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 Page 1


1     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
MARKING SCHEME  55/1/C 
Q. No. Expected Answer / Value Points Marks Total 
Marks 
SECTION-A 
SET1,Q1 
SET2,Q4 
SET3,Q5 
No work is done /
W = qV
AB 
= q x 0 = 0 
1 1 
SET1,Q2 
SET2,Q1 
SET3,Q3 
A diamagnetic specimen would move towards the weaker region of the field 
while a paramagnetic specimen would move towards the stronger region./ 
A diamagnetic specimen is repelled by a magnet while a paramagnetic 
specimen moves towards the magnet./
The paramagnetic get aligned along B and the diagrammatic perpendicular 
to the field. 
1 
1 
SET1,Q3 
SET2,Q5 
SET3,Q2 
Transmitter, Medium or Channel and Receiver. 1 1 
SET1,Q4 
SET2,Q3 
SET3,Q1 . 
It is due to least scattering of red light as it has the longest wavelength/
As per Rayleigh’s scattering, the amount of light scattered 
1 1 
SET1,Q5 
SET2,Q2 
SET3,Q4 
E = 2V ½ 
½ 1 
SECTION B 
SET1,Q6 
SET2,Q9 
SET3,Q8. 
Modulation index is the ratio of the amplitude of modulating signal to that of carrier 
wave 
Alternatively
c
m
A
A
 Reason- 
 To avoid distortion. 
Role- 
 A bandpass filter rejects low and high frequencies and allows a band of frequencies 
to pass through. 
1 
½ 
½ 2 
Definition-      1 
Reason-      ½ 
Role of bandpass filter-  ½ 
Page 2


1     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
MARKING SCHEME  55/1/C 
Q. No. Expected Answer / Value Points Marks Total 
Marks 
SECTION-A 
SET1,Q1 
SET2,Q4 
SET3,Q5 
No work is done /
W = qV
AB 
= q x 0 = 0 
1 1 
SET1,Q2 
SET2,Q1 
SET3,Q3 
A diamagnetic specimen would move towards the weaker region of the field 
while a paramagnetic specimen would move towards the stronger region./ 
A diamagnetic specimen is repelled by a magnet while a paramagnetic 
specimen moves towards the magnet./
The paramagnetic get aligned along B and the diagrammatic perpendicular 
to the field. 
1 
1 
SET1,Q3 
SET2,Q5 
SET3,Q2 
Transmitter, Medium or Channel and Receiver. 1 1 
SET1,Q4 
SET2,Q3 
SET3,Q1 . 
It is due to least scattering of red light as it has the longest wavelength/
As per Rayleigh’s scattering, the amount of light scattered 
1 1 
SET1,Q5 
SET2,Q2 
SET3,Q4 
E = 2V ½ 
½ 1 
SECTION B 
SET1,Q6 
SET2,Q9 
SET3,Q8. 
Modulation index is the ratio of the amplitude of modulating signal to that of carrier 
wave 
Alternatively
c
m
A
A
 Reason- 
 To avoid distortion. 
Role- 
 A bandpass filter rejects low and high frequencies and allows a band of frequencies 
to pass through. 
1 
½ 
½ 2 
Definition-      1 
Reason-      ½ 
Role of bandpass filter-  ½ 
2     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
SET1,Q7 
SET2,Q10 
SET3,Q6 
Face-AC 
Here ) 
     = 
on face AC is 30º which is less than .Hence the ray get replaced here. 
1 
½ 
½ 
2 
SET1,Q8 
SET2,Q6 
SET3,Q7 
Kinetic Energy for the second state- 
E
k
=  = =3.4X1.6X10
-19
J 
De Broglies wavelength ?= 
 = 0.067nm 
½ 
½ 
½ 
½ 2 
SET1,Q9 
SET2,Q8 
SET3,Q10 
The minimum energy, required to free the electron from the ground state of 
the hydrogen atom, is known as Ionization Energy. 
1 
Path of emergent ray  1 
Naming the face      ½ 
Justification      ½ 
Formulae of Kinetic energy and deBrogliea wavelength  ½ +½ 
Calculation and Result      ½+½ 
Definition      1 
Formula     ½ 
Calculation and Result ½ 
Page 3


