CBSE Physics Past year paper (Solutions) - 2016, Class 12 Class 12 Notes | EduRev

Physics Class 12

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Class 12 : CBSE Physics Past year paper (Solutions) - 2016, Class 12 Class 12 Notes | EduRev

 Page 1


  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
CBSE 
Class XII Physics 
Board Paper – 2016 Solution 
 
SECTION – A 
 
1. The frequencies of the side bands produced are 
?
?
? ? ?
? ? ?
?
??
?
3
Maximum frequency: 5 kHz 2 MHz
5 10 MHz 2 MHz
0.005 2 2.005 MHz
Minimum frequency: 2 MHz 5 kHz
2 MHz 0.005 MHz
1.995 MHz
 
 
2. A particle moves along a curved path if it moves in the presence of a magnetic 
field. 
Now, when there is an angular projection, i.e. when the velocity of the particle is 
at an angle with respect to the magnetic field, then the particle will move in a 
helical path. 
The vertical component (vsin ?) will be perpendicular to the field and hence will 
rotate the particle in a circular path. However, the horizontal component (vcos ?) 
will move the particle in a straight line path. So, the cumulative motion of the 
particle is a helix. 
 
3. The mobility of a charge carrier is defined as the drift velocity per unit electric 
field. 
??
?
?
?
?
? ? ?
d
d
v
E
eE
Now, v
m
Where,  is the relaxation time
e
m
 
 
4. The work done in moving a charge along any circular path is zero. So, to move a 
point charge Q around a circular arc of radius ‘r’ at the centre of which another 
point charge ‘q’ is located, no work has to be done. 
 
5. During fog, the light coming from an object is partly deflected by the particles of 
the fog and does not reach the eye of the observer. Thus, the objects are not 
seen clearly. 
The phenomenon is called scattering of light. 
 
Page 2


  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
CBSE 
Class XII Physics 
Board Paper – 2016 Solution 
 
SECTION – A 
 
1. The frequencies of the side bands produced are 
?
?
? ? ?
? ? ?
?
??
?
3
Maximum frequency: 5 kHz 2 MHz
5 10 MHz 2 MHz
0.005 2 2.005 MHz
Minimum frequency: 2 MHz 5 kHz
2 MHz 0.005 MHz
1.995 MHz
 
 
2. A particle moves along a curved path if it moves in the presence of a magnetic 
field. 
Now, when there is an angular projection, i.e. when the velocity of the particle is 
at an angle with respect to the magnetic field, then the particle will move in a 
helical path. 
The vertical component (vsin ?) will be perpendicular to the field and hence will 
rotate the particle in a circular path. However, the horizontal component (vcos ?) 
will move the particle in a straight line path. So, the cumulative motion of the 
particle is a helix. 
 
3. The mobility of a charge carrier is defined as the drift velocity per unit electric 
field. 
??
?
?
?
?
? ? ?
d
d
v
E
eE
Now, v
m
Where,  is the relaxation time
e
m
 
 
4. The work done in moving a charge along any circular path is zero. So, to move a 
point charge Q around a circular arc of radius ‘r’ at the centre of which another 
point charge ‘q’ is located, no work has to be done. 
 
5. During fog, the light coming from an object is partly deflected by the particles of 
the fog and does not reach the eye of the observer. Thus, the objects are not 
seen clearly. 
The phenomenon is called scattering of light. 
 
  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
SECTION – B 
 
6. The H
?
 line in the emission spectrum of the hydrogen atom is obtained when the 
electron transition takes place between n = 2 to n = 1. 
?
?
?
?
??
? ? ? ?
?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
??
? ? ? ? ?
? ? ? ?
th
n
2
21
22
21
2
21
19
21
19
21
We know that the energy in n level is
13.6
E
n
13.6 13.6
EE
21
13.6
E E 13.6
2
13
E E 13.6 1 13.6 10.2 eV
44
E E 10.2 1.6 10 J
E E 16.32 10 J
Now, this difference in energ
?
?
? ? ? ?
? ?
? ? ? ?
?
? ? ? ?
21
19
21
34
15
y gives the energy of the photon emitted
E E h
EE 16.32 10
h 6.63 10
2.46 10 Hz
 
