NEET 2020 Past Year Paper with Detailed Solutions

NEET 2020 Past Year Paper with Detailed Solutions | NEET Mock Test Series 2024 PDF Download

``` Page 2

As we know, for Bi-polar junction transistor
The length of each section of transistor is in order as shown above.
i.e., L C > L E > L B
and doping profile is Emitter > Collector > Base
Whereas, for transistor action Base-emitter junction is forward biased and Base-collector junction
is reversed biased.
Hence from this we can conclude that, for transistor action the base region must be very thin and
lightly doped.

Q 2. A spherical conductor of radius 10 cm has a charge of 3.2 × 10
–7
C distributed uniformly.
What is the magnitude of electric field at a point 15 cm from the centre of the sphere?
??
????
??
??
??
9 2 2
0
1
9 10 Nm / C
4

Option A 1.28 × 10
6
N/C
Option B 1.28 × 10
7
N/C
Option C 1.28 × 10
4
N/C
Option D 1.28 × 10
5
N/C
Correct Option D
Solution:
For a conducting sphere the electric field outside will be given as

Q 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 2 m is _______.
Option A 7.32 × 10
–7
Option B 6.00 × 10
–7
Option C 3.66 × 10
–7
Option D 1.83 × 10
–7
Correct Option C
Solution:
As we know,
Page 3

As we know, for Bi-polar junction transistor
The length of each section of transistor is in order as shown above.
i.e., L C > L E > L B
and doping profile is Emitter > Collector > Base
Whereas, for transistor action Base-emitter junction is forward biased and Base-collector junction
is reversed biased.
Hence from this we can conclude that, for transistor action the base region must be very thin and
lightly doped.

Q 2. A spherical conductor of radius 10 cm has a charge of 3.2 × 10
–7
C distributed uniformly.
What is the magnitude of electric field at a point 15 cm from the centre of the sphere?
??
????
??
??
??
9 2 2
0
1
9 10 Nm / C
4

Option A 1.28 × 10
6
N/C
Option B 1.28 × 10
7
N/C
Option C 1.28 × 10
4
N/C
Option D 1.28 × 10
5
N/C
Correct Option D
Solution:
For a conducting sphere the electric field outside will be given as

Q 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 2 m is _______.
Option A 7.32 × 10
–7
Option B 6.00 × 10
–7
Option C 3.66 × 10
–7
Option D 1.83 × 10
–7
Correct Option C
Solution:
As we know,

The limit of resolving power is given as

Q 4. Dimensions of stress are:
Option A [ML
0
T
–2
]
Option B [ML
–1
T
–2
]
Option C [MLT
–2
]
Option D [ML
2
T
–2
]
Correct Option B
Solution:
??
??
??
??
-2
2
-1 -2
Force
Stress=
Area
MLT
=
L
= ML T

Q 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 _______.
Option A 0.5 mm
Option B 1.0 mm
Option C 0.01 mm
Option D 0.25 mm
Correct Option A
Solution:
Least count of screw gauge
Pitch
=
Number of divisions on circular scale
Pitch
0.01 mm =
50
Pitch = 0.5 mm

Q 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 _______.
Page 4

As we know, for Bi-polar junction transistor
The length of each section of transistor is in order as shown above.
i.e., L C > L E > L B
and doping profile is Emitter > Collector > Base
Whereas, for transistor action Base-emitter junction is forward biased and Base-collector junction
is reversed biased.
Hence from this we can conclude that, for transistor action the base region must be very thin and
lightly doped.

Q 2. A spherical conductor of radius 10 cm has a charge of 3.2 × 10
–7
C distributed uniformly.
What is the magnitude of electric field at a point 15 cm from the centre of the sphere?
??
????
??
??
??
9 2 2
0
1
9 10 Nm / C
4

Option A 1.28 × 10
6
N/C
Option B 1.28 × 10
7
N/C
Option C 1.28 × 10
4
N/C
Option D 1.28 × 10
5
N/C
Correct Option D
Solution:
For a conducting sphere the electric field outside will be given as

Q 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 2 m is _______.
Option A 7.32 × 10
–7
Option B 6.00 × 10
–7
Option C 3.66 × 10
–7
Option D 1.83 × 10
–7
Correct Option C
Solution:
As we know,

The limit of resolving power is given as

Q 4. Dimensions of stress are:
Option A [ML
0
T
–2
]
Option B [ML
–1
T
–2
]
Option C [MLT
–2
]
Option D [ML
2
T
–2
]
Correct Option B
Solution:
??
??
??
??
-2
2
-1 -2
Force
Stress=
Area
MLT
=
L
= ML T

Q 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 _______.
Option A 0.5 mm
Option B 1.0 mm
Option C 0.01 mm
Option D 0.25 mm
Correct Option A
Solution:
Least count of screw gauge
Pitch
=
Number of divisions on circular scale
Pitch
0.01 mm =
50
Pitch = 0.5 mm

Q 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 _______.

Option A g/5
Option B g/10
Option C g
Option D g/2
Correct Option A
Solution:
For the given case, the acceleration of given system will be expressed as

Q 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 :
Option A 103 V
Option B 104 V
Option C 10 V
Option D 102 V
Correct Option B
Solution:
According to de Broglie equation, the wavelength of electron will be given as
?
o
-10
2
-11
4
12.27
? = A
V
12.27×10
V = = 10
1.227×10
V = 10 volts

Page 5

As we know, for Bi-polar junction transistor
The length of each section of transistor is in order as shown above.
i.e., L C > L E > L B
and doping profile is Emitter > Collector > Base
Whereas, for transistor action Base-emitter junction is forward biased and Base-collector junction
is reversed biased.
Hence from this we can conclude that, for transistor action the base region must be very thin and
lightly doped.

Q 2. A spherical conductor of radius 10 cm has a charge of 3.2 × 10
–7
C distributed uniformly.
What is the magnitude of electric field at a point 15 cm from the centre of the sphere?
??
????
??
??
??
9 2 2
0
1
9 10 Nm / C
4

Option A 1.28 × 10
6
N/C
Option B 1.28 × 10
7
N/C
Option C 1.28 × 10
4
N/C
Option D 1.28 × 10
5
N/C
Correct Option D
Solution:
For a conducting sphere the electric field outside will be given as

Q 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 2 m is _______.
Option A 7.32 × 10
–7
Option B 6.00 × 10
–7
Option C 3.66 × 10
–7
Option D 1.83 × 10
–7
Correct Option C
Solution:
As we know,

The limit of resolving power is given as

Q 4. Dimensions of stress are:
Option A [ML
0
T
–2
]
Option B [ML
–1
T
–2
]
Option C [MLT
–2
]
Option D [ML
2
T
–2
]
Correct Option B
Solution:
??
??
??
??
-2
2
-1 -2
Force
Stress=
Area
MLT
=
L
= ML T

Q 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 _______.
Option A 0.5 mm
Option B 1.0 mm
Option C 0.01 mm
Option D 0.25 mm
Correct Option A
Solution:
Least count of screw gauge
Pitch
=
Number of divisions on circular scale
Pitch
0.01 mm =
50
Pitch = 0.5 mm

Q 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 _______.

Option A g/5
Option B g/10
Option C g
Option D g/2
Correct Option A
Solution:
For the given case, the acceleration of given system will be expressed as

Q 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 :
Option A 103 V
Option B 104 V
Option C 10 V
Option D 102 V
Correct Option B
Solution:
According to de Broglie equation, the wavelength of electron will be given as
?
o
-10
2
-11
4
12.27
? = A
V
12.27×10
V = = 10
1.227×10
V = 10 volts

Q 8. In a certain region of space with volume 0.2 m
3
, the electric potential is found to be 5 V
throughout. The magnitude of electric field in this region is :
Option A 1 N/C
Option B 5 N/C
Option C zero
Option D 0.5 N/C
Correct Option C
Solution:
For the given case, since electric potential is found constant throughout the region.
Hence by using the equation of electric field.
i.e.,
dV
E = - = 0
dr

We can conclude that the magnitude of electric field in the given region will be zero

Q 9. A cylinder contains hydrogen gas at pressure of 249 kPa and temperature 27°C.
Its density is : (R = 8.3 J mol
–1
K
–1
)
Option A 0.1 kg/m
3

Option B 0.02 kg/m
3

Option C 0.5 kg/m
3

Option D 0.2 kg/m
3

Correct Option D
Solution:

Q 10. The mean free path for a gas, with molecular diameter d and number density n can be
expressed as :
Option A

Option B

Option C

Option D

