NEET 2020 Past Year Paper with Detailed Solutions | NEET Mock Test Series 2025 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
 rad 
Option B 6.00 × 10
–7
 rad 
Option C 3.66 × 10
–7
 rad 
Option D 1.83 × 10
–7
 rad 
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
 rad 
Option B 6.00 × 10
–7
 rad 
Option C 3.66 × 10
–7
 rad 
Option D 1.83 × 10
–7
 rad 
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
 rad 
Option B 6.00 × 10
–7
 rad 
Option C 3.66 × 10
–7
 rad 
Option D 1.83 × 10
–7
 rad 
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
 rad 
Option B 6.00 × 10
–7
 rad 
Option C 3.66 × 10
–7
 rad 
Option D 1.83 × 10
–7
 rad 
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