 (D) 3.42 eV 
Answer (D) 
Sol. Frequency of EM waves = 
15
6
10
2
×
p
 and 
15
9
10
2
×
p
 
 Energy of one photon of these waves 
 
15 15
6
 eV 4.14 10 10
2
-
? ?
= × ××
? ?
p ? ?
  
 and 
15 15
9
 eV 4.14 10 10
2
-
? ?
× ××
? ?
p ? ?
 
 = 3.95 eV and 5.93 eV 
 ? Energy of maximum energetic electrons 
  = 5.93 – 2.50 = 3.43 eV 
7. A capacitor is discharging through a resistor R. 
Consider in time t1, the energy stored in the 
capacitor reduces to half of its initial value and in 
time t2, the charge stored reduces to one eighth of 
its initial value. The ratio 
1
2
t
t
 will be 
 (A) 
1
2
 (B) 
1
3
 
 (C) 
1
4
 (D) 
1
6
 
Answer (D) 
Sol. For a discharging capacitor when energy reduces 
to half the charge would become 
1
2
times the 
initial value. 
 ? 
1
1/2
/
1
2
t
e
-t ? ?
=
? ?
? ?
 
 Similarly, 
2
3
/
1
2
t
e
-t ? ?
=
? ?
? ?
 
 ? 
1
2
1
6
t
t
= 
8. Starting with the same initial conditions, an ideal 
gas expands from volume V1 to V2 in three different 
ways. The work done by the gas is W1 if the process 
is purely isothermal, W2, if the process is purely 
adiabatic and W3 if the process is purely isobaric. 
Then, choose the correct option. 
 (A) W1 < W2 < W3 
 (B) W2 < W3 < W1 
 (C) W3 < W1 < W2 
 (D) W2 < W1 < W3 
Answer (D) 
Sol.  
 Comparing the area under the PV graph 
  A3 > A1 > A2 
 ? W3 > W1 > W2 
Page 3


 
  
PHYSICS 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. A small toy starts moving from the position of rest 
under a constant acceleration. If it travels a 
distance of 10 m in t s, the distance travelled by the 
toy in the next t s will be : 
 (A) 10 m (B) 20 m 
 (C) 30 m (D) 40 m 
Answer (C) 
Sol. 
2
1
10 m
2
at = 
 
2
1
(2 ) 40 m
2
at = 
 ? Distance travelled in next t s 
  = 40 – 10 = 30 m 
2. At what temperature a gold ring of diameter 
6.230 cm be heated so that it can be fitted on a 
wooden bangle of diameter 6.241 cm? Both the 
diameters have been measured at room 
temperature (27°C). 
 (Given: coefficient of linear thermal expansion of 
gold aL = 1.4 × 10
–5
 K
–1
) 
 (A) 125.7°C (B) 91.7°C 
 (C) 425.7°C (D) 152.7°C 
Answer (D) 
Sol. ?D = Da?T 
 
5
0.011
6.230 1.4 10
T
-
?=
××
 
 = 126.11°C 
 ? T f = T + ?T 
  = (27 + 126.11)°C 
  = 153.11°C 
 ? Option (D) is correct 
3. Two point charges Q each are placed at a distance 
d apart. A third point charge q is placed at a 
distance x from mid-point on the perpendicular 
bisector. The value of x at which charge q will 
experience the maximum Coulombs force is : 
 (A) x = d (B) 
2
=
d
x 
 (C) 
2
=
d
x (D) 
22
=
d
x 
Answer (D) 
Sol. Force experienced by the charge q 
 
3
2
2
2
2
kQqx
F
d
x
=
? ?
??
+ ? ? ??
?? ? ?
 
 For maximum Coulomb’s force for x 
 0
dF
dx
= 
 On solving 
22
d
x = 
4. The speed of light in media ‘A’ and ‘B’ are 
2.0 × 10
10
 cm/s and 1.5 × 10
10
 cm/s respectively. A 
ray of light enters from the medium B to A at an 
incident angle ‘?’. If the ray suffers total internal 
reflection, then 
 (A) 
1
3
sin
4
-
??
?=
??
??
 (B) 
1
2
sin
3
-
? ?
?>
? ?
? ?
 
 (C) 
1
3
sin
4
-
??
?<
??
??
 (D) 
1
3
sin
4
-
??
?>
??
??
 
Answer (D) 
Sol. 
8
8
3 10
1.5
2 10
A
×
µ= =
×
 
 
8
8
3 10
2
1.5 10
B
×
µ= =
×
 
 For TIR 
 ? > i c 
 
1
1.5
sin
2
-
??
?>
??
??
 
