CBSE Physics Past year paper (Solutions) - 2015, Class 12 | Additional Study Material for NEET PDF Download

 Page 1


  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
CBSE 
Class XII Physics 
Board Paper – 2015 Solution 
 
SECTION – A 
 
1. Mobility of the charge carrier is defined as the drift velocity of the charge carrier per 
unit electric field. 
It is denoted by ? and is given as 
 
?
??
d
E
 
The SI unit of ? is m
2 
V
-1 
s
-1
. 
 
2. In an LCR series circuit, power dissipated is given by 
 
2
P I Zcos ??
 
 
2 2 2
CL
P
cos
I (R (X X )
??
??
 
The quantity cos ? is the power factor. 
 
2 2 2
CL
L C L C
2
P
cos
I R (X X )
V V and X X
P
cos
IR
??
??
??
? ? ?
 
 
3. The focal length and the radius of curvature of a lens are equal if the lens is made of 
glass having a refractive index of 1.5. 
 
4. The relation for polarisation P of a dielectric material in the presence of an external 
electric field E is 
 
e
PE ??
 
where 
e
? is a constant characteristic of the dielectric and is known as the electric 
susceptibility of the dielectric medium. 
 
5. In forward biasing, the applied voltage mostly drops across the depletion region, and 
the voltage drop across the p-side and n-side of the junction is negligible. The direction 
of the applied voltage (V) is opposite to the built-in potential V o. As a result, the 
depletion layer width decreases and the barrier height is reduced. The effective barrier 
height under forward bias is (V o - V). 
 
  
Page 2


  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
CBSE 
Class XII Physics 
Board Paper – 2015 Solution 
 
SECTION – A 
 
1. Mobility of the charge carrier is defined as the drift velocity of the charge carrier per 
unit electric field. 
It is denoted by ? and is given as 
 
?
??
d
E
 
The SI unit of ? is m
2 
V
-1 
s
-1
. 
 
2. In an LCR series circuit, power dissipated is given by 
 
2
P I Zcos ??
 
 
2 2 2
CL
P
cos
I (R (X X )
??
??
 
The quantity cos ? is the power factor. 
 
2 2 2
CL
L C L C
2
P
cos
I R (X X )
V V and X X
P
cos
IR
??
??
??
? ? ?
 
 
3. The focal length and the radius of curvature of a lens are equal if the lens is made of 
glass having a refractive index of 1.5. 
 
4. The relation for polarisation P of a dielectric material in the presence of an external 
electric field E is 
 
e
PE ??
 
where 
e
? is a constant characteristic of the dielectric and is known as the electric 
susceptibility of the dielectric medium. 
 
5. In forward biasing, the applied voltage mostly drops across the depletion region, and 
the voltage drop across the p-side and n-side of the junction is negligible. The direction 
of the applied voltage (V) is opposite to the built-in potential V o. As a result, the 
depletion layer width decreases and the barrier height is reduced. The effective barrier 
height under forward bias is (V o - V). 
 
  
  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
SECTION – B  
6.  
(a)  
Analog form of communication Digital form of communication 
It makes use of analog signals whose 
amplitude varies continuously with 
time. 
It makes use of digital signals whose 
amplitude is of two levels, either low 
or high. 
The retrieved message in analog 
communication is not an exact copy 
of the original message. This is 
because the message suffers 
distortions over the medium and 
during its detection process and is 
distorted. 
The retrieved message in digital 
communication is always the exact 
copy of the original message and is 
practically free of distortion. 
It can broadcast only limited number 
of channels simultaneously. 
It can broadcast tremendously large 
number of channels simultaneously. 
 
(b) Applications of the Internet: 
1. E-mail: It allows the exchange of text or graphic material. One can write a letter 
and send it to the other person (recipient) through an internet service provider 
(ISP). The ISP works like the dispatching and receiving post office. 
2. File transfer: A program known as the file transfer program (FTP) allows the 
transfer of files or software from one computer to the other computer connected 
to the Internet. 
 
7. Ground state energy is E 0 = –13.6 eV 
1 2
1
1
-34
-31 -19
13.6 eV
Energy in the first excited state = E =  = -3.4 eV
(2)
We know that, the de Broglie wavelength  is given as
h
 = 
p
But, p = 2mE
h
  = 
2mE
6.63 10
        = 
2 (9.1 10 ) (3.4 1.6 10 )
     
?
?
?
??
?
? ? ? ? ?
-34
25
-10
6.63  10
   = 
9.95 10
     = 6.6 10 m
?
?
?
??
 
