Sample Solution Paper 3 - Physics, Class 12 NEET Notes | EduRev

Physics Class 12

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NEET : Sample Solution Paper 3 - Physics, Class 12 NEET Notes | EduRev

 Page 1


  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
CBSE Board 
Class XII – Physics 
Sample Paper – 3 Solution 
 
A1. q 1q 2 > 0 means that q 1 and q 2 have the same sign. Either both charges are positive or 
both are negative so the force between them will be repulsive. 
q 1q 2 < 0, then force will be attractive.   
 
A2. 
 
 
A3. The side bands are at 1000+10 = 1010kHz   
         and 1000 -10 = 990kHz.     
 
A4. The diode is reverse biased.   
        This is because the p side is at a lower voltage than the n side.   
 
A5.  1THz to 1000THz 
 
A6.  
(i) The flux is zero as the net charge enclosed by the Gaussian surface will be zero.  
(ii) If the Gaussian surface is made cubical, the flux is the same. It stays zero, this is 
because the flux depends only on the charge enclosed & is independent of the shape 
of the Gaussian surface.  
 
A7. The neutral temperature will be 
0
10
500
0.02
n
a
C
b
? ? ? ? ? .  
The inversion temperature is twice the neutral temperature, so it is 1000
0
C. 
 
A8. Total resistance of the circuit is 60+3 = 63 ohms. The current in the circuit is I = 3/63 = 
0.048 A. 
 
 
 
 
 
Page 2


  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
CBSE Board 
Class XII – Physics 
Sample Paper – 3 Solution 
 
A1. q 1q 2 > 0 means that q 1 and q 2 have the same sign. Either both charges are positive or 
both are negative so the force between them will be repulsive. 
q 1q 2 < 0, then force will be attractive.   
 
A2. 
 
 
A3. The side bands are at 1000+10 = 1010kHz   
         and 1000 -10 = 990kHz.     
 
A4. The diode is reverse biased.   
        This is because the p side is at a lower voltage than the n side.   
 
A5.  1THz to 1000THz 
 
A6.  
(i) The flux is zero as the net charge enclosed by the Gaussian surface will be zero.  
(ii) If the Gaussian surface is made cubical, the flux is the same. It stays zero, this is 
because the flux depends only on the charge enclosed & is independent of the shape 
of the Gaussian surface.  
 
A7. The neutral temperature will be 
0
10
500
0.02
n
a
C
b
? ? ? ? ? .  
The inversion temperature is twice the neutral temperature, so it is 1000
0
C. 
 
A8. Total resistance of the circuit is 60+3 = 63 ohms. The current in the circuit is I = 3/63 = 
0.048 A. 
 
 
 
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
A9. On introducing a dielectric in the capacitor, its capacitance will increase.  
Then, the total impedance of the circuit will decrease as 
2
2
1
ZR
C ?
??
??
??
??
.  
Hence, the current in the circuit increases and the brightness of the lamp increases.   
 
A10. Two crossed polaroids are placed perpendicular to each other. Let the intensity of 
incident light be I 0. The light transmitted by first Polaroid has intensity I 0/2 because if 
unpolarised light is incident on a polaroid the transmitted intensity is half the original 
intensity.   
The light transmitted by second Polaroid has intensity 
? ?
2 0
cos
2
I
? where ? is the angle 
between the axes of first and second Polaroid. 
 
Thus, the light transmitted by third Polaroid has intensity 
? ? ? ?
2 2 0 2 2 2 0 0 0
cos cos (90 ) cos (sin ) sin 2
2 2 8
I I I
? ? ? ? ? ? ? ? .   
This intensity will be maximum when 
0
45 ? ?  
 
A11. NAND gate is called a universal gate because it can be used to obtain other basic gates 
like AND, NOT and OR gates.   
         NAND gates can be combined as shown below to realize a basic OR gate. 
 
 
 
A12. Some characteristics of the carrier signal are varied in accordance with the 
modulating or message signal. This is called modulation.   
Amplitude modulation, frequency modulation and phase modulation of waves are the 
different types of modulation.    
 
A13. Nuclear density for iron will be 2.3x10
17
 kg/m
3
.   
Nuclear density is independent of mass number A, so iron also has the same nuclear 
density as hydrogen.   
 
