Physics Past Year Paper Anskey (Delhi Set -3) - 2014, Class 12, CBSE Class 12 Notes | EduRev

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Class 12 : Physics Past Year Paper Anskey (Delhi Set -3) - 2014, Class 12, CBSE Class 12 Notes | EduRev

 Page 1


MARKING SCHEME 
SET 55/1/3 (DELHI) 
Q.No. Expected Answer/Value Points Marks Total 
Marks 
1.
Drift velocity , is the directed velocity acquired by the electrons for a unit value
of the applied Electric field.
Current, 
½ 
½ 
1 
2.
Modulation index =
 = = 0.5 
½ 
½ 
1 
3. 
Perpendicular to the plane formed by  /
[Note: Give full credit for writing the expression only.] 
½ 
½ 1 
4. 
1 1 
5. 20cm 1 1 
6. X: Channel
It connects Transmitter and Receiver
½ 
½ 1 
7. Glass.
In glass there is no electromagnetic induction, due to presence of Earth’s magnetic
field, unlike in the case of metallic ball.
½ 
½ 1 
8. 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 
9. 
 = 
=>  = 12 
Potential energy 
½ 
½ 
Calculation of Potential energy of the dipole  2 
Page 2


MARKING SCHEME 
SET 55/1/3 (DELHI) 
Q.No. Expected Answer/Value Points Marks Total 
Marks 
1.
Drift velocity , is the directed velocity acquired by the electrons for a unit value
of the applied Electric field.
Current, 
½ 
½ 
1 
2.
Modulation index =
 = = 0.5 
½ 
½ 
1 
3. 
Perpendicular to the plane formed by  /
[Note: Give full credit for writing the expression only.] 
½ 
½ 1 
4. 
1 1 
5. 20cm 1 1 
6. X: Channel
It connects Transmitter and Receiver
½ 
½ 1 
7. Glass.
In glass there is no electromagnetic induction, due to presence of Earth’s magnetic
field, unlike in the case of metallic ball.
½ 
½ 1 
8. 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 
9. 
 = 
=>  = 12 
Potential energy 
½ 
½ 
Calculation of Potential energy of the dipole  2 
 Delhi  Set III 
U = -pE cos 
   = -12 x cos 60
0
 = -6J 
½ 
½ 2 
10. 
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 
11. 
(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 
12. 
(a) deutron 
?=
= = 
=> 
?
d
> ?
a
 
(b) deuteron 
 KE= qV 
 , 
for the same accelerating potential, we have 
½ 
½ 
½ 
½ 2 
Part (a) and its reason ½ + ½ 
Part (b) and its reason ½ + ½ 
Underlying principle 1 
Brief working  1 
Effect on glow of bulb in Part (i) 1 
Part (ii) 1 
Page 3


MARKING SCHEME 
SET 55/1/3 (DELHI) 
Q.No. Expected Answer/Value Points Marks Total 
Marks 
1.
Drift velocity , is the directed velocity acquired by the electrons for a unit value
of the applied Electric field.
Current, 
½ 
½ 
1 
2.
Modulation index =
 = = 0.5 
½ 
½ 
1 
3. 
Perpendicular to the plane formed by  /
[Note: Give full credit for writing the expression only.] 
½ 
½ 1 
4. 
1 1 
5. 20cm 1 1 
6. X: Channel
It connects Transmitter and Receiver
½ 
½ 1 
7. Glass.
In glass there is no electromagnetic induction, due to presence of Earth’s magnetic
field, unlike in the case of metallic ball.
½ 
½ 1 
8. 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 
9. 
 = 
=>  = 12 
Potential energy 
½ 
½ 
Calculation of Potential energy of the dipole  2 
 Delhi  Set III 
U = -pE cos 
   = -12 x cos 60
0
 = -6J 
½ 
½ 2 
10. 
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 
11. 
(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 
12. 
(a) deutron 
?=
= = 
=> 
?
d
> ?
a
 
(b) deuteron 
 KE= qV 
 , 
for the same accelerating potential, we have 
½ 
½ 
½ 
½ 2 
Part (a) and its reason ½ + ½ 
Part (b) and its reason ½ + ½ 
Underlying principle 1 
Brief working  1 
Effect on glow of bulb in Part (i) 1 
Part (ii) 1 
 Delhi  Set III 
13. 
A: Paramagnetic 
B: Diamagnetic 
Susceptibility 
For A:  positive 
For B:  negative 
½ 
½ 
½ 
½ 2 
14. 
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, ?  =0  , where  is the changes in potential 
Justification : Conservation of energy 
½ 
½ 
½ 
½ 2 
15. 
½ 
½ 
½ 
½ 2 
16. 
(a) power = , where n = no. of photons per second 
0909 photons/second 
 =9.1 photons/second 
½ 
½ 
(a) Estimation of number of photons per second 1 
(b) plot showing the variation  1 
Finding flux in two cases 1+1 
Identification of magnetic material ½ + ½ 
Susceptibility       ½ + ½ 
Statements of two Laws ½ + ½ 
Justification  ½ + ½ 
Page 4


