Physics Past Year Paper Anskey (Outside Delhi Set -2) - 2014, Class 12, CBSE Class 12 Notes | EduRev

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Class 12 : Physics Past Year Paper Anskey (Outside Delhi Set -2) - 2014, Class 12, CBSE Class 12 Notes | EduRev

 Page 1


Outside Delhi    SET II 
MARKING SCHEME 
SET 55/2 
Q. 
No. 
Expected Answer / Value Points Marks Total 
Marks 
1. Clockwise
1 
1 
2. 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 
3. 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 
Page 2


Outside Delhi    SET II 
MARKING SCHEME 
SET 55/2 
Q. 
No. 
Expected Answer / Value Points Marks Total 
Marks 
1. Clockwise
1 
1 
2. 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 
3. 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 
Outside Delhi    SET II 
4. They start from positive charges and end on negative charges
[Alternatively:
Electric field is conservative in nature.]
1 
1 
5. Converging lens
Light rays converge, on moving from denser to rarer medium.
Alternatively:
Note: If explained on the basis of lens makers formulae, award 1 mark.
½ 
½ 
1 
6. Metal A
? work	functions	  ! and !
"
#
 > !
"
½ 
½ 1 
7. Infrared radiation 1 1 
8. Neutrinos are neutral (chargeless), (almost) massless particles that hardly
interact with matter.
Alternatively
The neutrinos can penetrate large quantity of matter without any interaction
OR
Neutrinos are chargeless and (almost) massless particles.
1 
1 
9. 
I = AneV
d
?V
d 
 = 

.&
.'	(
)
*+
(,(
)
-
(
..(
)
*/
 = 5 X 10
-4
m/s
½ 
½ 
1 2 
10. 
 
According to  
Ampere’s  circuital Law 
?1
2 3
45
3
 =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 
½ 
½ 
Formula  ½ 
Calculation of drift velocity 1 ½ 
Statement of Ampere’s circuital law  ½ 
Showing inconsistency during the process of charging 1 
Displacement Current  ½ 
Page 3


Outside Delhi    SET II 
MARKING SCHEME 
SET 55/2 
Q. 
No. 
Expected Answer / Value Points Marks Total 
Marks 
1. Clockwise
1 
1 
2. 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 
3. 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 
Outside Delhi    SET II 
4. They start from positive charges and end on negative charges
[Alternatively:
Electric field is conservative in nature.]
1 
1 
5. Converging lens
Light rays converge, on moving from denser to rarer medium.
Alternatively:
Note: If explained on the basis of lens makers formulae, award 1 mark.
½ 
½ 
1 
6. Metal A
? work	functions	  ! and !
"
#
 > !
"
½ 
½ 1 
7. Infrared radiation 1 1 
8. Neutrinos are neutral (chargeless), (almost) massless particles that hardly
interact with matter.
Alternatively
The neutrinos can penetrate large quantity of matter without any interaction
OR
Neutrinos are chargeless and (almost) massless particles.
1 
1 
9. 
I = AneV
d
?V
d 
 = 

.&
.'	(
)
*+
(,(
)
-
(
..(
)
*/
 = 5 X 10
-4
m/s
½ 
½ 
1 2 
10. 
 
According to  
Ampere’s  circuital Law 
?1
2 3
45
3
 =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 
½ 
½ 
Formula  ½ 
Calculation of drift velocity 1 ½ 
Statement of Ampere’s circuital law  ½ 
Showing inconsistency during the process of charging 1 
Displacement Current  ½ 
Outside Delhi    SET II 
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 ( = 
6Ø
8
69
) 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 
?1
2 3
45
3
=  µ
0
 	
:
 + µ
0
 ?
"
4Ø
<
4=
[Note : If the student does the reasoning by using the (detailed) mathematics, 
relevant to displacement current, award full 2 marks ] 
½ 
½ 
2 
11. 
As per Rutherford’s  model 
>?

@
 = 


AB

CD

@

 !

 =


AB

CD

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


AB

CD

@
+ 



 !

 
       = - 



.


AB

CD

@
 = = - 


&B

CD

@
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=
EF

) 
>?

@
 = 


AB

CD

@

and mvr= 
EF



!



 
E

F

A

and m!

  = 


AB

G

 
½ 
½ 
½ 
½ 
½ 
½ 
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 4


Outside Delhi    SET II 
MARKING SCHEME 
SET 55/2 
Q. 
No. 
Expected Answer / Value Points Marks Total 
Marks 
1. Clockwise
1 
1 
2. 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 
3. 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 
Outside Delhi    SET II 
4. They start from positive charges and end on negative charges
[Alternatively:
Electric field is conservative in nature.]
1 
1 
5. Converging lens
Light rays converge, on moving from denser to rarer medium.
Alternatively:
Note: If explained on the basis of lens makers formulae, award 1 mark.
½ 
½ 
1 
6. Metal A
? work	functions	  ! and !
"
#
 > !
"
½ 
½ 1 
7. Infrared radiation 1 1 
8. Neutrinos are neutral (chargeless), (almost) massless particles that hardly
interact with matter.
Alternatively
The neutrinos can penetrate large quantity of matter without any interaction
OR
Neutrinos are chargeless and (almost) massless particles.
1 
1 
9. 
I = AneV
d
?V
d 
 = 

.&
.'	(
)
*+
(,(
)
-
(
..(
)
*/
 = 5 X 10
-4
m/s
½ 
½ 
1 2 
10. 
 
