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

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

 Page 1


Outside Delhi    SET III 
MARKING SCHEME 
SET 55/3 
Q. 
No. 
Expected Answer / Value Points Marks Total 
Marks 
1. Anticlockwise
1 
1 
2. Metal A
The minimum frequency, at which photoemission starts, is more for metal A
Alternatively: Work function of A is more.
½ 
½ 1 
3. Definition : One ampere is the value of steady current which when
maintained in each of the two very long, straight, parallel conductors of
negligible cross section and placed one metre apart in vaccum, would
produce on each of these conductors a force equal of 2 x 10
-7
 N/m of its
length.
Alternatively
If the student writes F = 








L
and says that when 	


	

1	
R= 1 meter and L = 1 meter, then
F= 2 x 10
-7
 N
Award full 1 mark
Alternatively
If the student draws any one of the two diagram, as shown ,
Award full 1 mark 
1 
1 
4. As a diverging lens
Light rays diverge on going from a rarer to a denser medium.  
[Alternatively  
Also accept the reason given on the basis of lens marker’s formula.] 
½ 
½ 
1 
5. At the point of intersection of the two field lines, there will be two directions
for the electric field. This is not acceptable.
1 1 
6. Short radio waves (or) microwaves 1 1 
7. Neutrinos are neutral (chargeless), (almost) massless particles that hardly 1 
Page 2


Outside Delhi    SET III 
MARKING SCHEME 
SET 55/3 
Q. 
No. 
Expected Answer / Value Points Marks Total 
Marks 
1. Anticlockwise
1 
1 
2. Metal A
The minimum frequency, at which photoemission starts, is more for metal A
Alternatively: Work function of A is more.
½ 
½ 1 
3. Definition : One ampere is the value of steady current which when
maintained in each of the two very long, straight, parallel conductors of
negligible cross section and placed one metre apart in vaccum, would
produce on each of these conductors a force equal of 2 x 10
-7
 N/m of its
length.
Alternatively
If the student writes F = 








L
and says that when 	


	

1	
R= 1 meter and L = 1 meter, then
F= 2 x 10
-7
 N
Award full 1 mark
Alternatively
If the student draws any one of the two diagram, as shown ,
Award full 1 mark 
1 
1 
4. As a diverging lens
Light rays diverge on going from a rarer to a denser medium.  
[Alternatively  
Also accept the reason given on the basis of lens marker’s formula.] 
½ 
½ 
1 
5. At the point of intersection of the two field lines, there will be two directions
for the electric field. This is not acceptable.
1 1 
6. Short radio waves (or) microwaves 1 1 
7. Neutrinos are neutral (chargeless), (almost) massless particles that hardly 1 
Outside Delhi    SET III 
interact with matter.  
Alternatively  
The neutrinos can penetrate large quantity of matter without any interaction 
OR 
Neutrinos are chargeless and (almost) massless particles. 1 
8. Any two of the following (or any other correct)  reasons :
i. AC can be transmitted with much lower energy losses as compared to
DC
ii. AC voltage can be adjusted (stepped up or stepped down) as per
requirement.
iii. AC current in a circuit can be controlled using (almost) wattless
devices like the choke coil.
iv. AC is easier to generate.
½ + ½ 
1 
9. 
 
According to  
Ampere’s  circuital Law 
?



 =µ
0
 I
 Applying ampere’s circuital law to fig (a) we see that, during charging, the 
right hand side in Ampere’s circuital law equals  µ
0
 I 
However on applying it to the surfaces of the fig (b) or fig (c), the right hand 
side is zero.  
Hence, there is a contradiction. 
We can remove the contradiction by assuming that there exists a current 
(associated with the changing electric field during charging), known as the 
displacement current. 
When this current ( = 
Ø


) is added on the right hand side, Ampere’s circuital
law, the inconsisitency disappears. 
It was, therefore necessary, to generalize the Ampere’s circuital law, as 
?



 =  µ
0
 	

 + µ
0
 ?