 
Page 3


  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
CBSE 
Class XII Physics 
Board Paper – 2015 Solution 
 
SECTION – A 
 
1. Mobility of the charge carrier is defined as the drift velocity of the charge carrier per 
unit electric field. 
It is denoted by ? and is given as 
 
?
??
d
E
 
The SI unit of ? is m
2 
V
-1 
s
-1
. 
 
2. In an LCR series circuit, power dissipated is given by 
 
2
P I Zcos ??
 
 
2 2 2
CL
P
cos
I (R (X X )
??
??
 
The quantity cos ? is the power factor. 
 
2 2 2
CL
L C L C
2
P
cos
I R (X X )
V V and X X
P
cos
IR
??
??
??
? ? ?
 
 
3. The focal length and the radius of curvature of a lens are equal if the lens is made of 
glass having a refractive index of 1.5. 
 
4. The relation for polarisation P of a dielectric material in the presence of an external 
electric field E is 
 
e
PE ??
 
where 
e
? is a constant characteristic of the dielectric and is known as the electric 
susceptibility of the dielectric medium. 
 
5. In forward biasing, the applied voltage mostly drops across the depletion region, and 
the voltage drop across the p-side and n-side of the junction is negligible. The direction 
of the applied voltage (V) is opposite to the built-in potential V o. As a result, the 
depletion layer width decreases and the barrier height is reduced. The effective barrier 
height under forward bias is (V o - V). 
 
  
  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
SECTION – B  
6.  
(a)  
Analog form of communication Digital form of communication 
It makes use of analog signals whose 
amplitude varies continuously with 
time. 
It makes use of digital signals whose 
amplitude is of two levels, either low 
or high. 
The retrieved message in analog 
communication is not an exact copy 
of the original message. This is 
because the message suffers 
distortions over the medium and 
during its detection process and is 
distorted. 
The retrieved message in digital 
communication is always the exact 
copy of the original message and is 
practically free of distortion. 
It can broadcast only limited number 
of channels simultaneously. 
It can broadcast tremendously large 
number of channels simultaneously. 
 
(b) Applications of the Internet: 
1. E-mail: It allows the exchange of text or graphic material. One can write a letter 
and send it to the other person (recipient) through an internet service provider 
(ISP). The ISP works like the dispatching and receiving post office. 
2. File transfer: A program known as the file transfer program (FTP) allows the 
transfer of files or software from one computer to the other computer connected 
to the Internet. 
 
7. Ground state energy is E 0 = –13.6 eV 
1 2
1
1
-34
-31 -19
13.6 eV
Energy in the first excited state = E =  = -3.4 eV
(2)
We know that, the de Broglie wavelength  is given as
h
 = 
p
But, p = 2mE
h
  = 
2mE
6.63 10
        = 
2 (9.1 10 ) (3.4 1.6 10 )
     
?
?
?
??
?
? ? ? ? ?
-34
25
-10
6.63  10
   = 
9.95 10
     = 6.6 10 m
?
?
?
??
 
 
  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
8. Bohr’s third postulate successfully explains emission lines. It states that ‘Whenever an 
electron jumps from one of its specified non-radiating orbit to another such orbit, it 
emits or absorbs a photon whose energy is equal to the energy difference between the 
initial and final states’. 
Thus, 
 
if
hc
E E h ? ? ? ?
?
 
 
The Rydberg formula for the Balmer series is given as 
 
22
fi
1 1 1
R     
nn
??
??
??
?
??
  
‘R’ is a constant called the Rydberg constant and its value is 1.03 × 10
7
 m
-1
. 
The H a-line of the Balmer series is obtained when an electron jumps to the second orbit 
(n f = 2) from the third orbit (n i = 3). 
 
7
22
1 1 1
1.03×10     
23
??
??
??
?
??
 
 
7
7
7
7
7
27
1 1 1
1.03×10 
49
1 9 4
1.03×10
94
15
1.03×10
36
36
5 1.03 10
6.99 10
699 10 10
699 nm
?
??
??
??
??
?
??
? ??
?
??
??
??
??
?
??
?
??
??
??
? ? ?
? ? ? ?
??
 
The value of ? lies in the visible region. 
 
 
 
 
 
 
 
 
 
 
 
 
 
Page 4


  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
CBSE 
Class XII Physics 
Board Paper – 2015 Solution 
 
SECTION – A 
 
1. Mobility of the charge carrier is defined as the drift velocity of the charge carrier per 
unit electric field. 
It is denoted by ? and is given as 
 
?
??
d
E
 
The SI unit of ? is m
2 
V
-1 
s
-1
. 
 
