Physics Past Year Paper Outside Delhi Anskey (North Zone All Set) - 2016, Class 12 Class 12 Notes | EduRev

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Class 12 : Physics Past Year Paper Outside Delhi Anskey (North Zone All Set) - 2016, Class 12 Class 12 Notes | EduRev

 Page 1


SET 55/1/1/N 
 Page 1 of 22  Final Draft  11/03/16 2:00 p.m. 
MARKING SCHEME 
Q. No. Expected Answer / Value Points Marks Total 
Marks 
Set1,Q1 
Set2,Q2 
Set3,Q4 
SECTION A 
Zero / No work done / None 1 1 
Set1,Q2 
Set2,Q5 
Set3,Q3 
Drift velocity per unit field 
(directly proportional to relaxation time) 
½ 
½ 1 
Set1,Q3 
Set2,Q4 
Set3,Q2 
Charged particle moves inclined to the magnetic field 
(angle between  and  is neither nor 0) 
(component of , parallel to  , is not zero.) 
1 1 
Set1,Q4 
Set2,Q1 
Set3,Q5 
(some) light gets deviated / scattered / absorbed 
Scattering of light 
½ 
½ 1 
Set1,Q5 
Set2,Q3 
Set3,Q1 
 = 2005 kHz ; 1995 kHz 
(Give full 1 mark if the student straightaway writes the answer as 2005 kHz 
and 1995 kHz)  
½ 
 ½ 1 
Set1,Q6 
Set2,Q8 
Set3,Q7 
SECTION B 
 ; 
Alternatively, 
(Award this 1 mark even if the student writes the formula for  directly as 
such) 
½ 
½ 
½ 
½ 2 
Formulae:   1 
Substitution and calculation: 1 
Page 2


SET 55/1/1/N 
 Page 1 of 22  Final Draft  11/03/16 2:00 p.m. 
MARKING SCHEME 
Q. No. Expected Answer / Value Points Marks Total 
Marks 
Set1,Q1 
Set2,Q2 
Set3,Q4 
SECTION A 
Zero / No work done / None 1 1 
Set1,Q2 
Set2,Q5 
Set3,Q3 
Drift velocity per unit field 
(directly proportional to relaxation time) 
½ 
½ 1 
Set1,Q3 
Set2,Q4 
Set3,Q2 
Charged particle moves inclined to the magnetic field 
(angle between  and  is neither nor 0) 
(component of , parallel to  , is not zero.) 
1 1 
Set1,Q4 
Set2,Q1 
Set3,Q5 
(some) light gets deviated / scattered / absorbed 
Scattering of light 
½ 
½ 1 
Set1,Q5 
Set2,Q3 
Set3,Q1 
 = 2005 kHz ; 1995 kHz 
(Give full 1 mark if the student straightaway writes the answer as 2005 kHz 
and 1995 kHz)  
½ 
 ½ 1 
Set1,Q6 
Set2,Q8 
Set3,Q7 
SECTION B 
 ; 
Alternatively, 
(Award this 1 mark even if the student writes the formula for  directly as 
such) 
½ 
½ 
½ 
½ 2 
Formulae:   1 
Substitution and calculation: 1 
SET 55/1/1/N 
 Page 2 of 22  Final Draft  11/03/16 2:00 p.m. 
Set1,Q7 
Set2,Q10 
Set3,Q8 
(i) 
 is more for  particle, we have 
 (Also, accept if the student writes ) 
(ii) K.E. = q V 
 q is less for proton, we have 
 =2) 
½ 
½ 
½ 
½ 2 
Set1,Q8 
Set2,Q9 
Set3,Q6 
When the electron jumps from the orbit with n=3 to n=2 
(Longest wavelength of the Balmer series / First line of the Balmer series) 
 =  = 
 =
 Hz 
 Hz. 
(If the student just writes , award ½ mark) 
(Alternatively, 
= =
) 
1 
½ 
½ 
½ 
½ 
2 
Formulae ½ + ½ 
Conclusions in the two cases ½ + ½ 
Indicating the transition 1 
Calculation of frequency 1 
Page 3