Ø
 
!
[Note : If the student does the reasoning by using the (detailed) mathematics, 
relevant to displacement current, award full 2 marks ] 
½ 
½ 
½ 
½ 
2 
Statement of Ampere’s circuital law  ½ 
Showing inconsistency during the process of charging 1 
Displacement Current  ½ 
Page 3


Outside Delhi    SET III 
MARKING SCHEME 
SET 55/3 
Q. 
No. 
Expected Answer / Value Points Marks Total 
Marks 
1. Anticlockwise
1 
1 
2. Metal A
The minimum frequency, at which photoemission starts, is more for metal A
Alternatively: Work function of A is more.
½ 
½ 1 
3. Definition : One ampere is the value of steady current which when
maintained in each of the two very long, straight, parallel conductors of
negligible cross section and placed one metre apart in vaccum, would
produce on each of these conductors a force equal of 2 x 10
-7
 N/m of its
length.
Alternatively
If the student writes F = 








L
and says that when 	


	

1	
R= 1 meter and L = 1 meter, then
F= 2 x 10
-7
 N
Award full 1 mark
Alternatively
If the student draws any one of the two diagram, as shown ,
Award full 1 mark 
1 
1 
4. As a diverging lens
Light rays diverge on going from a rarer to a denser medium.  
[Alternatively  
Also accept the reason given on the basis of lens marker’s formula.] 
½ 
½ 
1 
5. At the point of intersection of the two field lines, there will be two directions
for the electric field. This is not acceptable.
1 1 
6. Short radio waves (or) microwaves 1 1 
7. Neutrinos are neutral (chargeless), (almost) massless particles that hardly 1 
Outside Delhi    SET III 
interact with matter.  
Alternatively  
The neutrinos can penetrate large quantity of matter without any interaction 
OR 
Neutrinos are chargeless and (almost) massless particles. 1 
8. Any two of the following (or any other correct)  reasons :
i. AC can be transmitted with much lower energy losses as compared to
DC
ii. AC voltage can be adjusted (stepped up or stepped down) as per
requirement.
iii. AC current in a circuit can be controlled using (almost) wattless
devices like the choke coil.
iv. AC is easier to generate.
½ + ½ 
1 
9. 
 
According to  
Ampere’s  circuital Law 
?



 =µ
0
 I
 Applying ampere’s circuital law to fig (a) we see that, during charging, the 
right hand side in Ampere’s circuital law equals  µ
0
 I 
However on applying it to the surfaces of the fig (b) or fig (c), the right hand 
side is zero.  
Hence, there is a contradiction. 
We can remove the contradiction by assuming that there exists a current 
(associated with the changing electric field during charging), known as the 
displacement current. 
When this current ( = 
Ø


) is added on the right hand side, Ampere’s circuital
law, the inconsisitency disappears. 
It was, therefore necessary, to generalize the Ampere’s circuital law, as 
?



 =  µ
0
 	

 + µ
0
 ?

Ø
 
!
[Note : If the student does the reasoning by using the (detailed) mathematics, 
relevant to displacement current, award full 2 marks ] 
½ 
½ 
½ 
½ 
2 
Statement of Ampere’s circuital law  ½ 
Showing inconsistency during the process of charging 1 
Displacement Current  ½ 
Outside Delhi    SET III 
10. 
 
I = AneV
d
V
d 
 = 
.#
.$	%
&
'(
%
.)%
&
'*
%+%
&
,
 = 7.5 X 10
-4
m/s 
½ 
½ 
1 2 
11. 
 
The relation between V and I is  
V = E – Ir 
Hence, the graph, between V and I, has the form shown below. 
For point A, I=0, Hence, V
A
= E 
For point B, V=0, Hence, E=I
B
r 
Therefore, r = 
-

.
Alternatively:   emf (E) equals the intercept on the vertical axis. Internal 
resistance ( r) equals the negative of the slope of the graph. 
½ 
½ 
½ 
½ 
2 
12. 
Energy stored in a capacitor = 



/0 = 



10

 = 



2

3
 (any one )
Capacitance of the (parallel) combination = C+C=2C 
Here, total charge, Q, remains the same  
initial energy = 



2

3
And final energy = 



2

3
?
56789	:7:;<=
676689	:7:;<=
  =



[Note : If the student does the correct calculations by assuming the voltage 
across the  
½ 
½ 
½ 
½ 
Formula ½ 
Calculation of drift velocity 1 ½ 
Relation between V and I ½ 
Graph ½ 
Determination of emf and internal resistance      ½ + ½ 
Formula for energy stored ½ 
New value of capacitance ½ 
Calculation of ratio   1 
Page 4