 
 
 
 
Page 3


  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
CBSE Board 
Class XII – Physics 
Sample Paper – 3 Solution 
 
A1. q 1q 2 > 0 means that q 1 and q 2 have the same sign. Either both charges are positive or 
both are negative so the force between them will be repulsive. 
q 1q 2 < 0, then force will be attractive.   
 
A2. 
 
 
A3. The side bands are at 1000+10 = 1010kHz   
         and 1000 -10 = 990kHz.     
 
A4. The diode is reverse biased.   
        This is because the p side is at a lower voltage than the n side.   
 
A5.  1THz to 1000THz 
 
A6.  
(i) The flux is zero as the net charge enclosed by the Gaussian surface will be zero.  
(ii) If the Gaussian surface is made cubical, the flux is the same. It stays zero, this is 
because the flux depends only on the charge enclosed & is independent of the shape 
of the Gaussian surface.  
 
A7. The neutral temperature will be 
0
10
500
0.02
n
a
C
b
? ? ? ? ? .  
The inversion temperature is twice the neutral temperature, so it is 1000
0
C. 
 
A8. Total resistance of the circuit is 60+3 = 63 ohms. The current in the circuit is I = 3/63 = 
0.048 A. 
 
 
 
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
A9. On introducing a dielectric in the capacitor, its capacitance will increase.  
Then, the total impedance of the circuit will decrease as 
2
2
1
ZR
C ?
??
??
??
??
.  
Hence, the current in the circuit increases and the brightness of the lamp increases.   
 
A10. Two crossed polaroids are placed perpendicular to each other. Let the intensity of 
incident light be I 0. The light transmitted by first Polaroid has intensity I 0/2 because if 
unpolarised light is incident on a polaroid the transmitted intensity is half the original 
intensity.   
The light transmitted by second Polaroid has intensity 
? ?
2 0
cos
2
I
? where ? is the angle 
between the axes of first and second Polaroid. 
 
Thus, the light transmitted by third Polaroid has intensity 
? ? ? ?
2 2 0 2 2 2 0 0 0
cos cos (90 ) cos (sin ) sin 2
2 2 8
I I I
? ? ? ? ? ? ? ? .   
This intensity will be maximum when 
0
45 ? ?  
 
A11. NAND gate is called a universal gate because it can be used to obtain other basic gates 
like AND, NOT and OR gates.   
         NAND gates can be combined as shown below to realize a basic OR gate. 
 
 
 
A12. Some characteristics of the carrier signal are varied in accordance with the 
modulating or message signal. This is called modulation.   
Amplitude modulation, frequency modulation and phase modulation of waves are the 
different types of modulation.    
 
A13. Nuclear density for iron will be 2.3x10
17
 kg/m
3
.   
Nuclear density is independent of mass number A, so iron also has the same nuclear 
density as hydrogen.   
 
 
 
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
A14.  In amplitude modulation (AM), the amplitude of modulated (carrier) wave varies in 
accordance with amplitude of information (signal) wave. When amplitude of the 
information wave increases, the amplitude of modulated wave also increases and vice 
– versa.   
          In frequency modulation (FM), the frequency of modulated wave varies in accordance 
with the frequency of the signal wave. In this case the amplitude of modulated wave is 
fixed.   
 
A15. Ampere’s circuital law states that 
0
. ?
?
? Bdl i where I refers to the current passing 
through amperian loop S around the current element.    
The magnitude of the magnetic field due to a circular coil of radius R carrying a 
current i at its centre is 
0
2
i
B
R
?
?             
Magnetic field lines due to circular current carrying coil is represented as 
 
 
OR 
The capacitive reactance is 
C 6
C
11
X 212
2 fC 2 (50)(15.0x10 )
The rms current is 
V 220
i= 1.04A     (1/2)
X 212
The peak current 2 i 2(1.04) 1.47A   
?
? ? ? ?
??
??
??
 
If the frequency is doubled, the capacitive reactance is halved and the current gets doubled. 
 
 
 
 
 
 
 
Page 4


  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
CBSE Board 
Class XII – Physics 
Sample Paper – 3 Solution 
 
A1. q 1q 2 > 0 means that q 1 and q 2 have the same sign. Either both charges are positive or 
both are negative so the force between them will be repulsive. 
q 1q 2 < 0, then force will be attractive.   
 
A2. 
 
 
A3. The side bands are at 1000+10 = 1010kHz   
         and 1000 -10 = 990kHz.     
 
