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Physics Past Year Paper AnsKey (Delhi All Set) - 2016, Class 12, CBSE Class 12 Notes | EduRev

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Class 12 : Physics Past Year Paper AnsKey (Delhi All Set) - 2016, Class 12, CBSE Class 12 Notes | EduRev

``` Page 1

SET 55/1/1/D
Page 1 of 19  Final Draft  11/03/16 11:30a.m.
MARKING SCHEME
Q. No. Expected Answer / Value Points Marks Total
Marks
Set1,Q1
Set2,Q4
Set3,Q2
SECTION (A)
Positive 1 1
Set1,Q2
Set2,Q5
Set3,Q3
Electric flux remains unaffected.
[NOTE: (As per the Hindi translation), change in Electric field is being
asked, hence give credit if student writes answer as decreases] 1 1
Set1,Q3
Set2,Q1
Set3,Q5
A current carrying coil, in the presence of magnetic field, experiences a
torque, which produces proportionate deflection.
[Alternatively
( deflection)  ? a t  ( Torque)]
1 1
Set1,Q4
Set2,Q2
Set3,Q4
Due to their short wavelengths, (they are suitable for radar system used in
Set1,Q5
Set2,Q3
Set3,Q1
Quality factor   Q = ,
[Alternatively
Quality factor   Q = , Alternatively, It gives the sharpness of the
resonance circuit.]
It has no unit.
½
½ 1
Set1,Q6
Set2,Q9
Set3,Q7
SECTION (B)
(i) The loss of strength of a signal while propagating through a medium.
(ii) The process of retrieval of information, from the carrier wave, at the
1
1 2
Set1,Q7
Set2,Q10
Set3,Q8
Explanation of the terms
(i) Attenuation      1
(ii) Demodulation  1
Plotting of  graph ½ + ½
Identification of line representing lower mass ½
Reason  ½
Page 2

SET 55/1/1/D
Page 1 of 19  Final Draft  11/03/16 11:30a.m.
MARKING SCHEME
Q. No. Expected Answer / Value Points Marks Total
Marks
Set1,Q1
Set2,Q4
Set3,Q2
SECTION (A)
Positive 1 1
Set1,Q2
Set2,Q5
Set3,Q3
Electric flux remains unaffected.
[NOTE: (As per the Hindi translation), change in Electric field is being
asked, hence give credit if student writes answer as decreases] 1 1
Set1,Q3
Set2,Q1
Set3,Q5
A current carrying coil, in the presence of magnetic field, experiences a
torque, which produces proportionate deflection.
[Alternatively
( deflection)  ? a t  ( Torque)]
1 1
Set1,Q4
Set2,Q2
Set3,Q4
Due to their short wavelengths, (they are suitable for radar system used in
Set1,Q5
Set2,Q3
Set3,Q1
Quality factor   Q = ,
[Alternatively
Quality factor   Q = , Alternatively, It gives the sharpness of the
resonance circuit.]
It has no unit.
½
½ 1
Set1,Q6
Set2,Q9
Set3,Q7
SECTION (B)
(i) The loss of strength of a signal while propagating through a medium.
(ii) The process of retrieval of information, from the carrier wave, at the
1
1 2
Set1,Q7
Set2,Q10
Set3,Q8
Explanation of the terms
(i) Attenuation      1
(ii) Demodulation  1
Plotting of  graph ½ + ½
Identification of line representing lower mass ½
Reason  ½
SET 55/1/1/D
Page 2 of 19  Final Draft  11/03/16 11:30a.m.
As ? =
As the charge of two particles is same , therefore
i.e.    Slope a
Hence, particle with lower mass ( ) will have greater slope.
½ + ½
½
½ 2
Set1,Q8
Set2,Q6
Set3,Q10
Binding energy of nucleus with mass number 240,
= 240 × 7.6 MeV
Binding energy of two fragments
= 2 × 120 × 8.5 MeV
Energy released = 240 (8.5 â€“ 7.6) MeV
= 240 × 0.9
= 216 MeV
OR
Total Binding energy of Initial System
i.e.  = (2.23 + 2.23) MeV
= 4.46 MeV
Binding energy of Final System i.e.
= 7.73 MeV
Hence energy released = 7.73 MeV- 4.46 MeV
= 3.27 MeV
½
½
½
½
½
½
1
2
2
Calculation of Energy released  2
Calculation of Energy in the fusion Reaction  2
Page 3

SET 55/1/1/D
Page 1 of 19  Final Draft  11/03/16 11:30a.m.