SET 55/1/1/N 
 Page 1 of 22  Final Draft  11/03/16 2:00 p.m. 
MARKING SCHEME 
Q. No. Expected Answer / Value Points Marks Total 
Marks 
Set1,Q1 
Set2,Q2 
Set3,Q4 
SECTION A 
Zero / No work done / None 1 1 
Set1,Q2 
Set2,Q5 
Set3,Q3 
Drift velocity per unit field 
(directly proportional to relaxation time) 
½ 
½ 1 
Set1,Q3 
Set2,Q4 
Set3,Q2 
Charged particle moves inclined to the magnetic field 
(angle between  and  is neither nor 0) 
(component of , parallel to  , is not zero.) 
1 1 
Set1,Q4 
Set2,Q1 
Set3,Q5 
(some) light gets deviated / scattered / absorbed 
Scattering of light 
½ 
½ 1 
Set1,Q5 
Set2,Q3 
Set3,Q1 
 = 2005 kHz ; 1995 kHz 
(Give full 1 mark if the student straightaway writes the answer as 2005 kHz 
and 1995 kHz)  
½ 
 ½ 1 
Set1,Q6 
Set2,Q8 
Set3,Q7 
SECTION B 
 ; 
Alternatively, 
(Award this 1 mark even if the student writes the formula for  directly as 
such) 
½ 
½ 
½ 
½ 2 
Formulae:   1 
Substitution and calculation: 1 
SET 55/1/1/N 
 Page 2 of 22  Final Draft  11/03/16 2:00 p.m. 
Set1,Q7 
Set2,Q10 
Set3,Q8 
(i) 
 is more for  particle, we have 
 (Also, accept if the student writes ) 
(ii) K.E. = q V 
 q is less for proton, we have 
 =2) 
½ 
½ 
½ 
½ 2 
Set1,Q8 
Set2,Q9 
Set3,Q6 
When the electron jumps from the orbit with n=3 to n=2 
(Longest wavelength of the Balmer series / First line of the Balmer series) 
 =  = 
 =
 Hz 
 Hz. 
(If the student just writes , award ½ mark) 
(Alternatively, 
= =
) 
1 
½ 
½ 
½ 
½ 
2 
Formulae ½ + ½ 
Conclusions in the two cases ½ + ½ 
Indicating the transition 1 
Calculation of frequency 1 
SET 55/1/1/N 
 Page 3 of 22  Final Draft  11/03/16 2:00 p.m. 
OR 
 m 
½ 
½ 
1 2 
Set1,Q9 
Set2,Q6 
Set3,Q10 
If base band signal were to be transmitted directly 
1. The height of the antennae needed will be impractically large.
2. The effective power radiated would be too low.
3. There would be a high probability of different signals getting mixed
up with one another.
(Any two) 1+1 2 
Set1,Q10 
Set2,Q7 
Set3,Q9 
Since AQ = AR, we have 
QR|| BC 
 is the angle of minimum deviation. 
(Alternatively: Since AQ=AR, we get 
 is the angle of minimum deviation.) 
 =
 = 60 
or =60
0
½ 
½ 
½ 
½ 2 
Formula  1 
Calculation of 1 
Identifying that  is the angle of minimum deviation ½ 
Formula  ½ 
Calculation of    1 
Two Reasons  1+ 1 
Page 4