According to  
Ampere’s  circuital Law 
?1
2 3
45
3
 =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 
½ 
½ 
Formula  ½ 
Calculation of drift velocity 1 ½ 
Statement of Ampere’s circuital law  ½ 
Showing inconsistency during the process of charging 1 
Displacement Current  ½ 
Outside Delhi    SET II 
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 ( = 
6Ø
8
69
) 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 
?1
2 3
45
3
=  µ
0
 	
:
 + µ
0
 ?
"
4Ø
<
4=
[Note : If the student does the reasoning by using the (detailed) mathematics, 
relevant to displacement current, award full 2 marks ] 
½ 
½ 
2 
11. 
As per Rutherford’s  model 
>?

@
 = 


AB

CD

@

 !

 =


AB

CD

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


AB

CD

@
+ 



 !

 
       = - 



.


AB

CD

@
 = = - 


&B

CD

@
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=
EF

) 
>?

@
 = 


AB

CD

@

and mvr= 
EF



!



 
E

F

A

and m!

  = 


AB

G

 
½ 
½ 
½ 
½ 
½ 
½ 
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 II 
r = 
B

E

F

CD

>
Bohr’s radius (for n = 1) = H
"
 

/	IG


½ 
½ 
2 
12. 
[Note : If the student does not draw the diagram but just writes that the angle 
of incidence for both rays ‘2’ and ‘1’, on face AC = 45
o
 and says that it is less 
than critical angle for ray ‘1’ which therefore gets refracted and more than 
critical angle for ray ‘2’,which undergoes total internal reflection; Award ½ + 
½ marks] 
1+1 
2 
13. 
Transmitter: A transmitter processes the incoming message signal so as to 
make it suitable for transmission through a channel and subsequent reception. 
Modulator:  It is a device in which the amplitude/ (frequency/phase) of a 
high frequency carrier wave is made to change in accordance with the 
message signal through (appropriate) superposition. 
1 
1 
2 
Function of transmitter 1 
Function of modulator 1 
Tracing the path of the two rays 1 + 1 
Page 5


Outside Delhi    SET II 
MARKING SCHEME 
SET 55/2 
Q. 
No. 
Expected Answer / Value Points Marks Total 
Marks 
1. Clockwise
1 
1 
2. 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 
3. 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 
Outside Delhi    SET II 
4. They start from positive charges and end on negative charges
[Alternatively:
Electric field is conservative in nature.]
1 
1 
5. Converging lens
Light rays converge, on moving from denser to rarer medium.
Alternatively:
Note: If explained on the basis of lens makers formulae, award 1 mark.
½ 
½ 
1 
6. Metal A
? work	functions	  ! and !
"
#
 > !
"
½ 
½ 1 
7. Infrared radiation 1 1 
8. Neutrinos are neutral (chargeless), (almost) massless particles that hardly
interact with matter.
Alternatively
The neutrinos can penetrate large quantity of matter without any interaction
OR
Neutrinos are chargeless and (almost) massless particles.
1 
1 
9. 
I = AneV
d
?V
d 
 = 

.&
.'	(
)
*+
(,(
)
-
(
..(
)
*/
 = 5 X 10
-4
m/s
½ 
½ 
1 2 
10. 
 
According to  
Ampere’s  circuital Law 
?1
2 3
45
3
 =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 
½ 
½ 
Formula  ½ 
Calculation of drift velocity 1 ½ 
Statement of Ampere’s circuital law  ½ 
Showing inconsistency during the process of charging 1 
Displacement Current  ½ 
Outside Delhi    SET II 
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 ( = 
6Ø
8
69
) 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 
?1
2 3
45
3
=  µ
0
 	
:
 + µ
0
 ?
"
4Ø
<
4=
[Note : If the student does the reasoning by using the (detailed) mathematics, 
relevant to displacement current, award full 2 marks ] 
½ 
½ 
2 
11. 
As per Rutherford’s  model 
>?

@
 = 


AB

CD

@

 !

 =


AB

CD

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


AB

CD

@
+ 



 !

 
       = - 



.


AB

CD

@
 = = - 


&B

CD

@
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=
EF

) 
>?

@
 = 


AB

CD

@

and mvr= 
EF



!



 
E

F

A

and m!

  = 


AB

G

 
½ 
½ 
½ 
½ 
½ 
½ 
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 II 
r = 
B

E

F

CD

>
Bohr’s radius (for n = 1) = H
"
 

/	IG


½ 
½ 
2 
12. 
[Note : If the student does not draw the diagram but just writes that the angle 
of incidence for both rays ‘2’ and ‘1’, on face AC = 45
o
 and says that it is less 
than critical angle for ray ‘1’ which therefore gets refracted and more than 
critical angle for ray ‘2’,which undergoes total internal reflection; Award ½ + 
½ marks] 
1+1 
2 
13. 
Transmitter: A transmitter processes the incoming message signal so as to 
make it suitable for transmission through a channel and subsequent reception. 
Modulator:  It is a device in which the amplitude/ (frequency/phase) of a 
high frequency carrier wave is made to change in accordance with the 
message signal through (appropriate) superposition. 
1 
1 
2 
Function of transmitter 1 
Function of modulator 1 
Tracing the path of the two rays 1 + 1 
Outside Delhi    SET II 
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. 
Condition : The transistor must be operated close to the centre of its active 
region. 
Alternatively  
The base- emitter junction of the transistor must be (suitably) forward biased 
and the collector – emitter junction must be (suitably) reverse biased. 
1 ½ 
½ 2 
Diagrams  ½ + ½ 
Explanations ½ + ½ 
Circuit diagram 1 ½ 
Condition   ½ 
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