1     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
MARKING SCHEME  55/1/C 
Q. No. Expected Answer / Value Points Marks Total 
Marks 
SECTION-A 
SET1,Q1 
SET2,Q4 
SET3,Q5 
No work is done /
W = qV
AB 
= q x 0 = 0 
1 1 
SET1,Q2 
SET2,Q1 
SET3,Q3 
A diamagnetic specimen would move towards the weaker region of the field 
while a paramagnetic specimen would move towards the stronger region./ 
A diamagnetic specimen is repelled by a magnet while a paramagnetic 
specimen moves towards the magnet./
The paramagnetic get aligned along B and the diagrammatic perpendicular 
to the field. 
1 
1 
SET1,Q3 
SET2,Q5 
SET3,Q2 
Transmitter, Medium or Channel and Receiver. 1 1 
SET1,Q4 
SET2,Q3 
SET3,Q1 . 
It is due to least scattering of red light as it has the longest wavelength/
As per Rayleigh’s scattering, the amount of light scattered 
1 1 
SET1,Q5 
SET2,Q2 
SET3,Q4 
E = 2V ½ 
½ 1 
SECTION B 
SET1,Q6 
SET2,Q9 
SET3,Q8. 
Modulation index is the ratio of the amplitude of modulating signal to that of carrier 
wave 
Alternatively
c
m
A
A
 Reason- 
 To avoid distortion. 
Role- 
 A bandpass filter rejects low and high frequencies and allows a band of frequencies 
to pass through. 
1 
½ 
½ 2 
Definition-      1 
Reason-      ½ 
Role of bandpass filter-  ½ 
2     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
SET1,Q7 
SET2,Q10 
SET3,Q6 
Face-AC 
Here ) 
     = 
on face AC is 30º which is less than .Hence the ray get replaced here. 
1 
½ 
½ 
2 
SET1,Q8 
SET2,Q6 
SET3,Q7 
Kinetic Energy for the second state- 
E
k
=  = =3.4X1.6X10
-19
J 
De Broglies wavelength ?= 
 = 0.067nm 
½ 
½ 
½ 
½ 2 
SET1,Q9 
SET2,Q8 
SET3,Q10 
The minimum energy, required to free the electron from the ground state of 
the hydrogen atom, is known as Ionization Energy. 
1 
Path of emergent ray  1 
Naming the face      ½ 
Justification      ½ 
Formulae of Kinetic energy and deBrogliea wavelength  ½ +½ 
Calculation and Result      ½+½ 
Definition      1 
Formula     ½ 
Calculation and Result ½ 
3     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
m E e i
h
me
E
o
o
o
,. .
8
2 2
4
Therefore, Ionization Energy will become 200 times 
 OR 
2 2
1
2
1 1
R
For shortest wavelength, n = a 
Therefore, = => ?= =4x10
-7
m
½ 
½ 
1 
½ 
½ 
2 
2 
SET1,Q10 
SET2,Q7 
SET3,Q9 
a) Effective resistance of the circuit R
E
= 6?
Terminal potential difference across the cell, V=E-ir 
Also p.d. across 4? resistor =4X2V= 8V 
 Hence the volmeter gives the same reading in the two cases. 
b) In series -current same
In parallel – potential same
½ 
½ 
½ 
½ 2 
SECTION C 
SET1,Q11 
SET2,Q15 
SET3,Q22 
Surface with a constant value of potential at all points on the surface. ½ 
Formula      1 
Calculation and Result  ½+½ 
½
a) Relation for terminal potential ½ 
b) Justification ½ 
c) Explanation (parallel and series) ½ + ½ 
Definition-  ½ 
i.Diagram of Equipotential Surface ½ 
ii.Diagram and reason ½ +½ 
iii.Answer and Reason ½+½ 
Page 4