MARKING SCHEME
Q. No. Expected Answer / Value Points Marks Total
Marks
Set1,Q1
Set2,Q4
Set3,Q2
SECTION (A)
Positive 1 1
Set1,Q2
Set2,Q5
Set3,Q3
Electric flux remains unaffected.
[NOTE: (As per the Hindi translation), change in Electric field is being
asked, hence give credit if student writes answer as decreases] 1 1
Set1,Q3
Set2,Q1
Set3,Q5
A current carrying coil, in the presence of magnetic field, experiences a
torque, which produces proportionate deflection.
[Alternatively
( deflection)  ? a t  ( Torque)]
1 1
Set1,Q4
Set2,Q2
Set3,Q4
Due to their short wavelengths, (they are suitable for radar system used in
Set1,Q5
Set2,Q3
Set3,Q1
Quality factor   Q = ,
[Alternatively
Quality factor   Q = , Alternatively, It gives the sharpness of the
resonance circuit.]
It has no unit.
½
½ 1
Set1,Q6
Set2,Q9
Set3,Q7
SECTION (B)
(i) The loss of strength of a signal while propagating through a medium.
(ii) The process of retrieval of information, from the carrier wave, at the
1
1 2
Set1,Q7
Set2,Q10
Set3,Q8
Explanation of the terms
(i) Attenuation      1
(ii) Demodulation  1
Plotting of  graph ½ + ½
Identification of line representing lower mass ½
Reason  ½
SET 55/1/1/D
Page 2 of 19  Final Draft  11/03/16 11:30a.m.
As ? =
As the charge of two particles is same , therefore
i.e.    Slope a
Hence, particle with lower mass ( ) will have greater slope.
½ + ½
½
½ 2
Set1,Q8
Set2,Q6
Set3,Q10
Binding energy of nucleus with mass number 240,
= 240 × 7.6 MeV
Binding energy of two fragments
= 2 × 120 × 8.5 MeV
Energy released = 240 (8.5 â€“ 7.6) MeV
= 240 × 0.9
= 216 MeV
OR
Total Binding energy of Initial System
i.e.  = (2.23 + 2.23) MeV
= 4.46 MeV
Binding energy of Final System i.e.
= 7.73 MeV
Hence energy released = 7.73 MeV- 4.46 MeV
= 3.27 MeV
½
½
½
½
½
½
1
2
2
Calculation of Energy released  2
Calculation of Energy in the fusion Reaction  2
SET 55/1/1/D
Page 3 of 19  Final Draft  11/03/16 11:30a.m.
Set1,Q9
Set2,Q7
Set3,Q9
=  V
=
=  ?
=  ?
= 0.12 ?
½
½
½
½ 2
Set1,Q10
Set2,Q8
Set3,Q6
When unpolarised light is incident on the surface separating two media, the
reflected light gets (completely) polarized only when the reflected light and
refracted light become perpendicular to each other.
[ Alternatively
If the student draws the diagram, as shown, and
writes  as the polarizing angle, award this 1 mark.
If the student just writes award half mark
only.]
The refractive index of denser medium, with respect to rarer medium, is
given by
Since Refractive index (µ) of a transparent medium is different for different
colours, hence Brewster angle is different for different colours.
1
½
½ 2
Calculation of emf 1
Calculation of internal resistance  1
Statement of Brewsterâ€™s Law  1
Reason of different value      1
Page 4

SET 55/1/1/D
Page 1 of 19  Final Draft  11/03/16 11:30a.m.
MARKING SCHEME
Q. No. Expected Answer / Value Points Marks Total
Marks
Set1,Q1
Set2,Q4
Set3,Q2
SECTION (A)
Positive 1 1
Set1,Q2
Set2,Q5
Set3,Q3
Electric flux remains unaffected.
[NOTE: (As per the Hindi translation), change in Electric field is being
asked, hence give credit if student writes answer as decreases] 1 1
Set1,Q3
Set2,Q1
Set3,Q5
A current carrying coil, in the presence of magnetic field, experiences a
torque, which produces proportionate deflection.
[Alternatively
( deflection)  ? a t  ( Torque)]
1 1
Set1,Q4
Set2,Q2
Set3,Q4
Due to their short wavelengths, (they are suitable for radar system used in
Set1,Q5
Set2,Q3
Set3,Q1
Quality factor   Q = ,
[Alternatively
Quality factor   Q = , Alternatively, It gives the sharpness of the
resonance circuit.]