A4. The diode is reverse biased.   
        This is because the p side is at a lower voltage than the n side.   
 
A5.  1THz to 1000THz 
 
A6.  
(i) The flux is zero as the net charge enclosed by the Gaussian surface will be zero.  
(ii) If the Gaussian surface is made cubical, the flux is the same. It stays zero, this is 
because the flux depends only on the charge enclosed & is independent of the shape 
of the Gaussian surface.  
 
A7. The neutral temperature will be 
0
10
500
0.02
n
a
C
b
? ? ? ? ? .  
The inversion temperature is twice the neutral temperature, so it is 1000
0
C. 
 
A8. Total resistance of the circuit is 60+3 = 63 ohms. The current in the circuit is I = 3/63 = 
0.048 A. 
 
 
 
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
A9. On introducing a dielectric in the capacitor, its capacitance will increase.  
Then, the total impedance of the circuit will decrease as 
2
2
1
ZR
C ?
??
??
??
??
.  
Hence, the current in the circuit increases and the brightness of the lamp increases.   
 
A10. Two crossed polaroids are placed perpendicular to each other. Let the intensity of 
incident light be I 0. The light transmitted by first Polaroid has intensity I 0/2 because if 
unpolarised light is incident on a polaroid the transmitted intensity is half the original 
intensity.   
The light transmitted by second Polaroid has intensity 
? ?
2 0
cos
2
I
? where ? is the angle 
between the axes of first and second Polaroid. 
 
Thus, the light transmitted by third Polaroid has intensity 
? ? ? ?
2 2 0 2 2 2 0 0 0
cos cos (90 ) cos (sin ) sin 2
2 2 8
I I I
? ? ? ? ? ? ? ? .   
This intensity will be maximum when 
0
45 ? ?  
 
A11. NAND gate is called a universal gate because it can be used to obtain other basic gates 
like AND, NOT and OR gates.   
         NAND gates can be combined as shown below to realize a basic OR gate. 
 
 
 
A12. Some characteristics of the carrier signal are varied in accordance with the 
modulating or message signal. This is called modulation.   
Amplitude modulation, frequency modulation and phase modulation of waves are the 
different types of modulation.    
 
A13. Nuclear density for iron will be 2.3x10
17
 kg/m
3
.   
Nuclear density is independent of mass number A, so iron also has the same nuclear 
density as hydrogen.   
 
 
 
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
A14.  In amplitude modulation (AM), the amplitude of modulated (carrier) wave varies in 
accordance with amplitude of information (signal) wave. When amplitude of the 
information wave increases, the amplitude of modulated wave also increases and vice 
– versa.   
          In frequency modulation (FM), the frequency of modulated wave varies in accordance 
with the frequency of the signal wave. In this case the amplitude of modulated wave is 
fixed.   
 
A15. Ampere’s circuital law states that 
0
. ?
?
? Bdl i where I refers to the current passing 
through amperian loop S around the current element.    
The magnitude of the magnetic field due to a circular coil of radius R carrying a 
current i at its centre is 
0
2
i
B
R
?
?             
Magnetic field lines due to circular current carrying coil is represented as 
 
 
OR 
The capacitive reactance is 
C 6
C
11
X 212
2 fC 2 (50)(15.0x10 )
The rms current is 
V 220
i= 1.04A     (1/2)
X 212
The peak current 2 i 2(1.04) 1.47A   
?
? ? ? ?
??
??
??
 
If the frequency is doubled, the capacitive reactance is halved and the current gets doubled. 
 
 
 
 
 
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
A16. Force acting on charge q moving with velocity v in magnetic field B: 
          F q(v B) ??        
(i) Kinetic energy of the particle does not change as the force acting on it is always 
perpendicular to the velocity. A force that is perpendicular to the velocity cannot 
change the magnitude of the velocity. All that the force does is to change the 
direction of the particle keeping the magnitude of velocity unchanged.    
(ii) The instantaneous power is F.v, where F and v are the force and velocity vector 
respectively. Since the force and velocity are perpendicular to each other, so F.v = 0.  
 
A17. Intensity is I= 4I 0 cos
2
 F/2 
When path difference is ?, phase difference is 2p 
      I= 4I o cos
2
? = 4 I o = K                      (given)  
 
When path difference is ? = ? /3, then the phase difference will be 
             F'=2 ? ? / ? 
                 =2 ? × ?/3? = 2 ?/3  
Hence the intensity at a point where the path difference is ?/3, is 
              I'=4I 0 cos
2 
2 ? /6                 (since K = 4I 0) 
                = K cos
2 
?/3= K x {1/2}2 = (¼) K.  
 