 
OR 
 
The electron in the hydrogen atom jumps from n
2
 = ? to n
1
 = 1. 
Now, we know that 
??
??
??
??
??
?
??
??
? ? ? ? ? ?
??
??
??
? ? ? ? ? ? ?
?
? ? ?
22
12
77
8 10
7
o
1 1 1
R
nn
1 1 1
1.097 10 1.097 10
1
1
9.116 10 m 911.6 10 m
1.097 10
911.6 A
 
  
Page 3


  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
CBSE 
Class XII Physics 
Board Paper – 2016 Solution 
 
SECTION – A 
 
1. The frequencies of the side bands produced are 
?
?
? ? ?
? ? ?
?
??
?
3
Maximum frequency: 5 kHz 2 MHz
5 10 MHz 2 MHz
0.005 2 2.005 MHz
Minimum frequency: 2 MHz 5 kHz
2 MHz 0.005 MHz
1.995 MHz
 
 
2. A particle moves along a curved path if it moves in the presence of a magnetic 
field. 
Now, when there is an angular projection, i.e. when the velocity of the particle is 
at an angle with respect to the magnetic field, then the particle will move in a 
helical path. 
The vertical component (vsin ?) will be perpendicular to the field and hence will 
rotate the particle in a circular path. However, the horizontal component (vcos ?) 
will move the particle in a straight line path. So, the cumulative motion of the 
particle is a helix. 
 
3. The mobility of a charge carrier is defined as the drift velocity per unit electric 
field. 
??
?
?
?
?
? ? ?
d
d
v
E
eE
Now, v
m
Where,  is the relaxation time
e
m
 
 
4. The work done in moving a charge along any circular path is zero. So, to move a 
point charge Q around a circular arc of radius ‘r’ at the centre of which another 
point charge ‘q’ is located, no work has to be done. 
 
5. During fog, the light coming from an object is partly deflected by the particles of 
the fog and does not reach the eye of the observer. Thus, the objects are not 
seen clearly. 
The phenomenon is called scattering of light. 
 
  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
SECTION – B 
 
6. The H
?
 line in the emission spectrum of the hydrogen atom is obtained when the 
electron transition takes place between n = 2 to n = 1. 
?
?
?
?
??
? ? ? ?
?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
??
? ? ? ? ?
? ? ? ?
th
n
2
21
22
21
2
21
19
21
19
21
We know that the energy in n level is
13.6
E
n
13.6 13.6
EE
21
13.6
E E 13.6
2
13
E E 13.6 1 13.6 10.2 eV
44
E E 10.2 1.6 10 J
E E 16.32 10 J
Now, this difference in energ
?
?
? ? ? ?
? ?
? ? ? ?
?
? ? ? ?
21
19
21
34
15
y gives the energy of the photon emitted
E E h
EE 16.32 10
h 6.63 10
2.46 10 Hz
 
 
OR 
 
The electron in the hydrogen atom jumps from n
2
 = ? to n
1
 = 1. 
Now, we know that 
??
??
??
??
??
?
??
??
? ? ? ? ? ?
??
??
??
? ? ? ? ? ? ?
?
? ? ?
22
12
77
8 10
7
o
1 1 1
R
nn
1 1 1
1.097 10 1.097 10
1
1
9.116 10 m 911.6 10 m
1.097 10
911.6 A
 
  
  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
7. The drift speed is  
?
d
I
v
neA
 
Now, we know that the current is given from Ohm’s law as 
??
?
?
??
??
??
??
? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ?
d
d
d
d
28 19 4
5
V
I
R
V
v
RneA
V
RA
nev
V RA
nev l l
RA
Now, resistivity is given as 
l
V
nev l
5
8 10 1.6 10 2.5 10 0.1
1.56 10 m
 
 
8.  
(i) The de Broglie wavelength is related to the accelerating potential as 
?
??
? ? ? ?
?
?
??
? ? ?
??
?
? ? ? ?
?
??
? ? ? ?
p
pp
p
p p p p
p
h
2meV
h
2m e V
h
And,
2m e V
2m e V m e
h
h
2m e V m e
Now, mass and charge of particle is greater than that of proton.
 for the same potential difference.
 