MARKING SCHEME 
SET 55/1/3 (DELHI) 
Q.No. Expected Answer/Value Points Marks Total 
Marks 
1.
Drift velocity , is the directed velocity acquired by the electrons for a unit value
of the applied Electric field.
Current, 
½ 
½ 
1 
2.
Modulation index =
 = = 0.5 
½ 
½ 
1 
3. 
Perpendicular to the plane formed by  /
[Note: Give full credit for writing the expression only.] 
½ 
½ 1 
4. 
1 1 
5. 20cm 1 1 
6. X: Channel
It connects Transmitter and Receiver
½ 
½ 1 
7. Glass.
In glass there is no electromagnetic induction, due to presence of Earth’s magnetic
field, unlike in the case of metallic ball.
½ 
½ 1 
8. 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 
9. 
 = 
=>  = 12 
Potential energy 
½ 
½ 
Calculation of Potential energy of the dipole  2 
 Delhi  Set III 
U = -pE cos 
   = -12 x cos 60
0
 = -6J 
½ 
½ 2 
10. 
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 
11. 
(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 
12. 
(a) deutron 
?=
= = 
=> 
?
d
> ?
a
 
(b) deuteron 
 KE= qV 
 , 
for the same accelerating potential, we have 
½ 
½ 
½ 
½ 2 
Part (a) and its reason ½ + ½ 
Part (b) and its reason ½ + ½ 
Underlying principle 1 
Brief working  1 
Effect on glow of bulb in Part (i) 1 
Part (ii) 1 
 Delhi  Set III 
13. 
A: Paramagnetic 
B: Diamagnetic 
Susceptibility 
For A:  positive 
For B:  negative 
½ 
½ 
½ 
½ 2 
14. 
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, ?  =0  , where  is the changes in potential 
Justification : Conservation of energy 
½ 
½ 
½ 
½ 2 
15. 
½ 
½ 
½ 
½ 2 
16. 
(a) power = , where n = no. of photons per second 
0909 photons/second 
 =9.1 photons/second 
½ 
½ 
(a) Estimation of number of photons per second 1 
(b) plot showing the variation  1 
Finding flux in two cases 1+1 
Identification of magnetic material ½ + ½ 
Susceptibility       ½ + ½ 
Statements of two Laws ½ + ½ 
Justification  ½ + ½ 
 Delhi  Set III 
[Note: Even if the student doesn’t write the formula and calculates correctly, give 
full credit to this part] 
(b) 
1 2 
17. 
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 5


MARKING SCHEME 
SET 55/1/3 (DELHI) 
Q.No. Expected Answer/Value Points Marks Total 
Marks 
1.
Drift velocity , is the directed velocity acquired by the electrons for a unit value
of the applied Electric field.
Current, 
½ 
½ 
1 
2.
Modulation index =
 = = 0.5 
½ 
½ 
1 
3. 
Perpendicular to the plane formed by  /
[Note: Give full credit for writing the expression only.] 
½ 
½ 1 
4. 
1 1 
5. 20cm 1 1 
6. X: Channel
It connects Transmitter and Receiver
½ 
½ 1 
7. Glass.
In glass there is no electromagnetic induction, due to presence of Earth’s magnetic
field, unlike in the case of metallic ball.
½ 
½ 1 
8. 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 
9. 
 = 
=>  = 12 
Potential energy 
½ 
½ 
Calculation of Potential energy of the dipole  2 
 Delhi  Set III 
U = -pE cos 
   = -12 x cos 60
0
 = -6J 
½ 
½ 2 
10. 
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 
11. 
(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 
12. 
(a) deutron 
?=
= = 
=> 
?
d
> ?
a
 
(b) deuteron 
 KE= qV 
 , 
for the same accelerating potential, we have 
½ 
½ 
½ 
½ 2 
Part (a) and its reason ½ + ½ 
Part (b) and its reason ½ + ½ 
Underlying principle 1 
Brief working  1 
Effect on glow of bulb in Part (i) 1 
Part (ii) 1 
 Delhi  Set III 
13. 
A: Paramagnetic 
B: Diamagnetic 
Susceptibility 
For A:  positive 
For B:  negative 
½ 
½ 
½ 
½ 2 
14. 
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, ?  =0  , where  is the changes in potential 
Justification : Conservation of energy 
½ 
½ 
½ 
½ 2 
15. 
½ 
½ 
½ 
½ 2 
16. 
(a) power = , where n = no. of photons per second 
0909 photons/second 
 =9.1 photons/second 
½ 
½ 
(a) Estimation of number of photons per second 1 
(b) plot showing the variation  1 
Finding flux in two cases 1+1 
Identification of magnetic material ½ + ½ 
Susceptibility       ½ + ½ 
Statements of two Laws ½ + ½ 
Justification  ½ + ½ 
 Delhi  Set III 
[Note: Even if the student doesn’t write the formula and calculates correctly, give 
full credit to this part] 
(b) 
1 2 
17. 
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 III 
18. 
 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 
19. 
(a) Statement of law 
Expression of the law in integral form: 
 = 
(Award 1 mark if the student just writes the integral form of Ampere’s circuital 
law) 
(b)   B =  I 
Magnitude of net magnetic field inside  the combined system on the axis , 
 B =B
1 
- B
2
 
 I 
Also accept if the student writes B =  I 
(iii)Outside the combined system, the net magnetic field is zero. 
1 
½ 
½ 
½ 
½ 3 
20. 
Energy required to excite hydrogen atoms from ground state to the second  state 
= E
final
 – E
initial
 
 = -1.51 – (-13.6)eV = 12.09  eV 
½ 
Finding maximum energy level of hydrogen atoms ½ 
Calculation of wavelengths  2 ½ 
Explanation of the given statement 1 + 1 
(a) Statement of Ampere’s circuital Law 1 ½ 
(b) Calculation of net magnetic field 
(i) inside and (ii) outside 1 ½ 
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