 
1
3
sin
4
-
??
?>
??
??
 
 
  
5. In the following nuclear reaction, 
 
1 2 34
-
aß a ?
?? ? ?? ? ? ?? ? ?? ? DD D D D 
 Mass number of D is 182 and atomic number is 74. 
Mass number and atomic number of D4 respectively 
will be _____ 
 (A) 174 and 71 
 (B) 174 and 69 
 (C) 172 and 69 
 (D) 172 and 71 
Answer (A) 
Sol. Equivalent reaction can be written as 
 D ?? D4 + 2a + ß
–
 + ? 
 ? Mass number of D4 = Mass number of D – 2 × 4 
   = 182 – 8 = 174 
 ? Atomic number of D4 
  = Atomic number of D – 2 × 2 + 1 
  = 74 – 4 + 1 = 71 
6. The electric field at a point associated with a light 
wave is given by 
 E = 200[sin(6 × 10
15
)t + sin(9 × 10
15
)t] Vm
–1 
 Given : h = 4.14 × 10
–15
 eVs 
 If this light falls on a metal surface having a work 
function of 2.50 eV, the maximum kinetic energy of 
the photoelectrons will be 
 (A) 1.90 eV 
 (B) 3.27 eV 
 (C) 3.60 eV 
 (D) 3.42 eV 
Answer (D) 
Sol. Frequency of EM waves = 
15
6
10
2
×
p
 and 
15
9
10
2
×
p
 
 Energy of one photon of these waves 
 
15 15
6
 eV 4.14 10 10
2
-
? ?
= × ××
? ?
p ? ?
  
 and 
15 15
9
 eV 4.14 10 10
2
-
? ?
× ××
? ?
p ? ?
 
 = 3.95 eV and 5.93 eV 
 ? Energy of maximum energetic electrons 
  = 5.93 – 2.50 = 3.43 eV 
7. A capacitor is discharging through a resistor R. 
Consider in time t1, the energy stored in the 
capacitor reduces to half of its initial value and in 
time t2, the charge stored reduces to one eighth of 
its initial value. The ratio 
1
2
t
t
 will be 
 (A) 
1
2
 (B) 
1
3
 
 (C) 
1
4
 (D) 
1
6
 
Answer (D) 
Sol. For a discharging capacitor when energy reduces 
to half the charge would become 
1
2
times the 
initial value. 
 ? 
1
1/2
/
1
2
t
e
-t ? ?
=
? ?
? ?
 
 Similarly, 
2
3
/
1
2
t
e
-t ? ?
=
? ?
? ?
 
 ? 
1
2
1
6
t
t
= 
8. Starting with the same initial conditions, an ideal 
gas expands from volume V1 to V2 in three different 
ways. The work done by the gas is W1 if the process 
is purely isothermal, W2, if the process is purely 
adiabatic and W3 if the process is purely isobaric. 
Then, choose the correct option. 
 (A) W1 < W2 < W3 
 (B) W2 < W3 < W1 
 (C) W3 < W1 < W2 
 (D) W2 < W1 < W3 
Answer (D) 
Sol.  
 Comparing the area under the PV graph 
  A3 > A1 > A2 
 ? W3 > W1 > W2 
 
  
9. Two long current carrying conductors are placed 
parallel to each other at a distance of 8 cm between 
them. The magnitude of magnetic field produced at 
mid-point between the two conductors due to 
current flowing in them is 30 µT. The equal current 
flowing in the two conductors is: 
 (A) 30 A in the same direction 
 (B) 30 A in the opposite direction 
 (C) 60 A in the opposite direction 
 (D) 300 A in the opposite direction 
Answer (B) 
Sol. As Bnet ? 0 that is the wires are carrying current in 
opposite direction.  
 
6 0
2
2
30 10 T
2 (4 10 )
I
-
-
µ×
= ×
p×
 
 ? 
6
6
30 10
A
10
I
-
-
×
= = 30 A in opposite direction. 
10. The time period of a satellite revolving around earth 
in a given orbit is 7 hours. If the radius of orbit is 
increased to three times its previous value, then 
approximate new time period of the satellite will be 
 (A) 40 hours (B) 36 hours 
 (C) 30 hours (D) 25 hours 
Answer (B) 
Sol. 
3
2 22
21
1
R
TT
R
??
=
??
??
 
