Past Year Paper - Solutions, Physics (Set - 2), Delhi, 2014, Class 12, Physics | Additional Study Material for NEET PDF Download

 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 Earth’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 Earth’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. 
tpEsin? 
 
(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 Earth’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. 
tpEsin? 
 
(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 , '	˜ ? / 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.] 
 
 
 
 
 
 
 
 
2 
 
14.  
 
 
  
 
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 
 
    
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 Earth’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. 
tpEsin? 
 
(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 , '	˜ ? / 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.] 
 
 
 
 
 
 
 
 
2 
 
14.  
 
 
  
 
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 
 
    
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'	  
 
     = 210
&		
×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 Earth’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. 
tpEsin? 
 
(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 , '	˜ ? / 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.] 
 
 
 
 
 
 
 
 
2 
 
14.  
 
 
  
 
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 
 
    
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'	  
 
     = 210
&		
×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
610
>C
 
        
                  n  =  
.#	D	>#
EF
	
G.G	>#
EFH
G>#
IH
  
 
                          = 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 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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 Ampere’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 Ampere’s circuital Law    1 ½  
(b) Calculation of net magnetic field 
(i) inside and (ii) outside     1 ½