2. In an LCR series circuit, power dissipated is given by 
 
2
P I Zcos ??
 
 
2 2 2
CL
P
cos
I (R (X X )
??
??
 
The quantity cos ? is the power factor. 
 
2 2 2
CL
L C L C
2
P
cos
I R (X X )
V V and X X
P
cos
IR
??
??
??
? ? ?
 
 
3. The focal length and the radius of curvature of a lens are equal if the lens is made of 
glass having a refractive index of 1.5. 
 
4. The relation for polarisation P of a dielectric material in the presence of an external 
electric field E is 
 
e
PE ??
 
where 
e
? is a constant characteristic of the dielectric and is known as the electric 
susceptibility of the dielectric medium. 
 
5. In forward biasing, the applied voltage mostly drops across the depletion region, and 
the voltage drop across the p-side and n-side of the junction is negligible. The direction 
of the applied voltage (V) is opposite to the built-in potential V o. As a result, the 
depletion layer width decreases and the barrier height is reduced. The effective barrier 
height under forward bias is (V o - V). 
 
  
  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
SECTION – B  
6.  
(a)  
Analog form of communication Digital form of communication 
It makes use of analog signals whose 
amplitude varies continuously with 
time. 
It makes use of digital signals whose 
amplitude is of two levels, either low 
or high. 
The retrieved message in analog 
communication is not an exact copy 
of the original message. This is 
because the message suffers 
distortions over the medium and 
during its detection process and is 
distorted. 
The retrieved message in digital 
communication is always the exact 
copy of the original message and is 
practically free of distortion. 
It can broadcast only limited number 
of channels simultaneously. 
It can broadcast tremendously large 
number of channels simultaneously. 
 
(b) Applications of the Internet: 
1. E-mail: It allows the exchange of text or graphic material. One can write a letter 
and send it to the other person (recipient) through an internet service provider 
(ISP). The ISP works like the dispatching and receiving post office. 
2. File transfer: A program known as the file transfer program (FTP) allows the 
transfer of files or software from one computer to the other computer connected 
to the Internet. 
 
7. Ground state energy is E 0 = –13.6 eV 
1 2
1
1
-34
-31 -19
13.6 eV
Energy in the first excited state = E =  = -3.4 eV
(2)
We know that, the de Broglie wavelength  is given as
h
 = 
p
But, p = 2mE
h
  = 
2mE
6.63 10
        = 
2 (9.1 10 ) (3.4 1.6 10 )
     
?
?
?
??
?
? ? ? ? ?
-34
25
-10
6.63  10
   = 
9.95 10
     = 6.6 10 m
?
?
?
??
 
 
  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
8. Bohr’s third postulate successfully explains emission lines. It states that ‘Whenever an 
electron jumps from one of its specified non-radiating orbit to another such orbit, it 
emits or absorbs a photon whose energy is equal to the energy difference between the 
initial and final states’. 
Thus, 
 
if
hc
E E h ? ? ? ?
?
 
 
The Rydberg formula for the Balmer series is given as 
 
22
fi
1 1 1
R     
nn
??
??
??
?
??
  
‘R’ is a constant called the Rydberg constant and its value is 1.03 × 10
7
 m
-1
. 
The H a-line of the Balmer series is obtained when an electron jumps to the second orbit 
(n f = 2) from the third orbit (n i = 3). 
 
7
22
1 1 1
1.03×10     
23
??
??
??
?
??
 
 
7
7
7
7
7
27
1 1 1
1.03×10 
49
1 9 4
1.03×10
94
15
1.03×10
36
36
5 1.03 10
6.99 10
699 10 10
699 nm
?
??
??
??
??
?
??
? ??
?
??
??
??
??
?
??
?
??
??
??
? ? ?
? ? ? ?
??
 
The value of ? lies in the visible region. 
 
 
 
 
 
 
 
 
 
 
 
 
 
  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
9. Kirchhoff’s first rule:   
In any electrical network, the algebraic sum of currents meeting at a junction is always 
zero. 
 ?I = 0 
In the junction below, let I 1, I 2, I 3, I 4 and I 5 be the current in the conductors with 
directions as shown in the figure below. I 5 and I 3 are the currents which enter and 
currents I 1, I 2 and I 4 leave. 
 