 ? ( )
3/2
2
7 5.2 7
3
T= ט × 
  
2
36 hrs T ? 
11. The TV transmission tower at a particular station 
has a height of 125 m. For doubling the coverage 
of its range, the height of the tower should be 
increased by 
 (A) 125 m (B) 250 m 
 (C) 375 m (D) 500 m 
Answer (C) 
Sol. Range 2 Re = Rh 
 Let the height be h' to double the range so 
 2 2 Re ' = Rh 
 On solving h' = 4h 
 h' = 500 m 
 So ?h = 375 m 
12. The motion of a simple pendulum executing S.H.M. 
is represented by the following equation. 
 y = A sin(pt + f), where time is measured in second. 
 The length of pendulum is 
 (A) 97.23 cm (B) 25.3 cm 
 (C) 99.4 cm (D) 406.1 cm 
Answer (C) 
Sol. ?= p =
?
g
 
 So 
2
=
p
?
g
 
    ? 99.4 cm (Nearest value) 
13. A vessel contains 16 g of hydrogen and 128 g of 
oxygen at standard temperature and pressure. The 
volume of the vessel in cm
3
 is: 
 (A) 72 × 10
5
 (B) 32 × 10
5
 
 (C) 27 × 10
4
 (D) 54 × 10
4
 
Answer (C) 
Sol. Total number of moles are 
 
22
HO
nn n = + 
    
16 128
2 32
= + 
     = 12 moles 
 Using PV = nRT 
 =
nRT
V
P
 
     
3
5
12 8.31 273.15
m
10
××
= 
      = 0.27 m
3
 = 27 × 10
4
 cm
3 
14. Given below are two statements: 
 Statement I: The electric force changes the speed 
of the charged particle and hence changes its 
kinetic energy; whereas the magnetic force does 
not change the kinetic energy of the charged 
particle. 
 Statement II: The electric force accelerates the 
positively charged particle perpendicular to the 
direction of electric field. The magnetic force 
accelerates the moving charged particle along the 
direction of magnetic field. 
 In the light of the above statements, choose the 
most appropriate answer from the options given 
below: 
Page 4


 
  
PHYSICS 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. A small toy starts moving from the position of rest 
under a constant acceleration. If it travels a 
distance of 10 m in t s, the distance travelled by the 
toy in the next t s will be : 
 (A) 10 m (B) 20 m 
 (C) 30 m (D) 40 m 
Answer (C) 
Sol. 
2
1
10 m
2
at = 
 
2
1
(2 ) 40 m
2
at = 
 ? Distance travelled in next t s 
  = 40 – 10 = 30 m 
2. At what temperature a gold ring of diameter 
6.230 cm be heated so that it can be fitted on a 
wooden bangle of diameter 6.241 cm? Both the 
diameters have been measured at room 
temperature (27°C). 
 (Given: coefficient of linear thermal expansion of 
gold aL = 1.4 × 10
–5
 K
–1
) 
 (A) 125.7°C (B) 91.7°C 
 (C) 425.7°C (D) 152.7°C 
Answer (D) 
Sol. ?D = Da?T 
 
5
0.011
6.230 1.4 10
T
-
?=
××
 
 = 126.11°C 
 ? T f = T + ?T 
  = (27 + 126.11)°C 
  = 153.11°C 
 ? Option (D) is correct 
3. Two point charges Q each are placed at a distance 
d apart. A third point charge q is placed at a 
distance x from mid-point on the perpendicular 
bisector. The value of x at which charge q will 
experience the maximum Coulombs force is : 
 (A) x = d (B) 
2
=
d
x 
 (C) 
2
=
d
x (D) 
22
=
d
x 
Answer (D) 
Sol. Force experienced by the charge q 
 
3
2
2
2
2
kQqx
F
d
x
=
? ?
??
+ ? ? ??
?? ? ?
 
 For maximum Coulomb’s force for x 
 0
dF
dx
= 
 On solving 
22
d
x = 
4. The speed of light in media ‘A’ and ‘B’ are 
2.0 × 10
10
 cm/s and 1.5 × 10
10
 cm/s respectively. A 
ray of light enters from the medium B to A at an 
incident angle ‘?’. If the ray suffers total internal 
reflection, then 
 (A) 
1
3
sin
4
-
??
?=
??
??
 (B) 
1
2
sin
3
-
? ?
?>
? ?
? ?
 
 (C) 
1
3
sin
4
-
??
?<
??
??
 (D) 
1
3
sin
4
-
??
?>
??
??
 
Answer (D) 
Sol. 
8
8
3 10
1.5
2 10
A
×
µ= =
×
 
 
8
8
3 10
2
1.5 10
B
×
µ= =
×
 
 For TIR 
 ? > i c 
 
1
1.5
sin
2
-
??
?>
??
??
 
