Sample Solution Paper 1 - Physics, Class 12 | Additional Study Material for NEET PDF Download

 Page 1


  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
CBSE Board 
Class XII – Physics 
Sample Paper – 1 Solution 
 
 
A1. It is defined as the product of magnitude of charge and the distance between them, that 
is, 2qa and the direction is from –q to +q.   
 
A2. Electric flux: V.m; Electric field: V/m.   
 
A3. The input power is equal to the output power if 100% efficiency is assumed. So the 
power at the secondary will be 1kW.   
 
A4. X-ray, infrared, visible, microwaves   
 
A5. Frequency   
 
A6. It affects the stopping potential but not the saturation current.   
 
A7. Because the hole concentration is more in the p-region as compared to the n-region.  
 
A8. Noise refers to any unwanted signals that tend to disturb the transmission and 
processing of message signals in a communication system.   
 
A9. de Broglie wavelength, 
        
2
hh
p mK
???  
For the same kinetic energy, de Broglie wavelength depends on the mass of the 
particle.   
As alpha particle is the most massive of the three its de Broglie wavelength will be the 
shortest.   
 
A10. Yes, because the potentiometer does not draw any current from the circuit but it still 
measures the potential difference.   
An ideal voltmeter has infinite resistance, and hence it does not draw any current 
from the circuit.   
 
A11. The refractive index of the liquid must be equal to 1.47 for the lens to disappear.  
Now the focal length of the lens is 1/f = 0 or f tends to infinity. In other words, the lens 
in the liquid, acts like a plane sheet of glass.   
 
A12.  
           
20
2
2
0
mBsin      (1/ 2mark)
5 10 (m)(1.0T)sin 60
5 10 1000
m A.m    
(1.0)sin 60 3
? ? ?
??
?
??
 
 
Page 2


  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
CBSE Board 
Class XII – Physics 
Sample Paper – 1 Solution 
 
 
A1. It is defined as the product of magnitude of charge and the distance between them, that 
is, 2qa and the direction is from –q to +q.   
 
A2. Electric flux: V.m; Electric field: V/m.   
 
A3. The input power is equal to the output power if 100% efficiency is assumed. So the 
power at the secondary will be 1kW.   
 
A4. X-ray, infrared, visible, microwaves   
 
A5. Frequency   
 
A6. It affects the stopping potential but not the saturation current.   
 
A7. Because the hole concentration is more in the p-region as compared to the n-region.  
 
A8. Noise refers to any unwanted signals that tend to disturb the transmission and 
processing of message signals in a communication system.   
 
A9. de Broglie wavelength, 
        
2
hh
p mK
???  
For the same kinetic energy, de Broglie wavelength depends on the mass of the 
particle.   
As alpha particle is the most massive of the three its de Broglie wavelength will be the 
shortest.   
 
A10. Yes, because the potentiometer does not draw any current from the circuit but it still 
measures the potential difference.   
An ideal voltmeter has infinite resistance, and hence it does not draw any current 
from the circuit.   
 
A11. The refractive index of the liquid must be equal to 1.47 for the lens to disappear.  
Now the focal length of the lens is 1/f = 0 or f tends to infinity. In other words, the lens 
in the liquid, acts like a plane sheet of glass.   
 
A12.  
           
20
2
2
0
mBsin      (1/ 2mark)
5 10 (m)(1.0T)sin 60
5 10 1000
m A.m    
(1.0)sin 60 3
? ? ?
??
?
??
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
A13. The random thermal motion is reduced at low temperatures, so the tendency to 
disturb the alignment of dipoles is reduced at lower temperatures.   
However, in a diamagnetic sample the alignment is opposite to the magnetising field, 
so temperature does not affect diamagnetism.   
OR 
TV signals have a frequency between 100 and 200 MHz, so they are not reflected by 
the ionosphere. Thus their transmission via sky waves is not possible.   
     The range can be increased by using, i) taller antenna, and ii) geostationary satellites.  
 
