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# Physics Past Year Paper Anskey (Delhi Set -2) - 2014, Class 12, CBSE Class 12 Notes | EduRev

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## Class 12 : Physics Past Year Paper Anskey (Delhi Set -2) - 2014, Class 12, CBSE Class 12 Notes | EduRev

``` Page 1

Delhi Set II  FINAL print Draft  Page No. 1  12th March, 2014      3:20 pm
MARKING SCHEME
SET 55/1/2 (DELHI)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. Electrical conductivity is defined as current density per unit electric field
(Alternatively , Reciprocal of resistivity)
SI Unit : ohm
-1
m
-1
( any other correct SI unit)
½
½ 1
2.
Modulation index =




=


0.4
½
½ 1
3.
1 1
4. 20cm 1 1
5. 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
6.


	 	


Perpendicular to the plane formed by 	 	

/

		 	




[Note: Give full credit for writing the expression.]
½
½ 1
7. X: Channel
It connects the Transmitter to the Receiver
½
½ 1
8. Glass.
In glass there is no effect of electromagnetic induction, due to presence of Earthâ€™s
magnetic field, unlike in the case of metallic ball.
½
½ 1
9.
(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
Effect on glow of bulb in Part (i) 1
Part (ii) 1
Page 2

Delhi Set II  FINAL print Draft  Page No. 1  12th March, 2014      3:20 pm
MARKING SCHEME
SET 55/1/2 (DELHI)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. Electrical conductivity is defined as current density per unit electric field
(Alternatively , Reciprocal of resistivity)
SI Unit : ohm
-1
m
-1
( any other correct SI unit)
½
½ 1
2.
Modulation index =




=


0.4
½
½ 1
3.
1 1
4. 20cm 1 1
5. 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
6.


	 	


Perpendicular to the plane formed by 	 	

/

		 	




[Note: Give full credit for writing the expression.]
½
½ 1
7. X: Channel
It connects the Transmitter to the Receiver
½
½ 1
8. Glass.
In glass there is no effect of electromagnetic induction, due to presence of Earthâ€™s
magnetic field, unlike in the case of metallic ball.
½
½ 1
9.
(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
Effect on glow of bulb in Part (i) 1
Part (ii) 1
Delhi Set II  FINAL print Draft  Page No. 2  12th March, 2014      3:20 pm
10.
tpEsin?
(i) Shunt
8v3 = p	E	sin60
#
=\$%	×
v&

=> pE = 16
Potential energy, U = -pE cos '
= -16 x cos 60
0
= -8J
½
½
½
½ 2
11.
A: Paramagnetic
B: Diamagnetic
Susceptibility
For A:  positive
For B:  negative
½
½
½
½ 2
12.
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
13.
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 (.
1
1
Calculation of Potential energy of the dipole  2
Identification of magnetic material ½ + ½
Susceptibility       ½ + ½
Underlying principle 1
Brief working  1
Explanation of the given statement 1 + 1
Page 3

Delhi Set II  FINAL print Draft  Page No. 1  12th March, 2014      3:20 pm
MARKING SCHEME
SET 55/1/2 (DELHI)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. Electrical conductivity is defined as current density per unit electric field
(Alternatively , Reciprocal of resistivity)
SI Unit : ohm
-1
m
-1
( any other correct SI unit)
½
½ 1
2.
Modulation index =




=


0.4
½
½ 1
3.
1 1
4. 20cm 1 1
5. 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
6.


	 	


Perpendicular to the plane formed by 	 	

/

		 	




[Note: Give full credit for writing the expression.]
½
½ 1
7. X: Channel
It connects the Transmitter to the Receiver
½
½ 1
8. Glass.
In glass there is no effect of electromagnetic induction, due to presence of Earthâ€™s
magnetic field, unlike in the case of metallic ball.
½
½ 1
9.
(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
Effect on glow of bulb in Part (i) 1
Part (ii) 1
Delhi Set II  FINAL print Draft  Page No. 2  12th March, 2014      3:20 pm
10.
tpEsin?