 
(ii) The kinetic energy is related to accelerating potential as 
??
??
?
?
??
?
??
?
pp
pp
p
K.E. eV
K.E. e V
And, K.E. e V
K.E. e
K.E. e
Since, the charge on proton is less than that on alpha particle,
K.E. K.E. for the same potential difference.
 
 
Page 4


  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
CBSE 
Class XII Physics 
Board Paper – 2016 Solution 
 
SECTION – A 
 
1. The frequencies of the side bands produced are 
?
?
? ? ?
? ? ?
?
??
?
3
Maximum frequency: 5 kHz 2 MHz
5 10 MHz 2 MHz
0.005 2 2.005 MHz
Minimum frequency: 2 MHz 5 kHz
2 MHz 0.005 MHz
1.995 MHz
 
 
2. A particle moves along a curved path if it moves in the presence of a magnetic 
field. 
Now, when there is an angular projection, i.e. when the velocity of the particle is 
at an angle with respect to the magnetic field, then the particle will move in a 
helical path. 
The vertical component (vsin ?) will be perpendicular to the field and hence will 
rotate the particle in a circular path. However, the horizontal component (vcos ?) 
will move the particle in a straight line path. So, the cumulative motion of the 
particle is a helix. 
 
3. The mobility of a charge carrier is defined as the drift velocity per unit electric 
field. 
??
?
?
?
?
? ? ?
d
d
v
E
eE
Now, v
m
Where,  is the relaxation time
e
m
 
 
4. The work done in moving a charge along any circular path is zero. So, to move a 
point charge Q around a circular arc of radius ‘r’ at the centre of which another 
point charge ‘q’ is located, no work has to be done. 
 
5. During fog, the light coming from an object is partly deflected by the particles of 
the fog and does not reach the eye of the observer. Thus, the objects are not 
seen clearly. 
The phenomenon is called scattering of light. 
 
  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
SECTION – B 
 
6. The H
?
 line in the emission spectrum of the hydrogen atom is obtained when the 
electron transition takes place between n = 2 to n = 1. 
?
?
?
?
??
? ? ? ?
?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
??
? ? ? ? ?
? ? ? ?
th
n
2
21
22
21
2
21
19
21
19
21
We know that the energy in n level is
13.6
E
n
13.6 13.6
EE
21
13.6
E E 13.6
2
13
E E 13.6 1 13.6 10.2 eV
44
E E 10.2 1.6 10 J
E E 16.32 10 J
Now, this difference in energ
?
?
? ? ? ?
? ?
? ? ? ?
?
? ? ? ?
21
19
21
34
15
y gives the energy of the photon emitted
E E h
EE 16.32 10
h 6.63 10
2.46 10 Hz
 
 
OR 
 
The electron in the hydrogen atom jumps from n
2
 = ? to n
1
 = 1. 
Now, we know that 
??
??
??
??
??
?
??
??
? ? ? ? ? ?
??
??
??
? ? ? ? ? ? ?
?
? ? ?
22
12
77
8 10
7
o
1 1 1
R
nn
1 1 1
1.097 10 1.097 10
1
1
9.116 10 m 911.6 10 m
1.097 10
911.6 A
 
  
  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
7. The drift speed is  
?
d
I
v
neA
 
Now, we know that the current is given from Ohm’s law as 
??
?
?
??
??
??
??
? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ?
d
d
d
d
28 19 4
5
V
I
R
V
v
RneA
V
RA
nev
V RA
nev l l
RA
Now, resistivity is given as 
l
V
nev l
5
8 10 1.6 10 2.5 10 0.1
1.56 10 m
 
 
8.  
(i) The de Broglie wavelength is related to the accelerating potential as 
?
??
? ? ? ?
?
?
??
? ? ?
??
?
? ? ? ?
?
??
? ? ? ?
p
pp
p
p p p p
p
h
2meV
h
2m e V
h
And,
2m e V
2m e V m e
h
h
2m e V m e
Now, mass and charge of particle is greater than that of proton.
 for the same potential difference.
 