A14. Power factor is cos where P = VIcos ?? .   
If power factor is small, the current has to be increased to maintain the same power. 
This leads to greater 
2
IR losses in transmission.   
 
A15. Resolving power = 
2 sin n ?
?
 
i) The resolving power increases on decreasing the wavelength of light.   
ii) It does not change on decreasing the diameter of the objective.   
 
A16.  
        Energy of electrons  
        
3 19
3 19
18
34
8
18
E 30keV 30 10 1.6 10 joule
From Einstein equation,
Eh
E 30 10 1.6 10
7.27 10 Hz (1)
h 6.6 10
v c 3.0 10
0.0413 nm (1)
7.27 10
?
?
?
? ? ? ? ?
??
? ? ?
? ? ? ? ?
?
?
? ? ? ? ?
? ? ?
 
 
A17. Electrostatic potential at a point is defined as the amount of work done in bringing a 
unit positive charge from infinity to that point.   
           At a far off point lying on the axial line the electric potential is 
2
0
1
4
p
V
r ??
? . 
            So, 
2
1
V
r
? . (1) 
           On the equatorial plane V = 0 everywhere.   
 
A18. Each free electron accelerates due to the force that acts on it. This increases its drift 
speed until it collides with a positive ion of the metal. It loses some energy after each 
collision, but it again accelerates and then collides and once again loses some energy. 
This sequence of events continues.   
Hence, on an average the electrons acquire only a drift speed but they are not able to 
accelerate.   
In the presence of an electric field the path of the electrons is generally curved, rather 
than straight.    
 
 
Page 3


  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
CBSE Board 
Class XII – Physics 
Sample Paper – 1 Solution 
 
 
A1. It is defined as the product of magnitude of charge and the distance between them, that 
is, 2qa and the direction is from –q to +q.   
 
A2. Electric flux: V.m; Electric field: V/m.   
 
A3. The input power is equal to the output power if 100% efficiency is assumed. So the 
power at the secondary will be 1kW.   
 
A4. X-ray, infrared, visible, microwaves   
 
A5. Frequency   
 
A6. It affects the stopping potential but not the saturation current.   
 
A7. Because the hole concentration is more in the p-region as compared to the n-region.  
 
A8. Noise refers to any unwanted signals that tend to disturb the transmission and 
processing of message signals in a communication system.   
 
A9. de Broglie wavelength, 
        
2
hh
p mK
???  
For the same kinetic energy, de Broglie wavelength depends on the mass of the 
particle.   
As alpha particle is the most massive of the three its de Broglie wavelength will be the 
shortest.   
 
A10. Yes, because the potentiometer does not draw any current from the circuit but it still 
measures the potential difference.   
An ideal voltmeter has infinite resistance, and hence it does not draw any current 
from the circuit.   
 
A11. The refractive index of the liquid must be equal to 1.47 for the lens to disappear.  
Now the focal length of the lens is 1/f = 0 or f tends to infinity. In other words, the lens 
in the liquid, acts like a plane sheet of glass.   
 
A12.  
           
20
2
2
0
mBsin      (1/ 2mark)
5 10 (m)(1.0T)sin 60
5 10 1000
m A.m    
(1.0)sin 60 3
? ? ?
??
?
??
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
A13. The random thermal motion is reduced at low temperatures, so the tendency to 
disturb the alignment of dipoles is reduced at lower temperatures.   
However, in a diamagnetic sample the alignment is opposite to the magnetising field, 
so temperature does not affect diamagnetism.   
OR 
TV signals have a frequency between 100 and 200 MHz, so they are not reflected by 
the ionosphere. Thus their transmission via sky waves is not possible.   
     The range can be increased by using, i) taller antenna, and ii) geostationary satellites.  
 
A14. Power factor is cos where P = VIcos ?? .   
If power factor is small, the current has to be increased to maintain the same power. 
This leads to greater 
2
IR losses in transmission.   
 