(i) Shunt
8v3 = p	E	sin60
#
=\$%	×
v&

=> pE = 16
Potential energy, U = -pE cos '
= -16 x cos 60
0
= -8J
½
½
½
½ 2
11.
A: Paramagnetic
B: Diamagnetic
Susceptibility
For A:  positive
For B:  negative
½
½
½
½ 2
12.
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
13.
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 (.
1
1
Calculation of Potential energy of the dipole  2
Identification of magnetic material ½ + ½
Susceptibility       ½ + ½
Underlying principle 1
Brief working  1
Explanation of the given statement 1 + 1
Delhi Set II  FINAL print Draft  Page No. 3  12th March, 2014      3:20 pm
(Alternatively, if a student writes the conditions given below, give full credit.)
Condition for first minimum in single slit diffraction  is , '	Ëœ ? / a,
Whereas in case of two narrow slits separated by distance a, first maximum occurs
at angle 	'	Ëœ ? / a
[Note: Award 1 mark even if the candidate attempts this question partly.]
2
14.
Truth Table
Input Output
A B Yâ€™ 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
Truth Table  1
Names of gates used ½ + ½
Identification 1
Truth Table 1
Page 4

Delhi Set II  FINAL print Draft  Page No. 1  12th March, 2014      3:20 pm
MARKING SCHEME
SET 55/1/2 (DELHI)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. Electrical conductivity is defined as current density per unit electric field
(Alternatively , Reciprocal of resistivity)
SI Unit : ohm
-1
m
-1
( any other correct SI unit)
½
½ 1
2.
Modulation index =




=


0.4
½
½ 1
3.
1 1
4. 20cm 1 1
5. 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
6.


	 	


Perpendicular to the plane formed by 	 	

/

		 	




[Note: Give full credit for writing the expression.]
½
½ 1
7. X: Channel
It connects the Transmitter to the Receiver
½
½ 1
8. Glass.
In glass there is no effect of electromagnetic induction, due to presence of Earthâ€™s
magnetic field, unlike in the case of metallic ball.
½
½ 1
9.
(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
Effect on glow of bulb in Part (i) 1
Part (ii) 1
Delhi Set II  FINAL print Draft  Page No. 2  12th March, 2014      3:20 pm
10.
tpEsin?
(i) Shunt
8v3 = p	E	sin60
#
=\$%	×
v&

=> pE = 16
Potential energy, U = -pE cos '
= -16 x cos 60
0
= -8J
½
½
½
½ 2
11.
A: Paramagnetic
B: Diamagnetic
Susceptibility
For A:  positive
For B:  negative
½
½
½
½ 2
12.
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
13.
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 (.
1
1
Calculation of Potential energy of the dipole  2
Identification of magnetic material ½ + ½
Susceptibility       ½ + ½
Underlying principle 1
Brief working  1
Explanation of the given statement 1 + 1
Delhi Set II  FINAL print Draft  Page No. 3  12th March, 2014      3:20 pm
(Alternatively, if a student writes the conditions given below, give full credit.)
Condition for first minimum in single slit diffraction  is , '	Ëœ ? / a,
Whereas in case of two narrow slits separated by distance a, first maximum occurs
at angle 	'	Ëœ ? / a
[Note: Award 1 mark even if the candidate attempts this question partly.]
2
14.
Truth Table
Input Output
A B Yâ€™ 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
Truth Table  1
Names of gates used ½ + ½
Identification 1
Truth Table 1
Delhi Set II  FINAL print Draft  Page No. 4  12th March, 2014      3:20 pm
15.
(a) Proton
?=
*
+,-.
as mass of proton < mass of / particle and
?
= 2q
p
=>
(
1
> (
?

for the same accelerating potential.
(b) Alpha particle
K.E. = qV
We have
12

3
? (For same accelerating potential)Kinetic energy of proton < KE of / particle
½
½
½
½
2
16.
5 	%6	cos'
= 210
&
×4	×10
;
cos0
#

=80	<=
;>
?


5 =2×10
&
×4×10
;
cos60
#

= 40N=
;>
?

½
½
½
½ 2
17.
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
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
½ 2
18.