 
1
3
sin
4
-
??
?>
??
??
 
 
  
5. In the following nuclear reaction, 
 
1 2 34
-
aß a ?
?? ? ?? ? ? ?? ? ?? ? DD D D D 
 Mass number of D is 182 and atomic number is 74. 
Mass number and atomic number of D4 respectively 
will be _____ 
 (A) 174 and 71 
 (B) 174 and 69 
 (C) 172 and 69 
 (D) 172 and 71 
Answer (A) 
Sol. Equivalent reaction can be written as 
 D ?? D4 + 2a + ß
–
 + ? 
 ? Mass number of D4 = Mass number of D – 2 × 4 
   = 182 – 8 = 174 
 ? Atomic number of D4 
  = Atomic number of D – 2 × 2 + 1 
  = 74 – 4 + 1 = 71 
6. The electric field at a point associated with a light 
wave is given by 
 E = 200[sin(6 × 10
15
)t + sin(9 × 10
15
)t] Vm
–1 
 Given : h = 4.14 × 10
–15
 eVs 
 If this light falls on a metal surface having a work 
function of 2.50 eV, the maximum kinetic energy of 
the photoelectrons will be 
 (A) 1.90 eV 
 (B) 3.27 eV 
 (C) 3.60 eV 
 (D) 3.42 eV 
Answer (D) 
Sol. Frequency of EM waves = 
15
6
10
2
×
p
 and 
15
9
10
2
×
p
 
 Energy of one photon of these waves 
 
15 15
6
 eV 4.14 10 10
2
-
? ?
= × ××
? ?
p ? ?
  
 and 
15 15
9
 eV 4.14 10 10
2
-
? ?
× ××
? ?
p ? ?
 
 = 3.95 eV and 5.93 eV 
 ? Energy of maximum energetic electrons 
  = 5.93 – 2.50 = 3.43 eV 
7. A capacitor is discharging through a resistor R. 
Consider in time t1, the energy stored in the 
capacitor reduces to half of its initial value and in 
time t2, the charge stored reduces to one eighth of 
its initial value. The ratio 
1
2
t
t
 will be 
 (A) 
1
2
 (B) 
1
3
 
 (C) 
1
4
 (D) 
1
6
 
Answer (D) 
Sol. For a discharging capacitor when energy reduces 
to half the charge would become 
1
2
times the 
initial value. 
 ? 
1
1/2
/
1
2
t
e
-t ? ?
=
? ?
? ?
 
 Similarly, 
2
3
/
1
2
t
e
-t ? ?
=
? ?
? ?
 
 ? 
1
2
1
6
t
t
= 
8. Starting with the same initial conditions, an ideal 
gas expands from volume V1 to V2 in three different 
ways. The work done by the gas is W1 if the process 
is purely isothermal, W2, if the process is purely 
adiabatic and W3 if the process is purely isobaric. 
Then, choose the correct option. 
 (A) W1 < W2 < W3 
 (B) W2 < W3 < W1 
 (C) W3 < W1 < W2 
 (D) W2 < W1 < W3 
Answer (D) 
Sol.  
 Comparing the area under the PV graph 
  A3 > A1 > A2 
 ? W3 > W1 > W2 
 
  
9. Two long current carrying conductors are placed 
parallel to each other at a distance of 8 cm between 
them. The magnitude of magnetic field produced at 
mid-point between the two conductors due to 
current flowing in them is 30 µT. The equal current 
flowing in the two conductors is: 
 (A) 30 A in the same direction 
 (B) 30 A in the opposite direction 
 (C) 60 A in the opposite direction 
 (D) 300 A in the opposite direction 
Answer (B) 
Sol. As Bnet ? 0 that is the wires are carrying current in 
opposite direction.  
 
6 0
2
2
30 10 T
2 (4 10 )
I
-
-
µ×
= ×
p×
 
 ? 
6
6
30 10
A
10
I
-
-
×
= = 30 A in opposite direction. 
10. The time period of a satellite revolving around earth 
in a given orbit is 7 hours. If the radius of orbit is 
increased to three times its previous value, then 
approximate new time period of the satellite will be 
 (A) 40 hours (B) 36 hours 
 (C) 30 hours (D) 25 hours 
Answer (B) 
Sol. 
3
2 22
21
1
R
TT
R
??
=
??
??
 