SET 55/1/1/N 
 Page 1 of 22  Final Draft  11/03/16 2:00 p.m. 
MARKING SCHEME 
Q. No. Expected Answer / Value Points Marks Total 
Marks 
Set1,Q1 
Set2,Q2 
Set3,Q4 
SECTION A 
Zero / No work done / None 1 1 
Set1,Q2 
Set2,Q5 
Set3,Q3 
Drift velocity per unit field 
(directly proportional to relaxation time) 
½ 
½ 1 
Set1,Q3 
Set2,Q4 
Set3,Q2 
Charged particle moves inclined to the magnetic field 
(angle between  and  is neither nor 0) 
(component of , parallel to  , is not zero.) 
1 1 
Set1,Q4 
Set2,Q1 
Set3,Q5 
(some) light gets deviated / scattered / absorbed 
Scattering of light 
½ 
½ 1 
Set1,Q5 
Set2,Q3 
Set3,Q1 
 = 2005 kHz ; 1995 kHz 
(Give full 1 mark if the student straightaway writes the answer as 2005 kHz 
and 1995 kHz)  
½ 
 ½ 1 
Set1,Q6 
Set2,Q8 
Set3,Q7 
SECTION B 
 ; 
Alternatively, 
(Award this 1 mark even if the student writes the formula for  directly as 
such) 
½ 
½ 
½ 
½ 2 
Formulae:   1 
Substitution and calculation: 1 
SET 55/1/1/N 
 Page 2 of 22  Final Draft  11/03/16 2:00 p.m. 
Set1,Q7 
Set2,Q10 
Set3,Q8 
(i) 
 is more for  particle, we have 
 (Also, accept if the student writes ) 
(ii) K.E. = q V 
 q is less for proton, we have 
 =2) 
½ 
½ 
½ 
½ 2 
Set1,Q8 
Set2,Q9 
Set3,Q6 
When the electron jumps from the orbit with n=3 to n=2 
(Longest wavelength of the Balmer series / First line of the Balmer series) 
 =  = 
 =
 Hz 
 Hz. 
(If the student just writes , award ½ mark) 
(Alternatively, 
= =
) 
1 
½ 
½ 
½ 
½ 
2 
Formulae ½ + ½ 
Conclusions in the two cases ½ + ½ 
Indicating the transition 1 
Calculation of frequency 1 
SET 55/1/1/N 
 Page 3 of 22  Final Draft  11/03/16 2:00 p.m. 
OR 
 m 
½ 
½ 
1 2 
Set1,Q9 
Set2,Q6 
Set3,Q10 
If base band signal were to be transmitted directly 
1. The height of the antennae needed will be impractically large.
2. The effective power radiated would be too low.
3. There would be a high probability of different signals getting mixed
up with one another.
(Any two) 1+1 2 
Set1,Q10 
Set2,Q7 
Set3,Q9 
Since AQ = AR, we have 
QR|| BC 
 is the angle of minimum deviation. 
(Alternatively: Since AQ=AR, we get 
 is the angle of minimum deviation.) 
 =
 = 60 
or =60
0
½ 
½ 
½ 
½ 2 
Formula  1 
Calculation of 1 
Identifying that  is the angle of minimum deviation ½ 
Formula  ½ 
Calculation of    1 
Two Reasons  1+ 1 
SET 55/1/1/N 
 Page 4 of 22  Final Draft  11/03/16 2:00 p.m. 
SECTION C 
Set1,Q11 
Set2,Q17 
Set3,Q22 
We have by Gauss’s law 
Let Q be the total charge on the shell 
(i)  For the point M outside the shell, we have 
(ii) For the point N inside the shell, as charge enclosed inside the shell is 
zero. 
The graph is as shown 
½ 
½ 
½ 
½ 
½ 
½ 
3 
Statement of Gauss’s Law ½ 
Calculation of field 
(i) Outside the shell 1 
(ii) Inside the shell 1 
 Graph ½ 
Page 5


