NEET 2020 Paper
Date : 13th September 2020
Time : 02 : 00 pm - 5 : 00 pm
Subject : Physics
1. For transistor action, which of the following statements is correct ?
(1) Both emitter junction as well as the collector junction are forward biased.
(2) The base region must be very thin and lightly doped.
(3) Base, emitter and collector regions should have same doping concentrations.
(4) Base, emitter and collector regions should have same size.
Sol. (2)
For a transistor action, the base junction must be lightly doped so that the base region
is very thin.
2. A spherical conductor of radius 10 cm has a charge of 3.2 × 10-7 C distributed uni-
formly. what is the magnitude of electric field at a point 15 cm from the centre of the
sphere ?
1
9 109 Nm2 / C2
40
Sol. (4)
kQ 9 109 3.2 10 7
E= =
r2 15 15 10 4
3.2
E= 106 = 0.128 106
25
E = 1.28 ×105 N/C
3. Assume that light of wavelength 600 nm is coming from a star. The limit of resolution
of telescope whose objective has a diameter of 2m is:
(1) 7.32 × 10–7 rad (2) 6.00 × 10–7 rad (3) 3.66× 10–7 rad (4) 1.83× 10–7 rad
Sol. (3)
1.22 1.22 600 109
min =
d 2
min = 3.66 × 10–7 rad
4. Dimensions of stress are :
(1) [ML0T–2] (2) [ML–1T–2] (3) [MLT–2] (4) [ML2T–2]
Sol. (2)
F M1L1 T 2
Stress = = = ML–1T–2
A L2
13th September 2020 | Physics Page | 1
, NEET 2020 Paper
5. A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular
scale. The pitch of the screw gauge is :
(1) 0.5 mm (2) 1.0 mm (3) 0.01 mm (4) 0.25 mm
Sol. (1)
Pitch
Least Count =
no.of division on circular scale
pitch
= 0.01 =
50
Pitch = 0.01 × 50 = 0.5 mm
6. Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string
passes over a pulley which is frictionless (see figure). The acceleration of the system in
terms of acceleration due to gravity (g) is :
4 kg
6 kg
(1) g/5 (2) g/10 (3) g (4) g/2
Sol. (1)
6g – T = 6a
T – 4g = 4a
2g = 10a
g
a= m / s2
5
7. An electron is accelerated from rest through a potential difference of V volt. If the de
Broglie wavelength of the electron is 1.227 × 10–2 nm, the potential difference is :
(1) 103 V (2) 104 V (3) 10 V (4) 102V
Sol. 2
For e–, according to de-broglie's wavelength
12.27 12.27
= Å = × 10–10 = 1.227 × 10–2 × 10–9
v v
1 1
= v = 104 volt
v 100
13th September 2020 | Physics Page | 2
Date : 13th September 2020
Time : 02 : 00 pm - 5 : 00 pm
Subject : Physics
1. For transistor action, which of the following statements is correct ?
(1) Both emitter junction as well as the collector junction are forward biased.
(2) The base region must be very thin and lightly doped.
(3) Base, emitter and collector regions should have same doping concentrations.
(4) Base, emitter and collector regions should have same size.
Sol. (2)
For a transistor action, the base junction must be lightly doped so that the base region
is very thin.
2. A spherical conductor of radius 10 cm has a charge of 3.2 × 10-7 C distributed uni-
formly. what is the magnitude of electric field at a point 15 cm from the centre of the
sphere ?
1
9 109 Nm2 / C2
40
Sol. (4)
kQ 9 109 3.2 10 7
E= =
r2 15 15 10 4
3.2
E= 106 = 0.128 106
25
E = 1.28 ×105 N/C
3. Assume that light of wavelength 600 nm is coming from a star. The limit of resolution
of telescope whose objective has a diameter of 2m is:
(1) 7.32 × 10–7 rad (2) 6.00 × 10–7 rad (3) 3.66× 10–7 rad (4) 1.83× 10–7 rad
Sol. (3)
1.22 1.22 600 109
min =
d 2
min = 3.66 × 10–7 rad
4. Dimensions of stress are :
(1) [ML0T–2] (2) [ML–1T–2] (3) [MLT–2] (4) [ML2T–2]
Sol. (2)
F M1L1 T 2
Stress = = = ML–1T–2
A L2
13th September 2020 | Physics Page | 1
, NEET 2020 Paper
5. A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular
scale. The pitch of the screw gauge is :
(1) 0.5 mm (2) 1.0 mm (3) 0.01 mm (4) 0.25 mm
Sol. (1)
Pitch
Least Count =
no.of division on circular scale
pitch
= 0.01 =
50
Pitch = 0.01 × 50 = 0.5 mm
6. Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string
passes over a pulley which is frictionless (see figure). The acceleration of the system in
terms of acceleration due to gravity (g) is :
4 kg
6 kg
(1) g/5 (2) g/10 (3) g (4) g/2
Sol. (1)
6g – T = 6a
T – 4g = 4a
2g = 10a
g
a= m / s2
5
7. An electron is accelerated from rest through a potential difference of V volt. If the de
Broglie wavelength of the electron is 1.227 × 10–2 nm, the potential difference is :
(1) 103 V (2) 104 V (3) 10 V (4) 102V
Sol. 2
For e–, according to de-broglie's wavelength
12.27 12.27
= Å = × 10–10 = 1.227 × 10–2 × 10–9
v v
1 1
= v = 104 volt
v 100
13th September 2020 | Physics Page | 2