A15. Resolving power = 
2 sin n ?
?
 
i) The resolving power increases on decreasing the wavelength of light.   
ii) It does not change on decreasing the diameter of the objective.   
 
A16.  
        Energy of electrons  
        
3 19
3 19
18
34
8
18
E 30keV 30 10 1.6 10 joule
From Einstein equation,
Eh
E 30 10 1.6 10
7.27 10 Hz (1)
h 6.6 10
v c 3.0 10
0.0413 nm (1)
7.27 10
?
?
?
? ? ? ? ?
??
? ? ?
? ? ? ? ?
?
?
? ? ? ? ?
? ? ?
 
 
A17. Electrostatic potential at a point is defined as the amount of work done in bringing a 
unit positive charge from infinity to that point.   
           At a far off point lying on the axial line the electric potential is 
2
0
1
4
p
V
r ??
? . 
            So, 
2
1
V
r
? . (1) 
           On the equatorial plane V = 0 everywhere.   
 
A18. Each free electron accelerates due to the force that acts on it. This increases its drift 
speed until it collides with a positive ion of the metal. It loses some energy after each 
collision, but it again accelerates and then collides and once again loses some energy. 
This sequence of events continues.   
Hence, on an average the electrons acquire only a drift speed but they are not able to 
accelerate.   
In the presence of an electric field the path of the electrons is generally curved, rather 
than straight.    
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
A.19 
   
A coaxial cable is as shown in the figure above. It consists of an inner conductor that 
is made up of a copper wire and the outer conductor can be either a solid or braided 
mesh of fine wires. The outer conductor is externally covered with a polymer jacket 
for protection.   
Upper limit of the frequency that can be used is 20MHz.   
 
A20. A is a capacitor and B is an inductor.   
On decreasing the frequency of the applied voltage the inductive reactance decreases, 
so the current through the inductor will increase.   
For this change in the frequency, the capacitive reactance increases, so the current 
through the capacitor decreases.   
 
A21.  
        
0
E E sin t ?? 
     
0
0
0
0
0
0
0
dI
E L 0
dt
E dI E
sin t
dt L L
E
I sin t
L
E
=-( )cos t I cos t
L
E
where I
L
or,  I I sin( t )
2
??
? ? ?
??
? ? ? ?
?
?
?
?
? ? ?
?
 
Hence current lags behind the voltage by a phase angle of 
2
?
.  
 
 
 
 
 
 
 
 
 
Page 4


  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
CBSE Board 
Class XII – Physics 
Sample Paper – 1 Solution 
 
 
A1. It is defined as the product of magnitude of charge and the distance between them, that 
is, 2qa and the direction is from –q to +q.   
 
A2. Electric flux: V.m; Electric field: V/m.   
 
A3. The input power is equal to the output power if 100% efficiency is assumed. So the 
power at the secondary will be 1kW.   
 
A4. X-ray, infrared, visible, microwaves   
 
A5. Frequency   
 
A6. It affects the stopping potential but not the saturation current.   
 
A7. Because the hole concentration is more in the p-region as compared to the n-region.  
 
A8. Noise refers to any unwanted signals that tend to disturb the transmission and 
processing of message signals in a communication system.   
 
A9. de Broglie wavelength, 
        
2
hh
p mK
???  
For the same kinetic energy, de Broglie wavelength depends on the mass of the 
particle.   
As alpha particle is the most massive of the three its de Broglie wavelength will be the 
shortest.   
 
A10. Yes, because the potentiometer does not draw any current from the circuit but it still 
measures the potential difference.   
An ideal voltmeter has infinite resistance, and hence it does not draw any current 
from the circuit.   
 
A11. The refractive index of the liquid must be equal to 1.47 for the lens to disappear.  
Now the focal length of the lens is 1/f = 0 or f tends to infinity. In other words, the lens 
in the liquid, acts like a plane sheet of glass.   
 