According to the Kirchhoff’s law, we have 
 (–I 1) + (-I 2) + I 3 + (-I 4) + I 5 = 0 
 Or I 1 + I 2 + I 4 = I 3 + I 5  
Thus, at any junction of several circuit elements, the sum of currents entering the 
junction must equal the sum of currents leaving it. This is a consequence of charge 
conservation and the assumption that currents are steady, i.e. no charge piles up at the 
junction. 
Kirchhoff’s second rule: The algebraic sum of changes in potential around any closed 
loop involving resistors and cells in the loop is zero.  
OR 
The algebraic sum of the e.m.f. in any loop of a circuit is equal to the algebraic sum of 
the products of currents and resistances in it. 
Mathematically, the loop rule may be expressed as ? E = ? IR. 
 
10. Important characteristic features: 
1. Interference is the result of the interaction of light coming from two different waves 
originating from two coherent sources, whereas the diffraction pattern is the result 
of the interaction of light coming from different parts of the same wavefront. 
2. The fringes may or may not be of the same width in case of interference, while the 
fringes are always of varying width in diffraction. 
3. In interference, the fringes of minimum intensity are perfectly dark and all bright 
fringes are of the same intensity. In diffraction, the fringes of minimum intensity are 
not perfectly dark and the bright fringes are of varying intensity. 
4. There is a good contrast between the bright and dark fringes in the interference 
pattern. The contrast between the bright and dark fringes in the diffraction pattern 
is comparatively poor. 
  
Page 5


  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
CBSE 
Class XII Physics 
Board Paper – 2015 Solution 
 
SECTION – A 
 
1. Mobility of the charge carrier is defined as the drift velocity of the charge carrier per 
unit electric field. 
It is denoted by ? and is given as 
 
?
??
d
E
 
The SI unit of ? is m
2 
V
-1 
s
-1
. 
 
2. In an LCR series circuit, power dissipated is given by 
 
2
P I Zcos ??
 
 
2 2 2
CL
P
cos
I (R (X X )
??
??
 
The quantity cos ? is the power factor. 
 
2 2 2
CL
L C L C
2
P
cos
I R (X X )
V V and X X
P
cos
IR
??
??
??
? ? ?
 
 
3. The focal length and the radius of curvature of a lens are equal if the lens is made of 
glass having a refractive index of 1.5. 
 
4. The relation for polarisation P of a dielectric material in the presence of an external 
electric field E is 
 
e
PE ??
 
where 
e
? is a constant characteristic of the dielectric and is known as the electric 
susceptibility of the dielectric medium. 
 
5. In forward biasing, the applied voltage mostly drops across the depletion region, and 
the voltage drop across the p-side and n-side of the junction is negligible. The direction 
of the applied voltage (V) is opposite to the built-in potential V o. As a result, the 
depletion layer width decreases and the barrier height is reduced. The effective barrier 
height under forward bias is (V o - V). 
 
  
  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
SECTION – B  
6.  
(a)  
Analog form of communication Digital form of communication 
It makes use of analog signals whose 
amplitude varies continuously with 
time. 
It makes use of digital signals whose 
amplitude is of two levels, either low 
or high. 
The retrieved message in analog 
communication is not an exact copy 
of the original message. This is 
because the message suffers 
distortions over the medium and 
during its detection process and is 
distorted. 
The retrieved message in digital 
communication is always the exact 
copy of the original message and is 
practically free of distortion. 
It can broadcast only limited number 
of channels simultaneously. 
It can broadcast tremendously large 
number of channels simultaneously. 
 
(b) Applications of the Internet: 
1. E-mail: It allows the exchange of text or graphic material. One can write a letter 
and send it to the other person (recipient) through an internet service provider 
(ISP). The ISP works like the dispatching and receiving post office. 
2. File transfer: A program known as the file transfer program (FTP) allows the 
transfer of files or software from one computer to the other computer connected 
to the Internet. 
 
7. Ground state energy is E 0 = –13.6 eV 
1 2
1
1
-34
-31 -19
13.6 eV
Energy in the first excited state = E =  = -3.4 eV
(2)
We know that, the de Broglie wavelength  is given as
h
 = 
p
But, p = 2mE
h
  = 
2mE
6.63 10
        = 
2 (9.1 10 ) (3.4 1.6 10 )
     
?
?
?
??
?
? ? ? ? ?
-34
25
-10
6.63  10
   = 
9.95 10
     = 6.6 10 m
?
?
?
??
 
 
  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
8. Bohr’s third postulate successfully explains emission lines. It states that ‘Whenever an 
electron jumps from one of its specified non-radiating orbit to another such orbit, it 
emits or absorbs a photon whose energy is equal to the energy difference between the 
initial and final states’. 
Thus, 
 
if
hc
E E h ? ? ? ?
?
 