Outside Delhi    SET III 
MARKING SCHEME 
SET 55/3 
Q. 
No. 
Expected Answer / Value Points Marks Total 
Marks 
1. Anticlockwise
1 
1 
2. Metal A
The minimum frequency, at which photoemission starts, is more for metal A
Alternatively: Work function of A is more.
½ 
½ 1 
3. Definition : One ampere is the value of steady current which when
maintained in each of the two very long, straight, parallel conductors of
negligible cross section and placed one metre apart in vaccum, would
produce on each of these conductors a force equal of 2 x 10
-7
 N/m of its
length.
Alternatively
If the student writes F = 








L
and says that when 	


	

1	
R= 1 meter and L = 1 meter, then
F= 2 x 10
-7
 N
Award full 1 mark
Alternatively
If the student draws any one of the two diagram, as shown ,
Award full 1 mark 
1 
1 
4. As a diverging lens
Light rays diverge on going from a rarer to a denser medium.  
[Alternatively  
Also accept the reason given on the basis of lens marker’s formula.] 
½ 
½ 
1 
5. At the point of intersection of the two field lines, there will be two directions
for the electric field. This is not acceptable.
1 1 
6. Short radio waves (or) microwaves 1 1 
7. Neutrinos are neutral (chargeless), (almost) massless particles that hardly 1 
Outside Delhi    SET III 
interact with matter.  
Alternatively  
The neutrinos can penetrate large quantity of matter without any interaction 
OR 
Neutrinos are chargeless and (almost) massless particles. 1 
8. Any two of the following (or any other correct)  reasons :
i. AC can be transmitted with much lower energy losses as compared to
DC
ii. AC voltage can be adjusted (stepped up or stepped down) as per
requirement.
iii. AC current in a circuit can be controlled using (almost) wattless
devices like the choke coil.
iv. AC is easier to generate.
½ + ½ 
1 
9. 
 
According to  
Ampere’s  circuital Law 
?



 =µ
0
 I
 Applying ampere’s circuital law to fig (a) we see that, during charging, the 
right hand side in Ampere’s circuital law equals  µ
0
 I 
However on applying it to the surfaces of the fig (b) or fig (c), the right hand 
side is zero.  
Hence, there is a contradiction. 
We can remove the contradiction by assuming that there exists a current 
(associated with the changing electric field during charging), known as the 
displacement current. 
When this current ( = 
Ø


) is added on the right hand side, Ampere’s circuital
law, the inconsisitency disappears. 
It was, therefore necessary, to generalize the Ampere’s circuital law, as 
?



 =  µ
0
 	

 + µ
0
 ?

Ø
 
!
[Note : If the student does the reasoning by using the (detailed) mathematics, 
relevant to displacement current, award full 2 marks ] 
½ 
½ 
½ 
½ 
2 
Statement of Ampere’s circuital law  ½ 
Showing inconsistency during the process of charging 1 
Displacement Current  ½ 
Outside Delhi    SET III 
10. 
 
I = AneV
d
V
d 
 = 
.#
.$	%
&
'(
%
.)%
&
'*
%+%
&
,
 = 7.5 X 10
-4
m/s 
½ 
½ 
1 2 
11. 
 
The relation between V and I is  
V = E – Ir 
Hence, the graph, between V and I, has the form shown below. 
For point A, I=0, Hence, V
A
= E 
For point B, V=0, Hence, E=I
B
r 
Therefore, r = 
-

.
Alternatively:   emf (E) equals the intercept on the vertical axis. Internal 
resistance ( r) equals the negative of the slope of the graph. 
½ 
½ 
½ 
½ 
2 
12. 
Energy stored in a capacitor = 



/0 = 



10

 = 



2

3
 (any one )
Capacitance of the (parallel) combination = C+C=2C 
Here, total charge, Q, remains the same  
initial energy = 



2

3
And final energy = 



2

3
?
56789	:7:;<=
676689	:7:;<=
  =



[Note : If the student does the correct calculations by assuming the voltage 
across the  
½ 
½ 
½ 
½ 
Formula ½ 
Calculation of drift velocity 1 ½ 
Relation between V and I ½ 
Graph ½ 
Determination of emf and internal resistance      ½ + ½ 
Formula for energy stored ½ 
New value of capacitance ½ 
Calculation of ratio   1 
Outside Delhi    SET III 
(i) Parallel or (ii) Series combination  
to remain constant (=V) and obtain the answers 
as (i) 2:1  or (ii) 1:2 , award  full marks ] 2 
13. 
As per Rutherford’s  model 
>?