It has no unit.
½
½ 1
Set1,Q6
Set2,Q9
Set3,Q7
SECTION (B)
(i) The loss of strength of a signal while propagating through a medium.
(ii) The process of retrieval of information, from the carrier wave, at the
1
1 2
Set1,Q7
Set2,Q10
Set3,Q8
Explanation of the terms
(i) Attenuation      1
(ii) Demodulation  1
Plotting of  graph ½ + ½
Identification of line representing lower mass ½
Reason  ½
SET 55/1/1/D
Page 2 of 19  Final Draft  11/03/16 11:30a.m.
As ? =
As the charge of two particles is same , therefore
i.e.    Slope a
Hence, particle with lower mass ( ) will have greater slope.
½ + ½
½
½ 2
Set1,Q8
Set2,Q6
Set3,Q10
Binding energy of nucleus with mass number 240,
= 240 × 7.6 MeV
Binding energy of two fragments
= 2 × 120 × 8.5 MeV
Energy released = 240 (8.5 â€“ 7.6) MeV
= 240 × 0.9
= 216 MeV
OR
Total Binding energy of Initial System
i.e.  = (2.23 + 2.23) MeV
= 4.46 MeV
Binding energy of Final System i.e.
= 7.73 MeV
Hence energy released = 7.73 MeV- 4.46 MeV
= 3.27 MeV
½
½
½
½
½
½
1
2
2
Calculation of Energy released  2
Calculation of Energy in the fusion Reaction  2
SET 55/1/1/D
Page 3 of 19  Final Draft  11/03/16 11:30a.m.
Set1,Q9
Set2,Q7
Set3,Q9
=  V
=
=  ?
=  ?
= 0.12 ?
½
½
½
½ 2
Set1,Q10
Set2,Q8
Set3,Q6
When unpolarised light is incident on the surface separating two media, the
reflected light gets (completely) polarized only when the reflected light and
refracted light become perpendicular to each other.
[ Alternatively
If the student draws the diagram, as shown, and
writes  as the polarizing angle, award this 1 mark.
If the student just writes award half mark
only.]
The refractive index of denser medium, with respect to rarer medium, is
given by
Since Refractive index (µ) of a transparent medium is different for different
colours, hence Brewster angle is different for different colours.
1
½
½ 2
Calculation of emf 1
Calculation of internal resistance  1
Statement of Brewsterâ€™s Law  1
Reason of different value      1
SET 55/1/1/D
Page 4 of 19  Final Draft  11/03/16 11:30a.m.
Set1,Q11
Set2,Q14
Set3,Q12
SECTION (C)

Net Electric Field at point P =
dE = Electric field due to a small element having charge dq
=
Let  = Linear charge density
=
=
Hence E =  , where
=
=  , where total charge Q = ×2
At large distance i.e. x>>a
E
This is the Electric field due to a point charge at distance x.
(NOTE: Award two marks for this question, if a student attempts this
question but does not give the complete answer)
½
½
½
½
½
½
3
Set1,Q12
Set2,Q15
Set3,Q13

The three characteristic features which canâ€™t be explained by wave theory
are:
i. Kinetic energy of emitted electrons are found to be independent of
intensity of incident light.
1
Obtaining an expression for Electric field intensity  2
Showing behavior at large distance  1
Three Characteristic features 1+1+1
Page 5

SET 55/1/1/D
Page 1 of 19  Final Draft  11/03/16 11:30a.m.
MARKING SCHEME
Q. No. Expected Answer / Value Points Marks Total
Marks
Set1,Q1
Set2,Q4
Set3,Q2
SECTION (A)
Positive 1 1
Set1,Q2
Set2,Q5
Set3,Q3
Electric flux remains unaffected.
[NOTE: (As per the Hindi translation), change in Electric field is being
asked, hence give credit if student writes answer as decreases] 1 1
Set1,Q3
Set2,Q1
Set3,Q5
A current carrying coil, in the presence of magnetic field, experiences a
torque, which produces proportionate deflection.
[Alternatively
( deflection)  ? a t  ( Torque)]
1 1
Set1,Q4
Set2,Q2
Set3,Q4
Due to their short wavelengths, (they are suitable for radar system used in
Set1,Q5
Set2,Q3
Set3,Q1
Quality factor   Q = ,
[Alternatively
Quality factor   Q = , Alternatively, It gives the sharpness of the
resonance circuit.]
It has no unit.