A12.  
           
20
2
2
0
mBsin      (1/ 2mark)
5 10 (m)(1.0T)sin 60
5 10 1000
m A.m    
(1.0)sin 60 3
? ? ?
??
?
??
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
A13. The random thermal motion is reduced at low temperatures, so the tendency to 
disturb the alignment of dipoles is reduced at lower temperatures.   
However, in a diamagnetic sample the alignment is opposite to the magnetising field, 
so temperature does not affect diamagnetism.   
OR 
TV signals have a frequency between 100 and 200 MHz, so they are not reflected by 
the ionosphere. Thus their transmission via sky waves is not possible.   
     The range can be increased by using, i) taller antenna, and ii) geostationary satellites.  
 
A14. Power factor is cos where P = VIcos ?? .   
If power factor is small, the current has to be increased to maintain the same power. 
This leads to greater 
2
IR losses in transmission.   
 
A15. Resolving power = 
2 sin n ?
?
 
i) The resolving power increases on decreasing the wavelength of light.   
ii) It does not change on decreasing the diameter of the objective.   
 
A16.  
        Energy of electrons  
        
3 19
3 19
18
34
8
18
E 30keV 30 10 1.6 10 joule
From Einstein equation,
Eh
E 30 10 1.6 10
7.27 10 Hz (1)
h 6.6 10
v c 3.0 10
0.0413 nm (1)
7.27 10
?
?
?
? ? ? ? ?
??
? ? ?
? ? ? ? ?
?
?
? ? ? ? ?
? ? ?
 
 
A17. Electrostatic potential at a point is defined as the amount of work done in bringing a 
unit positive charge from infinity to that point.   
           At a far off point lying on the axial line the electric potential is 
2
0
1
4
p
V
r ??
? . 
            So, 
2
1
V
r
? . (1) 
           On the equatorial plane V = 0 everywhere.   
 
A18. Each free electron accelerates due to the force that acts on it. This increases its drift 
speed until it collides with a positive ion of the metal. It loses some energy after each 
collision, but it again accelerates and then collides and once again loses some energy. 
This sequence of events continues.   
Hence, on an average the electrons acquire only a drift speed but they are not able to 
accelerate.   
In the presence of an electric field the path of the electrons is generally curved, rather 
than straight.    
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
A.19 
   
A coaxial cable is as shown in the figure above. It consists of an inner conductor that 
is made up of a copper wire and the outer conductor can be either a solid or braided 
mesh of fine wires. The outer conductor is externally covered with a polymer jacket 
for protection.   
Upper limit of the frequency that can be used is 20MHz.   
 
A20. A is a capacitor and B is an inductor.   
On decreasing the frequency of the applied voltage the inductive reactance decreases, 
so the current through the inductor will increase.   
For this change in the frequency, the capacitive reactance increases, so the current 
through the capacitor decreases.   
 
A21.  
        
0
E E sin t ?? 
     
0
0
0
0
0
0
0
dI
E L 0
dt
E dI E
sin t
dt L L
E
I sin t
L
E
=-( )cos t I cos t
L
E
where I
L
or,  I I sin( t )
2
??
? ? ?
??
? ? ? ?
?
?
?
?
? ? ?
?
 
Hence current lags behind the voltage by a phase angle of 
2
?
.  
 
 
 
 
 
 
 
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
A22. A wire loop of area A is free to rotate about an axis which is perpendicular to a 
uniform magnetic field B. 
            
 
If the normal to the loop n makes an angle ? with B , then flux through the loop is 
BAcos ? ? ? . 
If this loop rotates with a constant angular velocity 
d
dt
?
?? , the flux through it changes at 
the rate 
? ?
??
? ? ? ? ? ? ? ?
0
dd
BAsin BA sin t C
dt dt
 
where C 0 is a constant     
? emf is induced between ends A and B given by 
? ? BA sin t Co ? ? ? ? ? 
? ?
m
V sin t Co ? ? ? ? 
m
V BA ?? = Peak value of emf generated.    
 