```

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FAQs on NEET 2020 Past Year Paper with Detailed Solutions - NEET Mock Test Series 2024

 1. What is NEET 2020?
Ans. NEET 2020 refers to the National Eligibility cum Entrance Test conducted in the year 2020. It is a national-level medical entrance examination in India for students aspiring to pursue undergraduate medical and dental courses in government or private colleges.
 2. How can I obtain the NEET 2020 Past Year Paper?
Ans. NEET 2020 Past Year Papers can be obtained through various online platforms or educational websites. These papers are available for free or at a nominal cost. Additionally, some coaching institutes and educational apps also provide access to NEET 2020 Past Year Papers.
 3. Why are detailed solutions important for NEET 2020 Past Year Papers?
Ans. Detailed solutions for NEET 2020 Past Year Papers are essential as they help students understand the correct approach and methodology to solve each question. These solutions provide step-by-step explanations, helping students identify their mistakes and learn from them. Detailed solutions also enhance the overall understanding of concepts and improve problem-solving skills.
 4. Can solving NEET 2020 Past Year Papers help in exam preparation?
Ans. Yes, solving NEET 2020 Past Year Papers is highly beneficial for exam preparation. These papers give students an insight into the exam pattern, marking scheme, and the types of questions asked. By solving these papers, students can practice time management, improve their speed and accuracy, and gain confidence. It also helps them identify their weak areas and focus on them during revision.
 5. Are the frequently asked questions (FAQs) in this article relevant to NEET 2020 Past Year Paper with Detailed Solutions?
Ans. Yes, the FAQs provided in this article are relevant to NEET 2020 Past Year Paper with Detailed Solutions. They address common queries related to the exam, accessing past year papers, the importance of detailed solutions, and the benefits of solving these papers. These FAQs aim to provide relevant information to students preparing for NEET 2020.

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