A18.  
          The de Broglie wavelength is 
e e p p
hh
p 2mK
For the electron, proton and -particle,  is same
m K m K m K cons tan t
??
? ? ?
??
? ? ?
     
       As mass of electron is minimum its kinetic energy will be maximum.     
       As mass of alpha-particle is maximum its kinetic energy is minimum.    
  
A19. In any radioactive sample, the number of nuclei undergoing the decay per unit time is 
proportional to the total number of nuclei in the sample.   
         Radioactive decay law is 
dN
N
dt
? ? ?  
Integrating the above expression gives  
? ?
? ?
t
0
0
1/2
1/2
N t N e
N
Put   N t and t T
2
0.693
T
??
?
??
?
?
         
 
Page 5


  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
CBSE Board 
Class XII – Physics 
Sample Paper – 3 Solution 
 
A1. q 1q 2 > 0 means that q 1 and q 2 have the same sign. Either both charges are positive or 
both are negative so the force between them will be repulsive. 
q 1q 2 < 0, then force will be attractive.   
 
A2. 
 
 
A3. The side bands are at 1000+10 = 1010kHz   
         and 1000 -10 = 990kHz.     
 
A4. The diode is reverse biased.   
        This is because the p side is at a lower voltage than the n side.   
 
A5.  1THz to 1000THz 
 
A6.  
(i) The flux is zero as the net charge enclosed by the Gaussian surface will be zero.  
(ii) If the Gaussian surface is made cubical, the flux is the same. It stays zero, this is 
because the flux depends only on the charge enclosed & is independent of the shape 
of the Gaussian surface.  
 
A7. The neutral temperature will be 
0
10
500
0.02
n
a
C
b
? ? ? ? ? .  
The inversion temperature is twice the neutral temperature, so it is 1000
0
C. 
 
A8. Total resistance of the circuit is 60+3 = 63 ohms. The current in the circuit is I = 3/63 = 
0.048 A. 
 
 
 
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
A9. On introducing a dielectric in the capacitor, its capacitance will increase.  
Then, the total impedance of the circuit will decrease as 
2
2
1
ZR
C ?
??
??
??
??
.  
Hence, the current in the circuit increases and the brightness of the lamp increases.   
 
A10. Two crossed polaroids are placed perpendicular to each other. Let the intensity of 
incident light be I 0. The light transmitted by first Polaroid has intensity I 0/2 because if 
unpolarised light is incident on a polaroid the transmitted intensity is half the original 
intensity.   
The light transmitted by second Polaroid has intensity 
? ?
2 0
cos
2
I
? where ? is the angle 
between the axes of first and second Polaroid. 
 
Thus, the light transmitted by third Polaroid has intensity 
? ? ? ?
2 2 0 2 2 2 0 0 0
cos cos (90 ) cos (sin ) sin 2
2 2 8
I I I
? ? ? ? ? ? ? ? .   
This intensity will be maximum when 
0
45 ? ?  
 
A11. NAND gate is called a universal gate because it can be used to obtain other basic gates 
like AND, NOT and OR gates.   
         NAND gates can be combined as shown below to realize a basic OR gate. 
 
 
 
A12. Some characteristics of the carrier signal are varied in accordance with the 
modulating or message signal. This is called modulation.   
Amplitude modulation, frequency modulation and phase modulation of waves are the 
different types of modulation.    
 
A13. Nuclear density for iron will be 2.3x10
17
 kg/m
3
.   
Nuclear density is independent of mass number A, so iron also has the same nuclear 
density as hydrogen.   
 
 
 
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
A14.  In amplitude modulation (AM), the amplitude of modulated (carrier) wave varies in 
accordance with amplitude of information (signal) wave. When amplitude of the 
information wave increases, the amplitude of modulated wave also increases and vice 
– versa.   
          In frequency modulation (FM), the frequency of modulated wave varies in accordance 
with the frequency of the signal wave. In this case the amplitude of modulated wave is 
fixed.   
 