 ? ( )
3/2
2
7 5.2 7
3
T= ט × 
  
2
36 hrs T ? 
11. The TV transmission tower at a particular station 
has a height of 125 m. For doubling the coverage 
of its range, the height of the tower should be 
increased by 
 (A) 125 m (B) 250 m 
 (C) 375 m (D) 500 m 
Answer (C) 
Sol. Range 2 Re = Rh 
 Let the height be h' to double the range so 
 2 2 Re ' = Rh 
 On solving h' = 4h 
 h' = 500 m 
 So ?h = 375 m 
12. The motion of a simple pendulum executing S.H.M. 
is represented by the following equation. 
 y = A sin(pt + f), where time is measured in second. 
 The length of pendulum is 
 (A) 97.23 cm (B) 25.3 cm 
 (C) 99.4 cm (D) 406.1 cm 
Answer (C) 
Sol. ?= p =
?
g
 
 So 
2
=
p
?
g
 
    ? 99.4 cm (Nearest value) 
13. A vessel contains 16 g of hydrogen and 128 g of 
oxygen at standard temperature and pressure. The 
volume of the vessel in cm
3
 is: 
 (A) 72 × 10
5
 (B) 32 × 10
5
 
 (C) 27 × 10
4
 (D) 54 × 10
4
 
Answer (C) 
Sol. Total number of moles are 
 
22
HO
nn n = + 
    
16 128
2 32
= + 
     = 12 moles 
 Using PV = nRT 
 =
nRT
V
P
 
     
3
5
12 8.31 273.15
m
10
××
= 
      = 0.27 m
3
 = 27 × 10
4
 cm
3 
14. Given below are two statements: 
 Statement I: The electric force changes the speed 
of the charged particle and hence changes its 
kinetic energy; whereas the magnetic force does 
not change the kinetic energy of the charged 
particle. 
 Statement II: The electric force accelerates the 
positively charged particle perpendicular to the 
direction of electric field. The magnetic force 
accelerates the moving charged particle along the 
direction of magnetic field. 
 In the light of the above statements, choose the 
most appropriate answer from the options given 
below: 
 
  
 (A) Both statement I and statement II are correct 
 (B) Both statement I and statement II are incorrect 
 (C) Statement I is correct but statement II is 
incorrect 
 (D) Statement I is incorrect but statement II is 
correct 
Answer (C) 
Sol. Electric field accelerates the particle in the direction 
of field 
( )
= =
??
?
F qE maand magnetic field 
accelerates the particle perpendicular to the field 
( )
= ×=
??
? ?
F qv B ma 
15. A block of mass 40 kg slides over a surface, when 
a mass of 4 kg is suspended through an 
inextensible massless string passing over 
frictionless pulley as shown below. 
 The coefficient of kinetic friction between the 
surface and block is 0.02. The acceleration of block 
is (Given g = 10 ms
–2
.) 
 
 (A) 1 ms
–2 
 (B) 1/5 ms
–2
 
 (C) 4/5 ms
–2 
 (D) 8/11 ms
–2
 
Answer (D) 
Sol.  
 
max
N = µ
r
f 
   = 0.02 × 400 
   = 8 N 
 Let the acceleration is a as shown then. 
 40 – T = 4a 
 T – 8 = 40a 
 ? 
2
32 8
m/s
44 11
= = a 
16. In the given figure, the block of mass m is dropped 
from the point ‘A’. The expression for kinetic energy 
of block when it reaches point ‘B’ is 
 
 (A) 
2
0
1
2
mgy (B) 
2
1
2
mgy 
 (C) mg(y – y0) (D) mgy0 
Answer (D) 
Sol. Loss is potential energy = gain in kinetic energy 
 – (mg(y – y0) – mgy) = KE – 0 
 ? KE = mgy0 
17. A block of mass M placed inside a box descends 
vertically with acceleration ‘a’. The block exerts a 
force equal to one-fourth of its weight on the floor of 
the box. 
 The value of ‘a’ will be 
 (A) 
4
g
 (B) 
2
g
 
 (C) 
3
4
g
 (D) g 
Answer (C) 
Sol.  
 Using Newton’s second law 
 
4
mg
mg ma -= 
 ? 
3
4
g
a = 
Page 5


 
  
PHYSICS 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. A small toy starts moving from the position of rest 
under a constant acceleration. If it travels a 
distance of 10 m in t s, the distance travelled by the 
toy in the next t s will be : 
 (A) 10 m (B) 20 m 
 (C) 30 m (D) 40 m 
Answer (C) 
Sol. 
2
1
10 m
2
at = 
 