;
 = 


@A

 
B:

;

 
 C

 =


@A

 
B:

;
 
Total energy = P.E +K.E. 
= - 


@A

  
B:

;
+ 



  C

= - 



.


@A

  
B:

;
  = = - 


DA

 
B:

;
 
Negative Sign  implies that  
Electron – nucleus form a bound system. 
Alternatively  
Electron – nucleus form an attractive system) 
OR 
 
 
For the electron, we have 
Bohr’s Postulate (mvr=
7E

) 
>?

;
 = 


@A

 
B:

;

 
and mvr= 
7E



C



	
7

E

@

and mC

  = 


@A

F

r = 
A

7

E

B:

>
Bohr’s radius (for n = 1) = G

H

/	IF


½ 
½ 
½ 
½ 
½ 
½ 
½ 
½ 
2 
2 
Derivation of energy expression 1 ½ 
Significance of negative sign  ½ 
Bohr’s Postulate ½ 
Derivation of radius of nth orbit 1 
Bohr’s radius  ½ 
Page 5


Outside Delhi    SET III 
MARKING SCHEME 
SET 55/3 
Q. 
No. 
Expected Answer / Value Points Marks Total 
Marks 
1. Anticlockwise
1 
1 
2. Metal A
The minimum frequency, at which photoemission starts, is more for metal A
Alternatively: Work function of A is more.
½ 
½ 1 
3. Definition : One ampere is the value of steady current which when
maintained in each of the two very long, straight, parallel conductors of
negligible cross section and placed one metre apart in vaccum, would
produce on each of these conductors a force equal of 2 x 10
-7
 N/m of its
length.
Alternatively
If the student writes F = 








L
and says that when 	


	

1	
R= 1 meter and L = 1 meter, then
F= 2 x 10
-7
 N
Award full 1 mark
Alternatively
If the student draws any one of the two diagram, as shown ,
Award full 1 mark 
1 
1 
4. As a diverging lens
Light rays diverge on going from a rarer to a denser medium.  
[Alternatively  
Also accept the reason given on the basis of lens marker’s formula.] 
½ 
½ 
1 
5. At the point of intersection of the two field lines, there will be two directions
for the electric field. This is not acceptable.
1 1 
6. Short radio waves (or) microwaves 1 1 
7. Neutrinos are neutral (chargeless), (almost) massless particles that hardly 1 
Outside Delhi    SET III 
interact with matter.  
Alternatively  
The neutrinos can penetrate large quantity of matter without any interaction 
OR 
Neutrinos are chargeless and (almost) massless particles. 1 
8. Any two of the following (or any other correct)  reasons :
i. AC can be transmitted with much lower energy losses as compared to
DC
ii. AC voltage can be adjusted (stepped up or stepped down) as per
requirement.
iii. AC current in a circuit can be controlled using (almost) wattless
devices like the choke coil.
iv. AC is easier to generate.
½ + ½ 
1 
9. 
 
According to  
Ampere’s  circuital Law 
?



 =µ
0
 I
 Applying ampere’s circuital law to fig (a) we see that, during charging, the 
right hand side in Ampere’s circuital law equals  µ
0
 I 
However on applying it to the surfaces of the fig (b) or fig (c), the right hand 
side is zero.  
Hence, there is a contradiction. 
We can remove the contradiction by assuming that there exists a current 
(associated with the changing electric field during charging), known as the 
displacement current. 
When this current ( = 
Ø


) is added on the right hand side, Ampere’s circuital
law, the inconsisitency disappears. 
It was, therefore necessary, to generalize the Ampere’s circuital law, as 
?



 =  µ
0
 	

 + µ
0
 ?

