Physics Past Year Paper Anskey (Outside Delhi Set -1) - 2014, Class 12, CBSE Class 12 Notes | EduRev

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Class 12 : Physics Past Year Paper Anskey (Outside Delhi Set -1) - 2014, Class 12, CBSE Class 12 Notes | EduRev

 Page 1


Outside Delhi    SET I 
MARKING SCHEME 
SET 55/1 
Q. No Expected Answer / Value Points Marks Total 
Marks 
1. 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 
2. X – 	  /  1 1 
3. Force decreases 1 1 
4. Intensity of radiation depends on the number of photons incident per unit area
per unit time.
[Note: Also accept the definition: ‘number of quanta of radiation per unit area
per unit time’. Also accept if the student writes:
All photons, of a particular frequency, have the same kinetic energy and
momentum, irrespective of the intensity of incident radiation.
Alternatively
The amount of light energy / Photon energy, incident per metre square per
second is called intensity of radiation
SI Unit : W/m
2
 or J/(s- m
2
)
½ 
½ 1 
5. Clockwise
Alternatively
1 
1 
Page 2


Outside Delhi    SET I 
MARKING SCHEME 
SET 55/1 
Q. No Expected Answer / Value Points Marks Total 
Marks 
1. 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 
2. X – 	  /  1 1 
3. Force decreases 1 1 
4. Intensity of radiation depends on the number of photons incident per unit area
per unit time.
[Note: Also accept the definition: ‘number of quanta of radiation per unit area
per unit time’. Also accept if the student writes:
All photons, of a particular frequency, have the same kinetic energy and
momentum, irrespective of the intensity of incident radiation.
Alternatively
The amount of light energy / Photon energy, incident per metre square per
second is called intensity of radiation
SI Unit : W/m
2
 or J/(s- m
2
)
½ 
½ 1 
5. Clockwise
Alternatively
1 
1 
Outside Delhi    SET I 
6. 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 
7. 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 
8. 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 
9. 
As per Rutherford’s  model 



 = 








 

 =







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







+ 



 

 
= - 



.







 = = - 


 




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=
!"

) 
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 3


Outside Delhi    SET I 
MARKING SCHEME 
SET 55/1 
Q. No Expected Answer / Value Points Marks Total 
Marks 
1. 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 
2. X – 	  /  1 1 
3. Force decreases 1 1 
4. Intensity of radiation depends on the number of photons incident per unit area
per unit time.
[Note: Also accept the definition: ‘number of quanta of radiation per unit area
per unit time’. Also accept if the student writes:
All photons, of a particular frequency, have the same kinetic energy and
momentum, irrespective of the intensity of incident radiation.
Alternatively
The amount of light energy / Photon energy, incident per metre square per
second is called intensity of radiation
SI Unit : W/m
2
 or J/(s- m
2
)
½ 
½ 1 
5. Clockwise
Alternatively
1 
1 
Outside Delhi    SET I 
6. 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 
7. 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 
8. 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 
9. 
As per Rutherford’s  model 



 = 








 

 =







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







+ 



 

 
= - 



.







 = = - 


 




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=
!"

) 
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 I 



 = 








and mvr= 
!"







	
!

"



and m

  = 




#

 
r = 


!

"




Bohr’s radius (for n = 1) = $
%
&

/	'#


½ 
½ 
½ 2 
10. 
Energy stored in a capacitor = 



() = 



*)

 = 



+

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



+

,
And final energy = 



+

,
?
./!01	!23
/!/4/01	!23
= 



[Note : If the student does the correct calculations by assuming the voltage 
across the  
(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 ] 
½ 
½ 
½ 
½ 
2 
11. 
 
According to  
Ampere’s  circuital Law 
?6
78
9:
8
 =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.  
½ 
½ 
½ 
Statement of Ampere’s circuital law  ½ 
Showing inconsistency during the process of charging 1 
Displacement Current  ½ 
Formula for energy stored ½ 
New value of capacitance ½ 
Calculation of ratio   1 
Page 4


Outside Delhi    SET I 
MARKING SCHEME 
SET 55/1 
Q. No Expected Answer / Value Points Marks Total 
Marks 
1. 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 
2. X – 	  /  1 1 
3. Force decreases 1 1 
4. Intensity of radiation depends on the number of photons incident per unit area
per unit time.
[Note: Also accept the definition: ‘number of quanta of radiation per unit area
per unit time’. Also accept if the student writes:
All photons, of a particular frequency, have the same kinetic energy and
momentum, irrespective of the intensity of incident radiation.
Alternatively
The amount of light energy / Photon energy, incident per metre square per
second is called intensity of radiation
SI Unit : W/m
2
 or J/(s- m
2
)
½ 
½ 1 
5. Clockwise
Alternatively
1 
1 
Outside Delhi    SET I 
6. 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 
7. 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 
8. 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 
9. 
As per Rutherford’s  model 



 = 








 

 =







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







+ 



 

 
= - 



.