1     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
MARKING SCHEME  55/1/C 
Q. No. Expected Answer / Value Points Marks Total 
Marks 
SECTION-A 
SET1,Q1 
SET2,Q4 
SET3,Q5 
No work is done /
W = qV
AB 
= q x 0 = 0 
1 1 
SET1,Q2 
SET2,Q1 
SET3,Q3 
A diamagnetic specimen would move towards the weaker region of the field 
while a paramagnetic specimen would move towards the stronger region./ 
A diamagnetic specimen is repelled by a magnet while a paramagnetic 
specimen moves towards the magnet./
The paramagnetic get aligned along B and the diagrammatic perpendicular 
to the field. 
1 
1 
SET1,Q3 
SET2,Q5 
SET3,Q2 
Transmitter, Medium or Channel and Receiver. 1 1 
SET1,Q4 
SET2,Q3 
SET3,Q1 . 
It is due to least scattering of red light as it has the longest wavelength/
As per Rayleigh’s scattering, the amount of light scattered 
1 1 
SET1,Q5 
SET2,Q2 
SET3,Q4 
E = 2V ½ 
½ 1 
SECTION B 
SET1,Q6 
SET2,Q9 
SET3,Q8. 
Modulation index is the ratio of the amplitude of modulating signal to that of carrier 
wave 
Alternatively
c
m
A
A
 Reason- 
 To avoid distortion. 
Role- 
 A bandpass filter rejects low and high frequencies and allows a band of frequencies 
to pass through. 
1 
½ 
½ 2 
Definition-      1 
Reason-      ½ 
Role of bandpass filter-  ½ 
2     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
SET1,Q7 
SET2,Q10 
SET3,Q6 
Face-AC 
Here ) 
     = 
on face AC is 30º which is less than .Hence the ray get replaced here. 
1 
½ 
½ 
2 
SET1,Q8 
SET2,Q6 
SET3,Q7 
Kinetic Energy for the second state- 
E
k
=  = =3.4X1.6X10
-19
J 
De Broglies wavelength ?= 
 = 0.067nm 
½ 
½ 
½ 
½ 2 
SET1,Q9 
SET2,Q8 
SET3,Q10 
The minimum energy, required to free the electron from the ground state of 
the hydrogen atom, is known as Ionization Energy. 
1 
Path of emergent ray  1 
Naming the face      ½ 
Justification      ½ 
Formulae of Kinetic energy and deBrogliea wavelength  ½ +½ 
Calculation and Result      ½+½ 
Definition      1 
Formula     ½ 
Calculation and Result ½ 
3     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
m E e i
h
me
E
o
o
o
,. .
8
2 2
4
Therefore, Ionization Energy will become 200 times 
 OR 
2 2
1
2
1 1
R
For shortest wavelength, n = a 
Therefore, = => ?= =4x10
-7
m
½ 
½ 
1 
½ 
½ 
2 
2 
SET1,Q10 
SET2,Q7 
SET3,Q9 
a) Effective resistance of the circuit R
E
= 6?
Terminal potential difference across the cell, V=E-ir 
Also p.d. across 4? resistor =4X2V= 8V 
 Hence the volmeter gives the same reading in the two cases. 
b) In series -current same
In parallel – potential same
½ 
½ 
½ 
½ 2 
SECTION C 
SET1,Q11 
SET2,Q15 
SET3,Q22 
Surface with a constant value of potential at all points on the surface. ½ 
Formula      1 
Calculation and Result  ½+½ 
½
a) Relation for terminal potential ½ 
b) Justification ½ 
c) Explanation (parallel and series) ½ + ½ 
Definition-  ½ 
i.Diagram of Equipotential Surface ½ 
ii.Diagram and reason ½ +½ 
iii.Answer and Reason ½+½ 
4     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
i. 
ii. 
 V / 
iii.No
If the field lines are tangential, work will be done in moving a charge on the 
surface which goes against the definition of equipotential surface. 
½ 
½ 
½ 
½ 
½ 3 
SET1,Q12 
SET2,Q14 
SET3,Q12 
i.When the pass axis of a  poloroid makes an angle  with the plane of
polarisation of polorised light of intensity I
o
  incident on it, then the intensity 
of the tramsmitted emergent light is given by I=
Note: If the student writes the formula I= and draws the 
1 
Statement      1 
Plotting the graph      1 
Calculating value of ( refractive index  1 
Page 5