Part (a) and its reason ½ + ½
Part (b) and its reason ½ + ½
Finding flux in the two cases 1+1
Statements of two Laws ½ + ½
Justification  ½ + ½
Page 5

Delhi Set II  FINAL print Draft  Page No. 1  12th March, 2014      3:20 pm
MARKING SCHEME
SET 55/1/2 (DELHI)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. Electrical conductivity is defined as current density per unit electric field
(Alternatively , Reciprocal of resistivity)
SI Unit : ohm
-1
m
-1
( any other correct SI unit)
½
½ 1
2.
Modulation index =




=


0.4
½
½ 1
3.
1 1
4. 20cm 1 1
5. 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
6.


	 	


Perpendicular to the plane formed by 	 	

/

		 	




[Note: Give full credit for writing the expression.]
½
½ 1
7. X: Channel
It connects the Transmitter to the Receiver
½
½ 1
8. Glass.
In glass there is no effect of electromagnetic induction, due to presence of Earthâ€™s
magnetic field, unlike in the case of metallic ball.
½
½ 1
9.
(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
Effect on glow of bulb in Part (i) 1
Part (ii) 1
Delhi Set II  FINAL print Draft  Page No. 2  12th March, 2014      3:20 pm
10.
tpEsin?
(i) Shunt
8v3 = p	E	sin60
#
=\$%	×
v&

=> pE = 16
Potential energy, U = -pE cos '
= -16 x cos 60
0
= -8J
½
½
½
½ 2
11.
A: Paramagnetic
B: Diamagnetic
Susceptibility
For A:  positive
For B:  negative
½
½
½
½ 2
12.
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
13.
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 (.
1
1
Calculation of Potential energy of the dipole  2
Identification of magnetic material ½ + ½
Susceptibility       ½ + ½
Underlying principle 1
Brief working  1
Explanation of the given statement 1 + 1
Delhi Set II  FINAL print Draft  Page No. 3  12th March, 2014      3:20 pm
(Alternatively, if a student writes the conditions given below, give full credit.)
Condition for first minimum in single slit diffraction  is , '	Ëœ ? / a,
Whereas in case of two narrow slits separated by distance a, first maximum occurs
at angle 	'	Ëœ ? / a
[Note: Award 1 mark even if the candidate attempts this question partly.]
2
14.
Truth Table
Input Output
A B Yâ€™ 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
Truth Table  1
Names of gates used ½ + ½
Identification 1
Truth Table 1
Delhi Set II  FINAL print Draft  Page No. 4  12th March, 2014      3:20 pm
15.
(a) Proton
?=
*
+,-.
as mass of proton < mass of / particle and
?
= 2q
p
=>
(
1
> (
?

for the same accelerating potential.
(b) Alpha particle
K.E. = qV
We have
12

3
? (For same accelerating potential)Kinetic energy of proton < KE of / particle
½
½
½
½
2
16.
5 	%6	cos'
= 210
&
×4	×10
;
cos0
#

=80	<=
;>
?


5 =2×10
&
×4×10
;
cos60
#

= 40N=
;>
?

½
½
½
½ 2
17.
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
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
½ 2
18.
Part (a) and its reason ½ + ½
Part (b) and its reason ½ + ½
Finding flux in the two cases 1+1
Statements of two Laws ½ + ½
Justification  ½ + ½
Delhi Set II     FINAL print Draft  Page No. 5  12th March, 2014      3:20 pm
(a) Power =	B, where n = no. of photons per second
2.0 x 10
-3
= 		6.6	10
;&C
610
>C

n  =
.#	D	>#
EF

G.G	>#
EFH
G>#
IH
= 0.050 x 10
17
= 5x 10
15
photons / second
[Note: Even if the student doesnâ€™t write the formula but calculates correctly, give
full credit to this part]
(b)
½
½
1
2
19.
(a) Statement of law
Expression of the law in integral form:
?

.K

= L
#
M
(Award 1 mark if the student just writes the integral form of Ampereâ€™s circuital
law)
(b)   B = µ
O
n I
Magnitude of net magnetic field inside  the combined system on the axis ,
1
½
½
(a) Estimation of no. of photons per second 1
(b) Plot showing the variation 1
(a) Statement of Ampereâ€™s circuital Law 1 ½
(b) Calculation of net magnetic field
(i) inside and (ii) outside 1 ½
```
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