A23.  
Principle: A beam of charged particles describe a circular path when subjected to a 
uniform magnetic field directed perpendicular to their plane of motion. The beam can 
be accelerated time and again by a high frequency electric field of correctly adjusted 
frequency applied between two dees.   
 (1) 
 
Construction: It consists of a two semi circular disc like hollow metal containers called 
dees D 1 and D 2 connected to high frequency alternating voltage. 
A magnetic field is applied perpendicular to the plane of the dees.  
 
 
 
 
 
 
 
Page 5


  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
CBSE Board 
Class XII – Physics 
Sample Paper – 1 Solution 
 
 
A1. It is defined as the product of magnitude of charge and the distance between them, that 
is, 2qa and the direction is from –q to +q.   
 
A2. Electric flux: V.m; Electric field: V/m.   
 
A3. The input power is equal to the output power if 100% efficiency is assumed. So the 
power at the secondary will be 1kW.   
 
A4. X-ray, infrared, visible, microwaves   
 
A5. Frequency   
 
A6. It affects the stopping potential but not the saturation current.   
 
A7. Because the hole concentration is more in the p-region as compared to the n-region.  
 
A8. Noise refers to any unwanted signals that tend to disturb the transmission and 
processing of message signals in a communication system.   
 
A9. de Broglie wavelength, 
        
2
hh
p mK
???  
For the same kinetic energy, de Broglie wavelength depends on the mass of the 
particle.   
As alpha particle is the most massive of the three its de Broglie wavelength will be the 
shortest.   
 
A10. Yes, because the potentiometer does not draw any current from the circuit but it still 
measures the potential difference.   
An ideal voltmeter has infinite resistance, and hence it does not draw any current 
from the circuit.   
 
A11. The refractive index of the liquid must be equal to 1.47 for the lens to disappear.  
Now the focal length of the lens is 1/f = 0 or f tends to infinity. In other words, the lens 
in the liquid, acts like a plane sheet of glass.   
 
A12.  
           
20
2
2
0
mBsin      (1/ 2mark)
5 10 (m)(1.0T)sin 60
5 10 1000
m A.m    
(1.0)sin 60 3
? ? ?
??
?
??
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
A13. The random thermal motion is reduced at low temperatures, so the tendency to 
disturb the alignment of dipoles is reduced at lower temperatures.   
However, in a diamagnetic sample the alignment is opposite to the magnetising field, 
so temperature does not affect diamagnetism.   
OR 
TV signals have a frequency between 100 and 200 MHz, so they are not reflected by 
the ionosphere. Thus their transmission via sky waves is not possible.   
     The range can be increased by using, i) taller antenna, and ii) geostationary satellites.  
 
A14. Power factor is cos where P = VIcos ?? .   
If power factor is small, the current has to be increased to maintain the same power. 
This leads to greater 
2
IR losses in transmission.   
 
A15. Resolving power = 
2 sin n ?
?
 
i) The resolving power increases on decreasing the wavelength of light.   
ii) It does not change on decreasing the diameter of the objective.   
 
A16.  
        Energy of electrons  
        
3 19
3 19
18
34
8
18
E 30keV 30 10 1.6 10 joule
From Einstein equation,
Eh
E 30 10 1.6 10
7.27 10 Hz (1)
h 6.6 10
v c 3.0 10
0.0413 nm (1)
7.27 10
?
?
?
? ? ? ? ?
??
? ? ?
? ? ? ? ?
?
?
? ? ? ? ?
? ? ?
 
 
A17. Electrostatic potential at a point is defined as the amount of work done in bringing a 
unit positive charge from infinity to that point.   
           At a far off point lying on the axial line the electric potential is 
2
0
1
4
p
V
r ??
? . 
            So, 
2
1
V
r
? . (1) 
           On the equatorial plane V = 0 everywhere.   
 
