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Page 1
Delhi SET I Page 1 of 20 Final Draft 12/3/2013 2: 00 pm
MARKING SCHEME
SET 55/1/1
Q. No. Expected Answer / Value Points Marks Total
Marks
1. Magnitude of the drift velocity of charge carrier per unit Electric field is
called mobility.
Alternatively, µ
=
E
v
d
| |
or
m
et
SI unit = m
2
/ (volt second) or ms
-1
N
-1
C
½
½
1
2.
Modulation Index =
c
m
a
a
= 1/2 = 0.5
½
½
1
3. If Electric field is not normal, it will have non-zero component along the
surface. In that case, work would be done in moving a charge on an
equipotential surface.
1
1
4. Glass.
In glass there is no effect of electromagnetic induction, due to presence of
Earth’s magnetic field, unlike in the case of metallic ball.
½
½
1
5.
1
1
6. 20cm 1 1
7.
Perpendicular to the plane formed by
/
[Note: Give full credit for writing the expression.]
½
½
1
8. X: Channel
It connects the Transmitter to the Receiver
½
½
1
Page 2
Delhi SET I Page 1 of 20 Final Draft 12/3/2013 2: 00 pm
MARKING SCHEME
SET 55/1/1
Q. No. Expected Answer / Value Points Marks Total
Marks
1. Magnitude of the drift velocity of charge carrier per unit Electric field is
called mobility.
Alternatively, µ
=
E
v
d
| |
or
m
et
SI unit = m
2
/ (volt second) or ms
-1
N
-1
C
½
½
1
2.
Modulation Index =
c
m
a
a
= 1/2 = 0.5
½
½
1
3. If Electric field is not normal, it will have non-zero component along the
surface. In that case, work would be done in moving a charge on an
equipotential surface.
1
1
4. Glass.
In glass there is no effect of electromagnetic induction, due to presence of
Earth’s magnetic field, unlike in the case of metallic ball.
½
½
1
5.
1
1
6. 20cm 1 1
7.
Perpendicular to the plane formed by
/
[Note: Give full credit for writing the expression.]
½
½
1
8. X: Channel
It connects the Transmitter to the Receiver
½
½
1
Delhi SET I Page 2 of 20 Final Draft 12/3/2013 2: 00 pm
9.
A: Paramagnetic
B: Diamagnetic
Susceptibility
For A: positive
For B: negative
½
½
½
½
2
10.
cos
= 5 10
10
cos0
!
50
!
=5 10
10
cos60
!
= 25 N
!
½
½
½
½
2
11.
In the first case, the overlapping of the contributions of the wavelets from
two halves of a single slit produces a minimum because corresponding
wavelets from two halves have a path difference of
2
?
.
In the second case, the overlapping of the wavefronts from the two slits
produces first maximum because these wavefronts have the path difference
of #.
(Alternatively, if a student writes the conditions given below, give full
credit.)
Condition for first minimum in single slit diffraction is , ˜ ? / a,
Whereas in case of two narrow slits separated by distance a, first maximum
occurs at angle ˜ ? / a
[Note: Award 1 mark even if the candidate attempts this question partly.]
1
1
2
Identification of magnetic material ½ + ½
Susceptibility ½ + ½
Finding flux in two cases 1+1
Explanation of the given statement 1 + 1
Page 3
Delhi SET I Page 1 of 20 Final Draft 12/3/2013 2: 00 pm
MARKING SCHEME
SET 55/1/1
Q. No. Expected Answer / Value Points Marks Total
Marks
1. Magnitude of the drift velocity of charge carrier per unit Electric field is
called mobility.
Alternatively, µ
=
E
v
d
| |
or
m
et
SI unit = m
2
/ (volt second) or ms
-1
N
-1
C
½
½
1
2.
Modulation Index =
c
m
a
a
= 1/2 = 0.5
½
½
1
3. If Electric field is not normal, it will have non-zero component along the
surface. In that case, work would be done in moving a charge on an
equipotential surface.
1
1
4. Glass.
In glass there is no effect of electromagnetic induction, due to presence of
Earth’s magnetic field, unlike in the case of metallic ball.
½
½
1
5.
1
1
6. 20cm 1 1
7.
Perpendicular to the plane formed by
/
[Note: Give full credit for writing the expression.]
½
½
1
8. X: Channel
It connects the Transmitter to the Receiver
½
½
1
Delhi SET I Page 2 of 20 Final Draft 12/3/2013 2: 00 pm
9.