2
1
(2 ) 40 m
2
at = 
 ? Distance travelled in next t s 
  = 40 – 10 = 30 m 
2. At what temperature a gold ring of diameter 
6.230 cm be heated so that it can be fitted on a 
wooden bangle of diameter 6.241 cm? Both the 
diameters have been measured at room 
temperature (27°C). 
 (Given: coefficient of linear thermal expansion of 
gold aL = 1.4 × 10
–5
 K
–1
) 
 (A) 125.7°C (B) 91.7°C 
 (C) 425.7°C (D) 152.7°C 
Answer (D) 
Sol. ?D = Da?T 
 
5
0.011
6.230 1.4 10
T
-
?=
××
 
 = 126.11°C 
 ? T f = T + ?T 
  = (27 + 126.11)°C 
  = 153.11°C 
 ? Option (D) is correct 
3. Two point charges Q each are placed at a distance 
d apart. A third point charge q is placed at a 
distance x from mid-point on the perpendicular 
bisector. The value of x at which charge q will 
experience the maximum Coulombs force is : 
 (A) x = d (B) 
2
=
d
x 
 (C) 
2
=
d
x (D) 
22
=
d
x 
Answer (D) 
Sol. Force experienced by the charge q 
 
3
2
2
2
2
kQqx
F
d
x
=
? ?
??
+ ? ? ??
?? ? ?
 
 For maximum Coulomb’s force for x 
 0
dF
dx
= 
 On solving 
22
d
x = 
4. The speed of light in media ‘A’ and ‘B’ are 
2.0 × 10
10
 cm/s and 1.5 × 10
10
 cm/s respectively. A 
ray of light enters from the medium B to A at an 
incident angle ‘?’. If the ray suffers total internal 
reflection, then 
 (A) 
1
3
sin
4
-
??
?=
??
??
 (B) 
1
2
sin
3
-
? ?
?>
? ?
? ?
 
 (C) 
1
3
sin
4
-
??
?<
??
??
 (D) 
1
3
sin
4
-
??
?>
??
??
 
Answer (D) 
Sol. 
8
8
3 10
1.5
2 10
A
×
µ= =
×
 
 
8
8
3 10
2
1.5 10
B
×
µ= =
×
 
 For TIR 
 ? > i c 
 
1
1.5
sin
2
-
??
?>
??
??
 
 
1
3
sin
4
-
??
?>
??
??
 
 
  
5. In the following nuclear reaction, 
 
1 2 34
-
aß a ?
?? ? ?? ? ? ?? ? ?? ? DD D D D 
 Mass number of D is 182 and atomic number is 74. 
Mass number and atomic number of D4 respectively 
will be _____ 
 (A) 174 and 71 
 (B) 174 and 69 
 (C) 172 and 69 
 (D) 172 and 71 
Answer (A) 
Sol. Equivalent reaction can be written as 
 D ?? D4 + 2a + ß
–
 + ? 
 ? Mass number of D4 = Mass number of D – 2 × 4 
   = 182 – 8 = 174 
 ? Atomic number of D4 
  = Atomic number of D – 2 × 2 + 1 
  = 74 – 4 + 1 = 71 
6. The electric field at a point associated with a light 
wave is given by 
 E = 200[sin(6 × 10
15
)t + sin(9 × 10
15
)t] Vm
–1 
 Given : h = 4.14 × 10
–15
 eVs 
 If this light falls on a metal surface having a work 
function of 2.50 eV, the maximum kinetic energy of 
the photoelectrons will be 
 (A) 1.90 eV 
 (B) 3.27 eV 
 (C) 3.60 eV 
 (D) 3.42 eV 
Answer (D) 
Sol. Frequency of EM waves = 
15
6
10
2
×
p
 and 
15
9
10
2
×
p
 
 Energy of one photon of these waves 
 
15 15
6
 eV 4.14 10 10
2
-
? ?
= × ××
? ?
p ? ?
  
 and 
15 15
9
 eV 4.14 10 10
2
-
? ?
× ××
? ?
p ? ?
 
 = 3.95 eV and 5.93 eV 
 ? Energy of maximum energetic electrons 
  = 5.93 – 2.50 = 3.43 eV 
7. A capacitor is discharging through a resistor R. 
Consider in time t1, the energy stored in the 
capacitor reduces to half of its initial value and in 
time t2, the charge stored reduces to one eighth of 
its initial value. The ratio 
1
2
t
t
 will be 
 (A) 
1
2
 (B) 
1
3
 
 (C) 
1
4
 (D) 
1
6
 
Answer (D) 
Sol. For a discharging capacitor when energy reduces 
to half the charge would become 
1
2
times the 
initial value. 
 ? 
1
1/2
/
1
2
t
e
-t ? ?
=
? ?
? ?
 