A15. Ampere’s circuital law states that 
0
. ?
?
? Bdl i where I refers to the current passing 
through amperian loop S around the current element.    
The magnitude of the magnetic field due to a circular coil of radius R carrying a 
current i at its centre is 
0
2
i
B
R
?
?             
Magnetic field lines due to circular current carrying coil is represented as 
 
 
OR 
The capacitive reactance is 
C 6
C
11
X 212
2 fC 2 (50)(15.0x10 )
The rms current is 
V 220
i= 1.04A     (1/2)
X 212
The peak current 2 i 2(1.04) 1.47A   
?
? ? ? ?
??
??
??
 
If the frequency is doubled, the capacitive reactance is halved and the current gets doubled. 
 
 
 
 
 
 
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
A16. Force acting on charge q moving with velocity v in magnetic field B: 
          F q(v B) ??        
(i) Kinetic energy of the particle does not change as the force acting on it is always 
perpendicular to the velocity. A force that is perpendicular to the velocity cannot 
change the magnitude of the velocity. All that the force does is to change the 
direction of the particle keeping the magnitude of velocity unchanged.    
(ii) The instantaneous power is F.v, where F and v are the force and velocity vector 
respectively. Since the force and velocity are perpendicular to each other, so F.v = 0.  
 
A17. Intensity is I= 4I 0 cos
2
 F/2 
When path difference is ?, phase difference is 2p 
      I= 4I o cos
2
? = 4 I o = K                      (given)  
 
When path difference is ? = ? /3, then the phase difference will be 
             F'=2 ? ? / ? 
                 =2 ? × ?/3? = 2 ?/3  
Hence the intensity at a point where the path difference is ?/3, is 
              I'=4I 0 cos
2 
2 ? /6                 (since K = 4I 0) 
                = K cos
2 
?/3= K x {1/2}2 = (¼) K.  
 
A18.  
          The de Broglie wavelength is 
e e p p
hh
p 2mK
For the electron, proton and -particle,  is same
m K m K m K cons tan t
??
? ? ?
??
? ? ?
     
       As mass of electron is minimum its kinetic energy will be maximum.     
       As mass of alpha-particle is maximum its kinetic energy is minimum.    
  
A19. In any radioactive sample, the number of nuclei undergoing the decay per unit time is 
proportional to the total number of nuclei in the sample.   
         Radioactive decay law is 
dN
N
dt
? ? ?  
Integrating the above expression gives  
? ?
? ?
t
0
0
1/2
1/2
N t N e
N
Put   N t and t T
2
0.693
T
??
?
??
?
?
         
 
  
 
CBSE XII  |  PHYSICS 
Sample Paper – 3 Solution 
 
     
A20.  
        1 amu = 931.5 MeV 
        amu m 004 . 0 ) 009 . 1 017 . 3 ( ) 015 . 2 ( 2 ? ? ? ? ?     
 
       Hence energy released per deuteron = (0.004 x 931.5)/2 = 1.863 MeV   
 
       The number of deuterons in 1kg =N A/2 =6.023 × 10
26
/2  
 
       Energy released = (3.01 × 10
26
)(1.863 × 10
6
)(1.6x10
-19
) J = 9.0×10
13
 J     
 
A21.  
(i) If the audio signal is directly transmitted, the size of antenna will be very large 
which is not practically feasible. This is because, the size of the antenna required is 
proportional to /4 ? .   
 
(ii) Effective power radiated by antenna is proportional to square of frequency. For an 
audio frequency wave the radiated power will be extremely small.  
 
(iii) iIf different programmes, audio frequencies are directly transmitted then what 
you will hear at the receiver will be a mixture of all these signals.  
 
A22. 
(i) E remains same as it depends on the charge on the plates and the medium between 
the plates. Q remains same as charge does not change on the plates.  
(ii) V = Ed, so as the distance is doubled V also doubles.  
(iii) C = Q/V and V is doubled whereas Q remains same. So C is halved.   
 
A23.  
C b 
=0.8
= /(1- )   
=0.8/(1-0.8)   =  4     
I = I = 6 4 = 24 mA    
?
? ? ?
? ? ? ?
 
OR 
Here, m = 3.2g = 3.2  10
-3
 kg  
E = 10
10 
NC
-1
  
Let n be the number of electrons removed from the coin. Then the charge on the coin is 
q = +n e  
When the coin just floats,  
Upward force of electric field = weight of the coin  
n q E = mg 
n = 
?
?
??
3
7
19 10
mg 3.2x10 x9.8
1.96x10
qE 1.6x10 x10
electrons  
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