½
½ 1
Set1,Q6
Set2,Q9
Set3,Q7
SECTION (B)
(i) The loss of strength of a signal while propagating through a medium.
(ii) The process of retrieval of information, from the carrier wave, at the
1
1 2
Set1,Q7
Set2,Q10
Set3,Q8
Explanation of the terms
(i) Attenuation      1
(ii) Demodulation  1
Plotting of  graph ½ + ½
Identification of line representing lower mass ½
Reason  ½
SET 55/1/1/D
Page 2 of 19  Final Draft  11/03/16 11:30a.m.
As ? =
As the charge of two particles is same , therefore
i.e.    Slope a
Hence, particle with lower mass ( ) will have greater slope.
½ + ½
½
½ 2
Set1,Q8
Set2,Q6
Set3,Q10
Binding energy of nucleus with mass number 240,
= 240 × 7.6 MeV
Binding energy of two fragments
= 2 × 120 × 8.5 MeV
Energy released = 240 (8.5 â€“ 7.6) MeV
= 240 × 0.9
= 216 MeV
OR
Total Binding energy of Initial System
i.e.  = (2.23 + 2.23) MeV
= 4.46 MeV
Binding energy of Final System i.e.
= 7.73 MeV
Hence energy released = 7.73 MeV- 4.46 MeV
= 3.27 MeV
½
½
½
½
½
½
1
2
2
Calculation of Energy released  2
Calculation of Energy in the fusion Reaction  2
SET 55/1/1/D
Page 3 of 19  Final Draft  11/03/16 11:30a.m.
Set1,Q9
Set2,Q7
Set3,Q9
=  V
=
=  ?
=  ?
= 0.12 ?
½
½
½
½ 2
Set1,Q10
Set2,Q8
Set3,Q6
When unpolarised light is incident on the surface separating two media, the
reflected light gets (completely) polarized only when the reflected light and
refracted light become perpendicular to each other.
[ Alternatively
If the student draws the diagram, as shown, and
writes  as the polarizing angle, award this 1 mark.
If the student just writes award half mark
only.]
The refractive index of denser medium, with respect to rarer medium, is
given by
Since Refractive index (µ) of a transparent medium is different for different
colours, hence Brewster angle is different for different colours.
1
½
½ 2
Calculation of emf 1
Calculation of internal resistance  1
Statement of Brewsterâ€™s Law  1
Reason of different value      1
SET 55/1/1/D
Page 4 of 19  Final Draft  11/03/16 11:30a.m.
Set1,Q11
Set2,Q14
Set3,Q12
SECTION (C)

Net Electric Field at point P =
dE = Electric field due to a small element having charge dq
=
Let  = Linear charge density
=
=
Hence E =  , where
=
=  , where total charge Q = ×2
At large distance i.e. x>>a
E
This is the Electric field due to a point charge at distance x.
(NOTE: Award two marks for this question, if a student attempts this
question but does not give the complete answer)
½
½
½
½
½
½
3
Set1,Q12
Set2,Q15
Set3,Q13

The three characteristic features which canâ€™t be explained by wave theory
are:
i. Kinetic energy of emitted electrons are found to be independent of
intensity of incident light.
1
Obtaining an expression for Electric field intensity  2
Showing behavior at large distance  1
Three Characteristic features 1+1+1
SET 55/1/1/D
Page 5 of 19  Final Draft  11/03/16 11:30a.m.
ii. Below a certain frequency (threshold) there is no photo-emission.
iii. Spontaneous emission of photo-electrons.
1
1 3
Set1,Q13
Set2,Q16
Set3,Q11
= q (
(Give Full credit of this part even if  a student writes:
F=  and
Force (F) acts perpendicular to the plane containing   and )
b)
Justification: Direction of force experienced by the particle will be according
to the Flemingâ€™s Left hand rule / (any other alternative correct rule.)
1
½ +
½+
½
½ 3
Set1,Q14
Set2,Q11
Set3,Q15
(i) Magnetic flux, linked with the secondary coil due to the unit current
flowing in the primary coil,
[Alternatively
Induced emf associated with the secondary coil, for a unit rate of
change of current in the primary coil. ]
[Also accept the Definition of Mutual Induction, as per the
Hindi translation of the question]
[i.e. the phenomenon of production of induced emf in one coil
due to change in current in neighbouring coil ]
1
a) Expression for the magnetic force 1
b) Trace of paths ½ + ½ + ½
Justification ½
(i) Definition of mutual inductance      1
(ii) Calculation of change of flux linkage  2
```
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