Physics Past Year Paper Anskey (Delhi Set -1) - 2014, Class 12, CBSE Class 12 Notes | EduRev

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Class 12 : Physics Past Year Paper Anskey (Delhi Set -1) - 2014, Class 12, CBSE Class 12 Notes | EduRev

 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
3456
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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