 Similarly, 
2
3
/
1
2
t
e
-t ? ?
=
? ?
? ?
 
 ? 
1
2
1
6
t
t
= 
8. Starting with the same initial conditions, an ideal 
gas expands from volume V1 to V2 in three different 
ways. The work done by the gas is W1 if the process 
is purely isothermal, W2, if the process is purely 
adiabatic and W3 if the process is purely isobaric. 
Then, choose the correct option. 
 (A) W1 < W2 < W3 
 (B) W2 < W3 < W1 
 (C) W3 < W1 < W2 
 (D) W2 < W1 < W3 
Answer (D) 
Sol.  
 Comparing the area under the PV graph 
  A3 > A1 > A2 
 ? W3 > W1 > W2 
 
  
9. Two long current carrying conductors are placed 
parallel to each other at a distance of 8 cm between 
them. The magnitude of magnetic field produced at 
mid-point between the two conductors due to 
current flowing in them is 30 µT. The equal current 
flowing in the two conductors is: 
 (A) 30 A in the same direction 
 (B) 30 A in the opposite direction 
 (C) 60 A in the opposite direction 
 (D) 300 A in the opposite direction 
Answer (B) 
Sol. As Bnet ? 0 that is the wires are carrying current in 
opposite direction.  
 
6 0
2
2
30 10 T
2 (4 10 )
I
-
-
µ×
= ×
p×
 
 ? 
6
6
30 10
A
10
I
-
-
×
= = 30 A in opposite direction. 
10. The time period of a satellite revolving around earth 
in a given orbit is 7 hours. If the radius of orbit is 
increased to three times its previous value, then 
approximate new time period of the satellite will be 
 (A) 40 hours (B) 36 hours 
 (C) 30 hours (D) 25 hours 
Answer (B) 
Sol. 
3
2 22
21
1
R
TT
R
??
=
??
??
 
 ? ( )
3/2
2
7 5.2 7
3
T= ט × 
  
2
36 hrs T ? 
11. The TV transmission tower at a particular station 
has a height of 125 m. For doubling the coverage 
of its range, the height of the tower should be 
increased by 
 (A) 125 m (B) 250 m 
 (C) 375 m (D) 500 m 
Answer (C) 
Sol. Range 2 Re = Rh 
 Let the height be h' to double the range so 
 2 2 Re ' = Rh 
 On solving h' = 4h 
 h' = 500 m 
 So ?h = 375 m 
12. The motion of a simple pendulum executing S.H.M. 
is represented by the following equation. 
 y = A sin(pt + f), where time is measured in second. 
 The length of pendulum is 
 (A) 97.23 cm (B) 25.3 cm 
 (C) 99.4 cm (D) 406.1 cm 
Answer (C) 
Sol. ?= p =
?
g
 
 So 
2
=
p
?
g
 
    ? 99.4 cm (Nearest value) 
13. A vessel contains 16 g of hydrogen and 128 g of 
oxygen at standard temperature and pressure. The 
volume of the vessel in cm
3
 is: 
 (A) 72 × 10
5
 (B) 32 × 10
5
 
 (C) 27 × 10
4
 (D) 54 × 10
4
 
Answer (C) 
Sol. Total number of moles are 
 
22
HO
nn n = + 
    
16 128
2 32
= + 
     = 12 moles 
 Using PV = nRT 
 =
nRT
V
P
 
     
3
5
12 8.31 273.15
m
10
××
= 
      = 0.27 m
3
 = 27 × 10
4
 cm
3 
14. Given below are two statements: 
 Statement I: The electric force changes the speed 
of the charged particle and hence changes its 
kinetic energy; whereas the magnetic force does 
not change the kinetic energy of the charged 
particle. 
 Statement II: The electric force accelerates the 
positively charged particle perpendicular to the 
direction of electric field. The magnetic force 
accelerates the moving charged particle along the 
direction of magnetic field. 
 In the light of the above statements, choose the 
most appropriate answer from the options given 
below: 
 
  
 (A) Both statement I and statement II are correct 
 (B) Both statement I and statement II are incorrect 
 (C) Statement I is correct but statement II is 
incorrect 
 (D) Statement I is incorrect but statement II is 
correct 
Answer (C) 
Sol. Electric field accelerates the particle in the direction 
of field 
( )
= =
??
?
F qE maand magnetic field 
accelerates the particle perpendicular to the field 
( )
= ×=
??
? ?
F qv B ma 
15. A block of mass 40 kg slides over a surface, when 
a mass of 4 kg is suspended through an 
inextensible massless string passing over 
frictionless pulley as shown below. 
 The coefficient of kinetic friction between the 
surface and block is 0.02. The acceleration of block 
is (Given g = 10 ms
–2
.) 
 