A: Paramagnetic
B: Diamagnetic
Susceptibility
For A: positive
For B: negative
½
½
½
½
2
10.
cos
= 5 10
10
cos0
!
50
!
=5 10
10
cos60
!
= 25 N
!
½
½
½
½
2
11.
In the first case, the overlapping of the contributions of the wavelets from
two halves of a single slit produces a minimum because corresponding
wavelets from two halves have a path difference of
2
?
.
In the second case, the overlapping of the wavefronts from the two slits
produces first maximum because these wavefronts have the path difference
of #.
(Alternatively, if a student writes the conditions given below, give full
credit.)
Condition for first minimum in single slit diffraction is , ˜ ? / a,
Whereas in case of two narrow slits separated by distance a, first maximum
occurs at angle ˜ ? / a
[Note: Award 1 mark even if the candidate attempts this question partly.]
1
1
2
Identification of magnetic material ½ + ½
Susceptibility ½ + ½
Finding flux in two cases 1+1
Explanation of the given statement 1 + 1
Delhi SET I Page 3 of 20 Final Draft 12/3/2013 2: 00 pm
12.
Truth Table
Input Output
A B Y’ Y
0 0 0 0
0 1 1 0
1 0 1 1
1 1 1 1
Gate R: OR Gate
S: AND Gate
OR
P: NAND Gate
Q: OR Gate
Truth Table
Input Output
A B X
0 0 1
1 0 1
0 1 1
1 1 1
1
½
½
½
½
1
2
2
13.
Junction rule: At any junction, the sum of the currents entering the junction
is equal to the sum of currents leaving the junction.
Alternatively, ? i =0
½
Statements of two Laws ½ + ½
Justification ½ + ½
Truth Table 1
Names of gates used ½ + ½
Identification 1
Truth Table 1
Page 4
Delhi SET I Page 1 of 20 Final Draft 12/3/2013 2: 00 pm
MARKING SCHEME
SET 55/1/1
Q. No. Expected Answer / Value Points Marks Total
Marks
1. Magnitude of the drift velocity of charge carrier per unit Electric field is
called mobility.
Alternatively, µ
=
E
v
d
| |
or
m
et
SI unit = m
2
/ (volt second) or ms
-1
N
-1
C
½
½
1
2.
Modulation Index =
c
m
a
a
= 1/2 = 0.5
½
½
1
3. If Electric field is not normal, it will have non-zero component along the
surface. In that case, work would be done in moving a charge on an
equipotential surface.
1
1
4. Glass.
In glass there is no effect of electromagnetic induction, due to presence of
Earth’s magnetic field, unlike in the case of metallic ball.
½
½
1
5.
1
1
6. 20cm 1 1
7.
Perpendicular to the plane formed by
/
[Note: Give full credit for writing the expression.]
½
½
1
8. X: Channel
It connects the Transmitter to the Receiver
½
½
1
Delhi SET I Page 2 of 20 Final Draft 12/3/2013 2: 00 pm
9.
A: Paramagnetic
B: Diamagnetic
Susceptibility
For A: positive
For B: negative
½
½
½
½
2
10.
cos
= 5 10
10
cos0
!
50
!
=5 10
10
cos60
!
= 25 N
!
½
½
½
½
2
11.
In the first case, the overlapping of the contributions of the wavelets from
two halves of a single slit produces a minimum because corresponding
wavelets from two halves have a path difference of
2
?
.
In the second case, the overlapping of the wavefronts from the two slits
produces first maximum because these wavefronts have the path difference
of #.
(Alternatively, if a student writes the conditions given below, give full
credit.)
Condition for first minimum in single slit diffraction is , ˜ ? / a,
Whereas in case of two narrow slits separated by distance a, first maximum
occurs at angle ˜ ? / a
[Note: Award 1 mark even if the candidate attempts this question partly.]
1
1
2
Identification of magnetic material ½ + ½
Susceptibility ½ + ½
Finding flux in two cases 1+1
Explanation of the given statement 1 + 1
Delhi SET I Page 3 of 20 Final Draft 12/3/2013 2: 00 pm
12.
Truth Table
Input Output
A B Y’ Y
0 0 0 0
0 1 1 0
1 0 1 1
1 1 1 1
Gate R: OR Gate
S: AND Gate
OR
P: NAND Gate
Q: OR Gate
Truth Table
Input Output
A B X
0 0 1
1 0 1
0 1 1
1 1 1
1
½
½
½
½
1
2
2
13.