 
(ii) The kinetic energy is related to accelerating potential as 
??
??
?
?
??
?
??
?
pp
pp
p
K.E. eV
K.E. e V
And, K.E. e V
K.E. e
K.E. e
Since, the charge on proton is less than that on alpha particle,
K.E. K.E. for the same potential difference.
 
 
  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
9. The angle of the prism is A = 60 ?. It is also given that AQ = AR. 
Therefore, the angles opposite to these two sides are also equal. 
 
 
 
? ? ?
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ?
12
12
AQR ARQ
Now, for a triangle
A AQR ARQ 180
AQR ARQ 60
r r 30            ( AQO ARO 90 )
r r 60
 
When r
1
 and r
2
 are equal, we have i = e. 
Now, according to Snell’s law, 
??
? ? ? ? ?
? ? ?
1
1
sini
sinr
3
sini sinr 3 sin30
2
i 60
 
Now, the angle of deviation ? is  
? ? ? ?
? ? ? ? ?
? ? ? ?
i e A
60 60 60
60
 
 
10. The baseband signal is not transmitted directly because 
(i) Its higher wavelength (low frequency) will require the size of the antenna or 
aerial used for transmission to be very high.  
(ii) For linear antenna (length l), the power radiated is proportional to
??
??
?
??
2
1
, and 
hence, the effective power radiated by a long wavelength baseband signal would 
be small. 
 
  
Page 5


  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
CBSE 
Class XII Physics 
Board Paper – 2016 Solution 
 
SECTION – A 
 
1. The frequencies of the side bands produced are 
?
?
? ? ?
? ? ?
?
??
?
3
Maximum frequency: 5 kHz 2 MHz
5 10 MHz 2 MHz
0.005 2 2.005 MHz
Minimum frequency: 2 MHz 5 kHz
2 MHz 0.005 MHz
1.995 MHz
 
 
2. A particle moves along a curved path if it moves in the presence of a magnetic 
field. 
Now, when there is an angular projection, i.e. when the velocity of the particle is 
at an angle with respect to the magnetic field, then the particle will move in a 
helical path. 
The vertical component (vsin ?) will be perpendicular to the field and hence will 
rotate the particle in a circular path. However, the horizontal component (vcos ?) 
will move the particle in a straight line path. So, the cumulative motion of the 
particle is a helix. 
 
3. The mobility of a charge carrier is defined as the drift velocity per unit electric 
field. 
??
?
?
?
?
? ? ?
d
d
v
E
eE
Now, v
m
Where,  is the relaxation time
e
m
 
 
4. The work done in moving a charge along any circular path is zero. So, to move a 
point charge Q around a circular arc of radius ‘r’ at the centre of which another 
point charge ‘q’ is located, no work has to be done. 
 
5. During fog, the light coming from an object is partly deflected by the particles of 
the fog and does not reach the eye of the observer. Thus, the objects are not 
seen clearly. 
The phenomenon is called scattering of light. 
 
  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
SECTION – B 
 
6. The H
?
 line in the emission spectrum of the hydrogen atom is obtained when the 
electron transition takes place between n = 2 to n = 1. 
?
?
?
?
??
? ? ? ?
?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
??
? ? ? ? ?
? ? ? ?
th
n
2
21
22
21
2
21
19
21
19
21
We know that the energy in n level is
13.6
E
n
13.6 13.6
EE
21
13.6
E E 13.6
2
13
E E 13.6 1 13.6 10.2 eV
44
E E 10.2 1.6 10 J
E E 16.32 10 J
Now, this difference in energ
?
?
? ? ? ?
? ?
? ? ? ?
?
? ? ? ?
21
19
21
34
15
y gives the energy of the photon emitted
E E h
EE 16.32 10
h 6.63 10
2.46 10 Hz
 