Ø
 
!
[Note : If the student does the reasoning by using the (detailed) mathematics, 
relevant to displacement current, award full 2 marks ] 
½ 
½ 
½ 
½ 
2 
Statement of Ampere’s circuital law  ½ 
Showing inconsistency during the process of charging 1 
Displacement Current  ½ 
Outside Delhi    SET III 
10. 
 
I = AneV
d
V
d 
 = 
.#
.$	%
&
'(
%
.)%
&
'*
%+%
&
,
 = 7.5 X 10
-4
m/s 
½ 
½ 
1 2 
11. 
 
The relation between V and I is  
V = E – Ir 
Hence, the graph, between V and I, has the form shown below. 
For point A, I=0, Hence, V
A
= E 
For point B, V=0, Hence, E=I
B
r 
Therefore, r = 
-

.
Alternatively:   emf (E) equals the intercept on the vertical axis. Internal 
resistance ( r) equals the negative of the slope of the graph. 
½ 
½ 
½ 
½ 
2 
12. 
Energy stored in a capacitor = 



/0 = 



10

 = 



2

3
 (any one )
Capacitance of the (parallel) combination = C+C=2C 
Here, total charge, Q, remains the same  
initial energy = 



2

3
And final energy = 



2

3
?
56789	:7:;<=
676689	:7:;<=
  =



[Note : If the student does the correct calculations by assuming the voltage 
across the  
½ 
½ 
½ 
½ 
Formula ½ 
Calculation of drift velocity 1 ½ 
Relation between V and I ½ 
Graph ½ 
Determination of emf and internal resistance      ½ + ½ 
Formula for energy stored ½ 
New value of capacitance ½ 
Calculation of ratio   1 
Outside Delhi    SET III 
(i) Parallel or (ii) Series combination  
to remain constant (=V) and obtain the answers 
as (i) 2:1  or (ii) 1:2 , award  full marks ] 2 
13. 
As per Rutherford’s  model 
>?

;
 = 


@A

 
B:

;

 
 C

 =


@A

 
B:

;
 
Total energy = P.E +K.E. 
= - 


@A

  
B:

;
+ 



  C

= - 



.


@A

  
B:

;
  = = - 


DA

 
B:

;
 
Negative Sign  implies that  
Electron – nucleus form a bound system. 
Alternatively  
Electron – nucleus form an attractive system) 
OR 
 
 
For the electron, we have 
Bohr’s Postulate (mvr=
7E

) 
>?

;
 = 


@A

 
B:

;

 
and mvr= 
7E



C



	
7

E

@

and mC

  = 


@A

F

r = 
A

7

E

B:

>
Bohr’s radius (for n = 1) = G

H

/	IF


½ 
½ 
½ 
½ 
½ 
½ 
½ 
½ 
2 
2 
Derivation of energy expression 1 ½ 
Significance of negative sign  ½ 
Bohr’s Postulate ½ 
Derivation of radius of nth orbit 1 
Bohr’s radius  ½ 
Outside Delhi    SET III 
14. 
A paramagnetic material tends to move from weaker to stronger regions of 
the magnetic field and hence increases the number of lines of magnetic field 
passing through it. 
[Alternatively: A paramagnetic material, dipole moments are induced in the 
direction of the field.] 
A diamagnetic material tends to move from stronger to weaker regions of the 
magnetic field and hence, decreases the number of lines of magnetic field 
passing through it. 
[Alternatively: A diamagnetic material, dipole moments are induced in the 
opposite direction of the field.] 
[Note: If the student just writes that a paramagnetic material has a small 
positive susceptibility (0< X < J ) and a diamagnetic material has a negative
susceptibility (-1 X <  ), award the ½  mark for the second part of the 
question.] 
½ 
½ 
½ 
½ 
2 
15. 
Working:  
During one half of the input AC, the diode is forward biased and a current  
flows through R
L
.  
During the other half  of the input AC, the diode is reverse biased and no 
current flows through the load R
L.
Hence, the given AC input is rectified 
[Note : If the student just draws the waveforms, for the input AC voltage and 
output voltage (without giving any explanation) 
(award ½ mark only for “working”) 
1 
½ 
½ 
2 
Diagrams  ½ + ½ 
Explanations ½ + ½ 
Circuit diagram 1 
Working 1 
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