SET 55/1/1/N 
 Page 1 of 22  Final Draft  11/03/16 2:00 p.m. 
MARKING SCHEME 
Q. No. Expected Answer / Value Points Marks Total 
Marks 
Set1,Q1 
Set2,Q2 
Set3,Q4 
SECTION A 
Zero / No work done / None 1 1 
Set1,Q2 
Set2,Q5 
Set3,Q3 
Drift velocity per unit field 
(directly proportional to relaxation time) 
½ 
½ 1 
Set1,Q3 
Set2,Q4 
Set3,Q2 
Charged particle moves inclined to the magnetic field 
(angle between  and  is neither nor 0) 
(component of , parallel to  , is not zero.) 
1 1 
Set1,Q4 
Set2,Q1 
Set3,Q5 
(some) light gets deviated / scattered / absorbed 
Scattering of light 
½ 
½ 1 
Set1,Q5 
Set2,Q3 
Set3,Q1 
 = 2005 kHz ; 1995 kHz 
(Give full 1 mark if the student straightaway writes the answer as 2005 kHz 
and 1995 kHz)  
½ 
 ½ 1 
Set1,Q6 
Set2,Q8 
Set3,Q7 
SECTION B 
 ; 
Alternatively, 
(Award this 1 mark even if the student writes the formula for  directly as 
such) 
½ 
½ 
½ 
½ 2 
Formulae:   1 
Substitution and calculation: 1 
SET 55/1/1/N 
 Page 2 of 22  Final Draft  11/03/16 2:00 p.m. 
Set1,Q7 
Set2,Q10 
Set3,Q8 
(i) 
 is more for  particle, we have 
 (Also, accept if the student writes ) 
(ii) K.E. = q V 
 q is less for proton, we have 
 =2) 
½ 
½ 
½ 
½ 2 
Set1,Q8 
Set2,Q9 
Set3,Q6 
When the electron jumps from the orbit with n=3 to n=2 
(Longest wavelength of the Balmer series / First line of the Balmer series) 
 =  = 
 =
 Hz 
 Hz. 
(If the student just writes , award ½ mark) 
(Alternatively, 
= =
) 
1 
½ 
½ 
½ 
½ 
2 
Formulae ½ + ½ 
Conclusions in the two cases ½ + ½ 
Indicating the transition 1 
Calculation of frequency 1 
SET 55/1/1/N 
 Page 3 of 22  Final Draft  11/03/16 2:00 p.m. 
OR 
 m 
½ 
½ 
1 2 
Set1,Q9 
Set2,Q6 
Set3,Q10 
If base band signal were to be transmitted directly 
1. The height of the antennae needed will be impractically large.
2. The effective power radiated would be too low.
3. There would be a high probability of different signals getting mixed
up with one another.
(Any two) 1+1 2 
Set1,Q10 
Set2,Q7 
Set3,Q9 
Since AQ = AR, we have 
QR|| BC 
 is the angle of minimum deviation. 
(Alternatively: Since AQ=AR, we get 
 is the angle of minimum deviation.) 
 =
 = 60 
or =60
0
½ 
½ 
½ 
½ 2 
Formula  1 
Calculation of 1 
Identifying that  is the angle of minimum deviation ½ 
Formula  ½ 
Calculation of    1 
Two Reasons  1+ 1 
SET 55/1/1/N 
 Page 4 of 22  Final Draft  11/03/16 2:00 p.m. 
SECTION C 
Set1,Q11 
Set2,Q17 
Set3,Q22 
We have by Gauss’s law 
Let Q be the total charge on the shell 
(i)  For the point M outside the shell, we have 
(ii) For the point N inside the shell, as charge enclosed inside the shell is 
zero. 
The graph is as shown 
½ 
½ 
½ 
½ 
½ 
½ 
3 
Statement of Gauss’s Law ½ 
Calculation of field 
(i) Outside the shell 1 
(ii) Inside the shell 1 
 Graph ½ 
SET 55/1/1/N 
 Page 5 of 22  Final Draft  11/03/16 2:00 p.m. 
Set1,Q12 
Set2,Q19 
Set3,Q21 
We have, for a single cell, 
=0.5
(Alternatively, 
And 
This gives A = 0.4A 
 = 0.25 
 ) 
( Note: If the student just draws the circuit diagram of the setup but does not 
do any calculations, award 1 mark only.) 
1 
½ 
½ 
½ 
½ 
½ 
½ 
½ 
½ 
½ 
½ 3 
Formulae  1 
Calculation of r 2 
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