Junction rule: At any junction, the sum of the currents entering the junction
is equal to the sum of currents leaving the junction.
Alternatively, ? i =0
½
Statements of two Laws ½ + ½
Justification ½ + ½
Truth Table 1
Names of gates used ½ + ½
Identification 1
Truth Table 1
Delhi SET I Page 4 of 20 Final Draft 12/3/2013 2: 00 pm
Justification : Conservation of charge
Loop rule: The Algebraic sum of changes in the potential around any closed
loop involving resistors and cells in the loop is zero.
Alternatively, ? ? V =0 , where ? V is the changes in potential
Justification : Conservation of energy
½
½
½
2
14.
(i) Reactance of the capacitor will decrease, resulting in increase of the
current in the circuit. Therefore the bulb will glow brighter.
(ii) Increased resistance will decrease the current in the circuit, which
will decrease glow of the bulb.
[Note : Do not deduct any mark for not giving the reasons.]
1
1
2
15.
It makes use of the principle that the energy of the charged particles / ions
can be made to increase in presence of crossed Electric and magnetic fields.
A normal Magnetic field acts on the charged particle and makes them move
in a circular path .While moving from one dee to another; particle is acted
upon by the alternating electric field, and is accelerated by this field, which
increases the energy of the particle.
1
1
2
16.
tpEsin?
4v3 = pEsin60
0
v
? pE = 8
Potential energy
½
½
½
Calculation of Potential energy of the dipole 2
Underlying principle 1
Brief working 1
Effect on glow of bulb in Part (i) 1
Part (ii) 1
Page 5
Delhi SET I Page 1 of 20 Final Draft 12/3/2013 2: 00 pm
MARKING SCHEME
SET 55/1/1
Q. No. Expected Answer / Value Points Marks Total
Marks
1. Magnitude of the drift velocity of charge carrier per unit Electric field is
called mobility.
Alternatively, µ
=
E
v
d
| |
or
m
et
SI unit = m
2
/ (volt second) or ms
-1
N
-1
C
½
½
1
2.
Modulation Index =
c
m
a
a
= 1/2 = 0.5
½
½
1
3. If Electric field is not normal, it will have non-zero component along the
surface. In that case, work would be done in moving a charge on an
equipotential surface.
1
1
4. Glass.
In glass there is no effect of electromagnetic induction, due to presence of
Earth’s magnetic field, unlike in the case of metallic ball.
½
½
1
5.
1
1
6. 20cm 1 1
7.
Perpendicular to the plane formed by
/
[Note: Give full credit for writing the expression.]
½
½
1
8. X: Channel
It connects the Transmitter to the Receiver
½
½
1
Delhi SET I Page 2 of 20 Final Draft 12/3/2013 2: 00 pm
9.
A: Paramagnetic
B: Diamagnetic
Susceptibility
For A: positive
For B: negative
½
½
½
½
2
10.
cos
= 5 10
10
cos0
!
50
!
=5 10
10
cos60
!
= 25 N
!
½
½
½
½
2
11.
In the first case, the overlapping of the contributions of the wavelets from
two halves of a single slit produces a minimum because corresponding
wavelets from two halves have a path difference of
2
?
.
In the second case, the overlapping of the wavefronts from the two slits
produces first maximum because these wavefronts have the path difference
of #.
(Alternatively, if a student writes the conditions given below, give full
credit.)
Condition for first minimum in single slit diffraction is , ˜ ? / a,
Whereas in case of two narrow slits separated by distance a, first maximum
occurs at angle ˜ ? / a
[Note: Award 1 mark even if the candidate attempts this question partly.]
1
1
2
Identification of magnetic material ½ + ½
Susceptibility ½ + ½
Finding flux in two cases 1+1
Explanation of the given statement 1 + 1
Delhi SET I Page 3 of 20 Final Draft 12/3/2013 2: 00 pm
12.
Truth Table
Input Output
A B Y’ Y
0 0 0 0
0 1 1 0
1 0 1 1
1 1 1 1
Gate R: OR Gate
S: AND Gate
OR
P: NAND Gate
Q: OR Gate
Truth Table
Input Output
A B X
0 0 1
1 0 1
0 1 1
1 1 1
1
½
½
½
½
1
2
2
13.
Junction rule: At any junction, the sum of the currents entering the junction
is equal to the sum of currents leaving the junction.
