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 
aircraft navigation). 1 1 
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 
receiver.
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 
aircraft navigation). 1 1 
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 
receiver.
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 – 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 
aircraft navigation). 1 1 
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 
receiver.
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 – 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 Brewster’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 
aircraft navigation). 1 1 
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 
receiver.
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 – 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 Brewster’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 can’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 
aircraft navigation). 1 1 
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 
receiver.
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 – 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 Brewster’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 can’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 Fleming’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 ] 
(ii) Change of flux linkage 
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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