 
OR 
 
The electron in the hydrogen atom jumps from n
2
 = ? to n
1
 = 1. 
Now, we know that 
??
??
??
??
??
?
??
??
? ? ? ? ? ?
??
??
??
? ? ? ? ? ? ?
?
? ? ?
22
12
77
8 10
7
o
1 1 1
R
nn
1 1 1
1.097 10 1.097 10
1
1
9.116 10 m 911.6 10 m
1.097 10
911.6 A
 
  
  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
7. The drift speed is  
?
d
I
v
neA
 
Now, we know that the current is given from Ohm’s law as 
??
?
?
??
??
??
??
? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ?
d
d
d
d
28 19 4
5
V
I
R
V
v
RneA
V
RA
nev
V RA
nev l l
RA
Now, resistivity is given as 
l
V
nev l
5
8 10 1.6 10 2.5 10 0.1
1.56 10 m
 
 
8.  
(i) The de Broglie wavelength is related to the accelerating potential as 
?
??
? ? ? ?
?
?
??
? ? ?
??
?
? ? ? ?
?
??
? ? ? ?
p
pp
p
p p p p
p
h
2meV
h
2m e V
h
And,
2m e V
2m e V m e
h
h
2m e V m e
Now, mass and charge of particle is greater than that of proton.
 for the same potential difference.
 
 
(ii) The kinetic energy is related to accelerating potential as 
??
??
?
?
??
?
??
?
pp
pp
p
K.E. eV
K.E. e V
And, K.E. e V
K.E. e
K.E. e
Since, the charge on proton is less than that on alpha particle,
K.E. K.E. for the same potential difference.
 
 
  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
9. The angle of the prism is A = 60 ?. It is also given that AQ = AR. 
Therefore, the angles opposite to these two sides are also equal. 
 
 
 
? ? ?
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ?
12
12
AQR ARQ
Now, for a triangle
A AQR ARQ 180
AQR ARQ 60
r r 30            ( AQO ARO 90 )
r r 60
 
When r
1
 and r
2
 are equal, we have i = e. 
Now, according to Snell’s law, 
??
? ? ? ? ?
? ? ?
1
1
sini
sinr
3
sini sinr 3 sin30
2
i 60
 
Now, the angle of deviation ? is  
? ? ? ?
? ? ? ? ?
? ? ? ?
i e A
60 60 60
60
 
 
10. The baseband signal is not transmitted directly because 
(i) Its higher wavelength (low frequency) will require the size of the antenna or 
aerial used for transmission to be very high.  
(ii) For linear antenna (length l), the power radiated is proportional to
??
??
?
??
2
1
, and 
hence, the effective power radiated by a long wavelength baseband signal would 
be small. 
 
  
  
 
CBSE XII  | PHYSICS 
Board Paper – Solution 
 
     
SECTION – C  
11.  
(i) The distance between the object and the screen (image) is 100 cm. 
 
 
Now, for the first position of the lens: 
?
??
? ? ? ? ?
? ? ?
uu
v 100 u
1 1 1 1 1
                    ...... (1)
f 100 u u 100 u u
 
Now, for the second position of the lens: 
? ?
??
? ? ?
? ? ? ? ?
? ? ? ??
u u 20
v 100 u 20
1 1 1 1 1
                  ...... (2)
f 80 u 80 u u 20 u 20
 
Therefore, from equations (1) and (2), we get 
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ?
? ? ?
? ? ? ?
? ? ?
? ? ? ? ?
??
? ? ?
??
??
? ? ?
? ? ? ? ?
? ? ? ? ?
??
??
? ? ? ? ? ?
22
1 1 1 1
100 u u 80 u u 20
1 1 1 1
100 u 80 u u 20 u
80 u 100 u u u 20
100 u 80 u u u 20
20 20
100 u 80 u u u 20
u u 20 100 u 80 u
u 20u 8000 180u u
200u 8000
u 40 cm
v 100 u 100 40 60 cm
Therefore, from the lens for
? ? ? ? ? ?
?
?
? ? ?
??
mula
1 1 1 1 1 1 1
f v u 60 40 60 40
1 2 3 5
f 120 120
f 24 cm
 
 
 
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