Alternatively, ? i =0
½
Statements of two Laws ½ + ½
Justification ½ + ½
Truth Table 1
Names of gates used ½ + ½
Identification 1
Truth Table 1
Delhi SET I Page 4 of 20 Final Draft 12/3/2013 2: 00 pm
Justification : Conservation of charge
Loop rule: The Algebraic sum of changes in the potential around any closed
loop involving resistors and cells in the loop is zero.
Alternatively, ? ? V =0 , where ? V is the changes in potential
Justification : Conservation of energy
½
½
½
2
14.
(i) Reactance of the capacitor will decrease, resulting in increase of the
current in the circuit. Therefore the bulb will glow brighter.
(ii) Increased resistance will decrease the current in the circuit, which
will decrease glow of the bulb.
[Note : Do not deduct any mark for not giving the reasons.]
1
1
2
15.
It makes use of the principle that the energy of the charged particles / ions
can be made to increase in presence of crossed Electric and magnetic fields.
A normal Magnetic field acts on the charged particle and makes them move
in a circular path .While moving from one dee to another; particle is acted
upon by the alternating electric field, and is accelerated by this field, which
increases the energy of the particle.
1
1
2
16.
tpEsin?
4v3 = pEsin60
0
v
? pE = 8
Potential energy
½
½
½
Calculation of Potential energy of the dipole 2
Underlying principle 1
Brief working 1
Effect on glow of bulb in Part (i) 1
Part (ii) 1
Delhi SET I Page 5 of 20 Final Draft 12/3/2013 2: 00 pm
U = -pE cos
= -8 x cos 60
0
= -4J
[Give full credit to alternative methods of finding Potential energy.]
½
2
17.
(a) de Broglie wavelength is given by
?=
2
3 456
As mass of proton < mass of deuteron and q
p
= q
d
and v is same
=>
?p > ?
d
for same accelerating potential.
(b) Momentum =
2
7
? ?
p
> ?
d
? momentum of proton will be less , than that of deuteron
½
½
½
½
2
18.
(a) Power =
h, where n = no. of photons per second
2.0 x 10
-3
=
× 6.6 × 10
<
× 6 × 10
<
n =
. =
>?
@.@ ×
>?A
×@×
BA
= 0.050 x 10
17
= 5x 10
15
photons / second
[Note: Even if the student doesn’t write the formula but calculates correctly,
give full credit to this part]
(b)
½
½
Part (a) and reason ½ + ½
Part (b) and reason ½ + ½
(a) Estimation of no. of photons per second 1
(b) Plot showing the variation 1
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FAQs on Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics - Additional Study Material for NEET
| 1. What is the importance of solving past year papers in preparing for the Class 12 Physics exam? |  |
Ans. Solving past year papers is important as it helps students familiarize themselves with the exam pattern, types of questions asked, and the level of difficulty. It allows them to practice time management and identify their strengths and weaknesses in different topics of the Physics syllabus. Additionally, it helps students gain confidence and reduces exam anxiety.
| 2. How can solving past year papers help in improving my physics problem-solving skills? | |
Ans. Solving past year papers requires applying concepts and solving problems from various topics of Physics. By regularly practicing these papers, students can enhance their problem-solving skills, develop a deeper understanding of concepts, and learn different approaches to tackle complex problems. It also enables students to identify common mistakes and improve their overall accuracy.
| 3. Can solving past year papers help me predict the important topics for the Class 12 Physics exam? | |
Ans. Yes, solving past year papers can provide insights into the important topics for the Class 12 Physics exam. By analyzing the frequency of questions from different topics in previous years' papers, students can identify the areas that are frequently tested and allocate their study time accordingly. However, it is important to note that the exam pattern may change, and all topics should be thoroughly covered.
| 4. How should I use the solutions of past year papers to maximize my preparation for the Class 12 Physics exam? | |
Ans. To maximize preparation, students should solve the past year papers first without referring to the solutions. After completing the paper, they can cross-check their answers with the provided solutions. If any mistakes are made, they should understand the correct approach and rectify their errors. It is important to review the solutions in detail, especially for questions that were incorrectly answered or took a significant amount of time to solve.
| 5. Are the solutions provided for past year papers in the same language as the article title? | |
Ans. Yes, the solutions provided for past year papers are in the same language as the article title, which is Delhi 2014 Class 12 Physics paper. This ensures that students can easily understand and follow the solutions while preparing for the exam. The language used in the solutions is kept at a level similar to the complexity of the exam and the text.
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