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 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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FAQs on CBSE Physics Past year paper (Solutions) - 2016, Class 12 - Additional Study Material for NEET

1. What is the format of the CBSE Physics exam for Class 12 in 2016?
Ans. The CBSE Physics exam for Class 12 in 2016 had a total of 26 questions divided into two sections - A and B. Section A had 8 questions of 1 mark each, 10 questions of 2 marks each, and 3 questions of 3 marks each. Section B had 5 questions of 5 marks each. The total marks for the exam were 70.
2. How can I access the CBSE Physics past year paper solutions for Class 12, 2016?
Ans. The CBSE Physics past year paper solutions for Class 12, 2016 can be accessed through various online platforms. You can search for them on educational websites, forums, or even on the official CBSE website. These solutions are available in PDF format and can be downloaded for free.
3. What are some important topics to focus on while preparing for the CBSE Physics exam for Class 12, 2016?
Ans. Some important topics to focus on while preparing for the CBSE Physics exam for Class 12, 2016 include electrostatics, current electricity, magnetic effects of current and magnetism, electromagnetic induction and alternating currents, optics, dual nature of matter and radiation, atoms and nuclei, and electronic devices. These topics carry significant weightage in the exam and require thorough understanding and practice.
4. Are the CBSE Physics past year paper solutions for Class 12, 2016 helpful for exam preparation?
Ans. Yes, the CBSE Physics past year paper solutions for Class 12, 2016 are extremely helpful for exam preparation. By going through these solutions, you can understand the question pattern, marking scheme, and the approach to solving different types of questions. It also helps in identifying the areas where you need to improve and provides a good practice for time management during the actual exam.
5. Can I expect similar questions from the CBSE Physics past year paper for Class 12, 2016 in the upcoming exams?
Ans. While it is not guaranteed that the exact same questions will be repeated, studying the CBSE Physics past year paper for Class 12, 2016 can give you an idea of the type of questions and their level of difficulty that can be expected in the upcoming exams. It helps in familiarizing yourself with the exam pattern and allows you to practice solving similar questions, which can enhance your preparation and boost your confidence.
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Objective type Questions

,

Viva Questions

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CBSE Physics Past year paper (Solutions) - 2016

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