A18. Each free electron accelerates due to the force that acts on it. This increases its drift 
speed until it collides with a positive ion of the metal. It loses some energy after each 
collision, but it again accelerates and then collides and once again loses some energy. 
This sequence of events continues.   
Hence, on an average the electrons acquire only a drift speed but they are not able to 
accelerate.   
In the presence of an electric field the path of the electrons is generally curved, rather 
than straight.    
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
A.19 
   
A coaxial cable is as shown in the figure above. It consists of an inner conductor that 
is made up of a copper wire and the outer conductor can be either a solid or braided 
mesh of fine wires. The outer conductor is externally covered with a polymer jacket 
for protection.   
Upper limit of the frequency that can be used is 20MHz.   
 
A20. A is a capacitor and B is an inductor.   
On decreasing the frequency of the applied voltage the inductive reactance decreases, 
so the current through the inductor will increase.   
For this change in the frequency, the capacitive reactance increases, so the current 
through the capacitor decreases.   
 
A21.  
        
0
E E sin t ?? 
     
0
0
0
0
0
0
0
dI
E L 0
dt
E dI E
sin t
dt L L
E
I sin t
L
E
=-( )cos t I cos t
L
E
where I
L
or,  I I sin( t )
2
??
? ? ?
??
? ? ? ?
?
?
?
?
? ? ?
?
 
Hence current lags behind the voltage by a phase angle of 
2
?
.  
 
 
 
 
 
 
 
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
A22. A wire loop of area A is free to rotate about an axis which is perpendicular to a 
uniform magnetic field B. 
            
 
If the normal to the loop n makes an angle ? with B , then flux through the loop is 
BAcos ? ? ? . 
If this loop rotates with a constant angular velocity 
d
dt
?
?? , the flux through it changes at 
the rate 
? ?
??
? ? ? ? ? ? ? ?
0
dd
BAsin BA sin t C
dt dt
 
where C 0 is a constant     
? emf is induced between ends A and B given by 
? ? BA sin t Co ? ? ? ? ? 
? ?
m
V sin t Co ? ? ? ? 
m
V BA ?? = Peak value of emf generated.    
 
A23.  
Principle: A beam of charged particles describe a circular path when subjected to a 
uniform magnetic field directed perpendicular to their plane of motion. The beam can 
be accelerated time and again by a high frequency electric field of correctly adjusted 
frequency applied between two dees.   
 (1) 
 
Construction: It consists of a two semi circular disc like hollow metal containers called 
dees D 1 and D 2 connected to high frequency alternating voltage. 
A magnetic field is applied perpendicular to the plane of the dees.  
 
 
 
 
 
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 1 Solution 
 
     
A24.  
       Given: n i = 2 x 10
8
 /m
3 
       On doping n h = 4 x 10
10
 /m
3 
(i) On doping hole concentration will increase, hence, it is p-type semiconductor.  
(ii) Since 
2
i e h
n n n ? 
2 82
10
6 -3
(2 10 )
4 10
10 m
i
e
h
e
n
n
n
n
?
??
?
??
  
(iii) Energy gap decreases with doping because acceptor level gets created between 
the valance band and conduction band.   
OR 
(a) It is referring to the RMS value. V rms=220V  
So, the peak value for voltage is
0
2 311 V
rms
VV ??
.   
(b) The instantaneous power is always
( ) ( ) ( ) P t V t I t ?
, and in this case, it is 
       
2
0
( ) cos( )cos( 2)
c
V
P t t t
X
? ? ? ? ? ? ? ? ?
       
           The average power is,  
 
2
0
cos 0
22
c
c
V
P
X
? ??
??
??
??
    
 
A25. 
(a)  
 
  
 
      Here,      O = objective lens 
  E = Eye piece 
  AB = object, A” B” = final image 
  
 Magnifying power (
m
) of a compound microscope is given by