 (A) 1 ms
–2 
 (B) 1/5 ms
–2
 
 (C) 4/5 ms
–2 
 (D) 8/11 ms
–2
 
Answer (D) 
Sol.  
 
max
N = µ
r
f 
   = 0.02 × 400 
   = 8 N 
 Let the acceleration is a as shown then. 
 40 – T = 4a 
 T – 8 = 40a 
 ? 
2
32 8
m/s
44 11
= = a 
16. In the given figure, the block of mass m is dropped 
from the point ‘A’. The expression for kinetic energy 
of block when it reaches point ‘B’ is 
 
 (A) 
2
0
1
2
mgy (B) 
2
1
2
mgy 
 (C) mg(y – y0) (D) mgy0 
Answer (D) 
Sol. Loss is potential energy = gain in kinetic energy 
 – (mg(y – y0) – mgy) = KE – 0 
 ? KE = mgy0 
17. A block of mass M placed inside a box descends 
vertically with acceleration ‘a’. The block exerts a 
force equal to one-fourth of its weight on the floor of 
the box. 
 The value of ‘a’ will be 
 (A) 
4
g
 (B) 
2
g
 
 (C) 
3
4
g
 (D) g 
Answer (C) 
Sol.  
 Using Newton’s second law 
 
4
mg
mg ma -= 
 ? 
3
4
g
a = 
 
  
18. If the electric potential at any point (x, y, z)m in 
space is given by V = 3x
2
 volt.  
 The electric field at the point (1, 0, 3)m will be 
 (A) 3 Vm
–1
, directed along positive x-axis 
 (B) 3 Vm
–1
, directed along negative x-axis 
 (C) 6 Vm
–1
, directed along positive x-axis 
 (D) 6 Vm
–1
, directed along negative x-axis 
Answer (D) 
Sol. 
ˆ
dV
Ei
dx
= -
?
 
 
ˆ
6 E xi = -
?
 
 So, E
?
at (1, 0, 3) is 
 
ˆ
6 Vm Ei = -
?
 
19. The combination of two identical cells, whether 
connected in series or parallel combination 
provides the same current through an external 
resistance of 2 ?. The value of internal resistance 
of each cell is 
 (A) 2 ? (B) 4 ? 
 (C) 6 ? (D) 8 ? 
Answer (A) 
Sol.  
  
 From diagram 
 
2
2
p
E
i
r
=
+
 and 
2
22
s
E
i
r
=
+
 
 given i p = i s 
 
11
1
2
2
r
r
=
+
+
 
 12
2
r
r += + 
 r = 2 ? 
20. A person can throw a ball upto a maximum range 
of 100 m. How high above the ground he can throw 
the same ball? 
 (A) 25 m (B) 50 m 
 (C) 100 m (D) 200 m 
Answer (B) 
Sol. 
2
max
100 m
u
R
g
= = 
 So, 
2
max
50 m
2
u
H
g
= = 
SECTION - B 
Numerical Value Type Questions: This section 
contains 10 questions. In Section B, attempt any five 
questions out of 10. The answer to each question is a 
NUMERICAL VALUE. For each question, enter the 
correct numerical value (in decimal notation, 
truncated/rounded-off to the second decimal place; e.g. 
06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the 
mouse and the on-screen virtual numeric keypad in the 
place designated to enter the answer. 
1. The vernier constant of Vernier callipers is 0.1 mm 
and it has zero error of (–0.05) cm. While measuring 
diameter of a sphere, the main scale reading is 
1.7 cm and coinciding vernier division is 5. The 
corrected diameter will be ________× 10
–2
 cm. 
Answer (180) 
Sol. Since zero error is negative, we will add 0.05 cm. 
 ?  Corrected reading = 1.7 cm + 5 × 0.1 mm + 0.05 cm 
   = 180 × 10
–2
 cm 
2. A small spherical ball of radius 0.1 mm and density 
10
4
 kg m
–3
 falls freely under gravity through a 
distance h before entering a tank of water. If, after 
entering the water the velocity of ball does not 
change and it continue to fall with same constant 
velocity inside water, then the value of h will be 
______ m. 
 (Given g = 10 ms
–2
, viscosity of water = 1.0 × 10
–5
 
N-sm
–2
). 
Answer (20)