 = = - 


 




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=
!"

) 
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 I 



 = 








and mvr= 
!"







	
!

"



and m

  = 




#

 
r = 


!

"




Bohr’s radius (for n = 1) = $
%
&

/	'#


½ 
½ 
½ 2 
10. 
Energy stored in a capacitor = 



() = 



*)

 = 



+

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



+

,
And final energy = 



+

,
?
./!01	!23
/!/4/01	!23
= 



[Note : If the student does the correct calculations by assuming the voltage 
across the  
(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 ] 
½ 
½ 
½ 
½ 
2 
11. 
 
According to  
Ampere’s  circuital Law 
?6
78
9:
8
 =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.  
½ 
½ 
½ 
Statement of Ampere’s circuital law  ½ 
Showing inconsistency during the process of charging 1 
Displacement Current  ½ 
Formula for energy stored ½ 
New value of capacitance ½ 
Calculation of ratio   1 
Outside Delhi    SET I 
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
=
;4
) 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 
?6
78
9:
8
 =  µ
0
 	
>
 + µ
0
 ?
%
9Ø
@
9A
[Note : If the student does the reasoning by using the (detailed) mathematics, 
relevant to displacement current, award full 2 marks ] 
½ 
2 
12. 
 
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 = 
B

C
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 
13. 
1 
Relation between V and I ½ 
Graph ½ 
Determination of emf and internal resistance      ½ + ½ 
Circuit diagram 1 
Working 1 
Page 5


Outside Delhi    SET I 
MARKING SCHEME 
SET 55/1 
Q. No Expected Answer / Value Points Marks Total 
Marks 
1. 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 
2. X – 	  /  1 1 
3. Force decreases 1 1 
4. Intensity of radiation depends on the number of photons incident per unit area
per unit time.
[Note: Also accept the definition: ‘number of quanta of radiation per unit area
per unit time’. Also accept if the student writes:
All photons, of a particular frequency, have the same kinetic energy and
momentum, irrespective of the intensity of incident radiation.
Alternatively
The amount of light energy / Photon energy, incident per metre square per
second is called intensity of radiation
SI Unit : W/m
2
 or J/(s- m
2
)
½ 
½ 1 
5. Clockwise
Alternatively
1 
1 
Outside Delhi    SET I 
6. 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 
7. 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 
8. 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 
9. 
As per Rutherford’s  model 



 = 








 

 =







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







+ 



 

 
= - 



.







 = = - 


 




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=
!"

) 
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 I 



 = 








and mvr= 
!"







	
!

"



and m

  = 




#

 
r = 


!

"




Bohr’s radius (for n = 1) = $
%
&

/	'#


½ 
½ 
½ 2 
10. 
Energy stored in a capacitor = 



() = 



*)

 = 



+

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



+

,
And final energy = 



+

,
?
./!01	!23
/!/4/01	!23
= 



[Note : If the student does the correct calculations by assuming the voltage 
across the  
(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 ] 
½ 
½ 
½ 
½ 
2 
11. 
 
According to  
Ampere’s  circuital Law 
?6
78
9:
8
 =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.  
½ 
½ 
½ 
Statement of Ampere’s circuital law  ½ 
Showing inconsistency during the process of charging 1 
Displacement Current  ½ 
Formula for energy stored ½ 
New value of capacitance ½ 
Calculation of ratio   1 
Outside Delhi    SET I 
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
=
;4
) 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 
?6
78
9:
8
 =  µ
0
 	
>
 + µ
0
 ?
%
9Ø
@
9A
[Note : If the student does the reasoning by using the (detailed) mathematics, 
relevant to displacement current, award full 2 marks ] 
½ 
2 
12. 
 
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 = 
B

C
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 
13. 
1 
Relation between V and I ½ 
Graph ½ 
Determination of emf and internal resistance      ½ + ½ 
Circuit diagram 1 
Working 1 
Outside Delhi   SET I  
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”) 
½ 
½ 
2 
14. 
I =  neA V
d
 
V
d
 = 

DEF
 = 

.G
H	I
J
K
I
.LI
J
MN
I
.J	I
J
MO
	
 m/s 
 = 1.048 I10
QR
m/s ( 1mm/s) 
½ 
½ 
1 2 
15. 
[Note : If the student just writes (without drawing any diagram) that angle of 
incidence for both rays ‘1’ and ‘2’ on face AC equals  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 only  ½ + ½ marks.] 
1 
1 
2 
16. 
Transducer : Any device that converts one form of energy to another. 
Repeater : A repeater accepts the signal from the transmitter, amplifies and 
retransmits it to the receiver. 
1 
1 2 
17. 
Function of Transducer 1 
Function of Repeater  1 
Diagrams  ½ + ½ 
Explanations ½ + ½ 
Tracing of Path of Ray 1 1 
Tracing of Path of Ray 2 1 
Formula  ½ 
Substitution and calculation ½ + 1 
Read More
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