 
The Rydberg formula for the Balmer series is given as 
 
22
fi
1 1 1
R     
nn
??
??
??
?
??
  
‘R’ is a constant called the Rydberg constant and its value is 1.03 × 10
7
 m
-1
. 
The H a-line of the Balmer series is obtained when an electron jumps to the second orbit 
(n f = 2) from the third orbit (n i = 3). 
 
7
22
1 1 1
1.03×10     
23
??
??
??
?
??
 
 
7
7
7
7
7
27
1 1 1
1.03×10 
49
1 9 4
1.03×10
94
15
1.03×10
36
36
5 1.03 10
6.99 10
699 10 10
699 nm
?
??
??
??
??
?
??
? ??
?
??
??
??
??
?
??
?
??
??
??
? ? ?
? ? ? ?
??
 
The value of ? lies in the visible region. 
 
 
 
 
 
 
 
 
 
 
 
 
 
  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
9. Kirchhoff’s first rule:   
In any electrical network, the algebraic sum of currents meeting at a junction is always 
zero. 
 ?I = 0 
In the junction below, let I 1, I 2, I 3, I 4 and I 5 be the current in the conductors with 
directions as shown in the figure below. I 5 and I 3 are the currents which enter and 
currents I 1, I 2 and I 4 leave. 
 
According to the Kirchhoff’s law, we have 
 (–I 1) + (-I 2) + I 3 + (-I 4) + I 5 = 0 
 Or I 1 + I 2 + I 4 = I 3 + I 5  
Thus, at any junction of several circuit elements, the sum of currents entering the 
junction must equal the sum of currents leaving it. This is a consequence of charge 
conservation and the assumption that currents are steady, i.e. no charge piles up at the 
junction. 
Kirchhoff’s second rule: The algebraic sum of changes in potential around any closed 
loop involving resistors and cells in the loop is zero.  
OR 
The algebraic sum of the e.m.f. in any loop of a circuit is equal to the algebraic sum of 
the products of currents and resistances in it. 
Mathematically, the loop rule may be expressed as ? E = ? IR. 
 
10. Important characteristic features: 
1. Interference is the result of the interaction of light coming from two different waves 
originating from two coherent sources, whereas the diffraction pattern is the result 
of the interaction of light coming from different parts of the same wavefront. 
2. The fringes may or may not be of the same width in case of interference, while the 
fringes are always of varying width in diffraction. 
3. In interference, the fringes of minimum intensity are perfectly dark and all bright 
fringes are of the same intensity. In diffraction, the fringes of minimum intensity are 
not perfectly dark and the bright fringes are of varying intensity. 
4. There is a good contrast between the bright and dark fringes in the interference 
pattern. The contrast between the bright and dark fringes in the diffraction pattern 
is comparatively poor. 
  
  
 
CBSE XII  | PHYSICS 
Board Paper – 2015 Solution 
 
     
OR 
 
Telescope Microscope 
It is used for observing distant images of 
heavenly bodies such as stars and 
planets. 
It is used for observing magnifying 
images of tiny objects. 
The objective lens has a large focal 
length and large aperture. 
The objective lens has a small focal 
length and short aperture. 
The eye lens used has small focal length 
and small aperture. 
The eye lens used has moderate 
focal length and large aperture. 
The distance between the objective lens 
and eye lens is adjusted to focus the 
object situated at infinity.  
The objective and eye lens are kept 
at a fixed distance apart, whereas 
the distance of the objective lens 
from the object is adjusted to focus 
an object. 
 
SECTION – C  
11.  
(i) Keeping the anode potential and the frequency of the incident radiation constant, if 
the intensity of the incident light is increased, the photoelectric current or the anode 
current increases linearly. This is because photoelectric current is directly 
proportional to the number of photoelectrons emitted per second which is directly 
proportional to the intensity of the incident radiation. 
(ii) For photoelectric emission to occur, there is a minimum cut off frequency of the 
incident radiation called the threshold frequency below which no photoelectric 
emission occurs. This frequency is independent of the intensity of the incident light. 
With an increase in the frequency of the incident radiation, the kinetic energy of the 
photoelectrons ejected increases, whereas it is independent of the number of 
photoelectrons ejected. Hence, with the increase in the frequency of incident 
radiation, there will not be any change in the anode current. 
(iii) With an increase in the accelerating potential, the photoelectric current increases 
first, reaches maximum when all the electrons gets collected at the positive 
potential plate and then remains constant. The maximum value of the anode 
current is called the saturation current.