1     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
MARKING SCHEME  55/1/C 
Q. No. Expected Answer / Value Points Marks Total 
Marks 
SECTION-A 
SET1,Q1 
SET2,Q4 
SET3,Q5 
No work is done /
W = qV
AB 
= q x 0 = 0 
1 1 
SET1,Q2 
SET2,Q1 
SET3,Q3 
A diamagnetic specimen would move towards the weaker region of the field 
while a paramagnetic specimen would move towards the stronger region./ 
A diamagnetic specimen is repelled by a magnet while a paramagnetic 
specimen moves towards the magnet./
The paramagnetic get aligned along B and the diagrammatic perpendicular 
to the field. 
1 
1 
SET1,Q3 
SET2,Q5 
SET3,Q2 
Transmitter, Medium or Channel and Receiver. 1 1 
SET1,Q4 
SET2,Q3 
SET3,Q1 . 
It is due to least scattering of red light as it has the longest wavelength/
As per Rayleigh’s scattering, the amount of light scattered 
1 1 
SET1,Q5 
SET2,Q2 
SET3,Q4 
E = 2V ½ 
½ 1 
SECTION B 
SET1,Q6 
SET2,Q9 
SET3,Q8. 
Modulation index is the ratio of the amplitude of modulating signal to that of carrier 
wave 
Alternatively
c
m
A
A
 Reason- 
 To avoid distortion. 
Role- 
 A bandpass filter rejects low and high frequencies and allows a band of frequencies 
to pass through. 
1 
½ 
½ 2 
Definition-      1 
Reason-      ½ 
Role of bandpass filter-  ½ 
2     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
SET1,Q7 
SET2,Q10 
SET3,Q6 
Face-AC 
Here ) 
     = 
on face AC is 30º which is less than .Hence the ray get replaced here. 
1 
½ 
½ 
2 
SET1,Q8 
SET2,Q6 
SET3,Q7 
Kinetic Energy for the second state- 
E
k
=  = =3.4X1.6X10
-19
J 
De Broglies wavelength ?= 
 = 0.067nm 
½ 
½ 
½ 
½ 2 
SET1,Q9 
SET2,Q8 
SET3,Q10 
The minimum energy, required to free the electron from the ground state of 
the hydrogen atom, is known as Ionization Energy. 
1 
Path of emergent ray  1 
Naming the face      ½ 
Justification      ½ 
Formulae of Kinetic energy and deBrogliea wavelength  ½ +½ 
Calculation and Result      ½+½ 
Definition      1 
Formula     ½ 
Calculation and Result ½ 
3     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
m E e i
h
me
E
o
o
o
,. .
8
2 2
4
Therefore, Ionization Energy will become 200 times 
 OR 
2 2
1
2
1 1
R
For shortest wavelength, n = a 
Therefore, = => ?= =4x10
-7
m
½ 
½ 
1 
½ 
½ 
2 
2 
SET1,Q10 
SET2,Q7 
SET3,Q9 
a) Effective resistance of the circuit R
E
= 6?
Terminal potential difference across the cell, V=E-ir 
Also p.d. across 4? resistor =4X2V= 8V 
 Hence the volmeter gives the same reading in the two cases. 
b) In series -current same
In parallel – potential same
½ 
½ 
½ 
½ 2 
SECTION C 
SET1,Q11 
SET2,Q15 
SET3,Q22 
Surface with a constant value of potential at all points on the surface. ½ 
Formula      1 
Calculation and Result  ½+½ 
½
a) Relation for terminal potential ½ 
b) Justification ½ 
c) Explanation (parallel and series) ½ + ½ 
Definition-  ½ 
i.Diagram of Equipotential Surface ½ 
ii.Diagram and reason ½ +½ 
iii.Answer and Reason ½+½ 
4     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
i. 
ii. 
 V / 
iii.No
If the field lines are tangential, work will be done in moving a charge on the 
surface which goes against the definition of equipotential surface. 
½ 
½ 
½ 
½ 
½ 3 
SET1,Q12 
SET2,Q14 
SET3,Q12 
i.When the pass axis of a  poloroid makes an angle  with the plane of
polarisation of polorised light of intensity I
o
  incident on it, then the intensity 
of the tramsmitted emergent light is given by I=
Note: If the student writes the formula I= and draws the 
1 
Statement      1 
Plotting the graph      1 
Calculating value of ( refractive index  1 
5     FOURTH  DRAFT     1O
th 
March/2016 4:00 PM 
diagram give 1mark. 
i. 
iii.
 =  = 
1 
½ 
½ 3 
SET1,Q13 
SET2,Q13 
SET3,Q14 
(i) For material B 
From the graph for the same value of ‘ ‘, stopping potential is more for material ‘B’/
is higher for lower value of ] 
(ii) No 
As slope is given by  which is constant. 
1 
½ 
½ 
½ 
½ 3 
  Sketch of the Graph      1 
(i) Stopping Potential and Reason  ½+ ½ 
(ii) Dependence of Slope and Explanation   ½+ ½

FAQs on Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12

1. What is the purpose of the Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12?

Ans. The purpose of the Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12 is to provide students with a reference guide that contains the answers to the questions asked in the Physics exam conducted in the Central Zone outside Delhi in the year 2016. It helps students in assessing their performance, understanding the correct answers, and preparing for future exams.

2. How can the Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12 benefit students?

Ans. The Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12 can benefit students in several ways. It provides them with a comprehensive solution to the exam questions, which helps in understanding the concepts and improving their knowledge. It also serves as a valuable resource for practice and revision, allowing students to gauge their understanding of the subject and identify areas that require further attention.

3. Can the Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12 be used as a study material?

Ans. Yes, the Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12 can be used as a study material. It contains the answers to the questions asked in the exam, which helps students in understanding the correct approach to solving them. By studying the solutions provided, students can gain a deeper understanding of the concepts and improve their problem-solving skills.

4. How can students effectively use the Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12 for exam preparation?

Ans. Students can effectively use the Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12 for exam preparation by following these steps: 1. Start by attempting the exam questions on your own without referring to the anskey. 2. After attempting each question, compare your answers with the solutions provided in the anskey. 3. Analyze the mistakes made and review the corresponding concepts to understand the correct approach. 4. Work on the areas where you faced difficulties and practice similar questions to improve your skills. 5. Once you have reviewed all the questions, attempt the paper again without referring to the anskey to assess your progress.

5. How can the Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12 help in identifying important topics for the exam?

Ans. The Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12 can help in identifying important topics for the exam by analyzing the frequency of certain concepts or types of questions in the paper. By reviewing the anskey, students can observe patterns and trends in the questions asked, which can give them insights into the topics that are more likely to be tested in the future. This allows students to prioritize their study efforts and focus on the areas that carry more weightage in the exam.

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