Courses

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

Created by: Ambition Institute

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

``` Page 1

Compartment Page No. 1 20
th
July, 2014 final
MARKING SCHEME
SET 55/1/2 (Compartment)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. When a constant current flows through a wire, the Potential difference, between
any two points on the wire of uniform cross section, is directly proportional to the
length of the wire between these points.
Alternatively:
V or  = constant
1
1
2. Due to the heating effect of eddy currents set up in the metallic piece. 1 1
3.
Effective power
(Alternatively:  Effective power radiated decreases with an increase in
wavelength.)
1
1
4.
1
1
5. i. The two point charges( and ) should be of opposite nature.
ii. Magnitude of charge must be greater than that of charge
½
½ 1
6. Two monochromatic sources, which produce light waves, having a constant phase
difference, are known as coherent sources.
1
1
7. Ferromagnetic material 1 1
8. Random motion of free electrons gets directed towards the point at a higher
potential.
Alternatively:
Random motion becomes  a (partially) directed motion.
1
1
9

Work done in bringing the charge  from infinity to position
Hence, total work done in assembling the  two charges
OR
½
½ +½
½ 2
Obtaining the expression for total work done  2
Page 2

Compartment Page No. 1 20
th
July, 2014 final
MARKING SCHEME
SET 55/1/2 (Compartment)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. When a constant current flows through a wire, the Potential difference, between
any two points on the wire of uniform cross section, is directly proportional to the
length of the wire between these points.
Alternatively:
V or  = constant
1
1
2. Due to the heating effect of eddy currents set up in the metallic piece. 1 1
3.
Effective power
(Alternatively:  Effective power radiated decreases with an increase in
wavelength.)
1
1
4.
1
1
5. i. The two point charges( and ) should be of opposite nature.
ii. Magnitude of charge must be greater than that of charge
½
½ 1
6. Two monochromatic sources, which produce light waves, having a constant phase
difference, are known as coherent sources.
1
1
7. Ferromagnetic material 1 1
8. Random motion of free electrons gets directed towards the point at a higher
potential.
Alternatively:
Random motion becomes  a (partially) directed motion.
1
1
9

Work done in bringing the charge  from infinity to position
Hence, total work done in assembling the  two charges
OR
½
½ +½
½ 2
Obtaining the expression for total work done  2
Compartment Page No. 2 20
th
July, 2014 final

Work done in moving a unit  positive charge along distance
=
= V – (V +
= -
E= -
(i) Electric field is in the direction in which the potential decreases steepest.
(ii) Magnitude of Electric field is given by the change in the magnitude of
potential per unit displacement, normal to the equipotential surface at the
point.
½
½
½
½
2
10.
Charge on inner surface  :  - Q
Charge on outer surface  :  + Q
Electric field at point
E   =
½
½
1
2
11.

Equivalent magnetic moment of the coil
=IA
=I b
( = unit vector  to the plane of the coil)
Torque =
=  I b
= 0
(as  are parallel or antiparallel, to each other)
[Note: Also give credit, when student obtains the relation
, and substitutes  and writes
½
½
½
½
2
12.
i. emf induced
e =
=  0.2 x 20 x 10
-2
x 15
= 0.6 volt
ii. Current in the loop
i =
=    = 0.12 A
½
½
½
½ 2
Calculation of
i. emf induced in the arm PQ 1
ii. Current induced in the loop 1
Derivation  of relation between Electric field and potential gradient  1
Two important conclusions  ½ + ½
Charges on the inner and outer surfaces ½ + ½
Expression for electric field  1
Obtaining the expression for the torque 2
Page 3

Compartment Page No. 1 20
th
July, 2014 final
MARKING SCHEME
SET 55/1/2 (Compartment)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. When a constant current flows through a wire, the Potential difference, between
any two points on the wire of uniform cross section, is directly proportional to the
length of the wire between these points.
Alternatively:
V or  = constant
1
1
2. Due to the heating effect of eddy currents set up in the metallic piece. 1 1
3.
Effective power
(Alternatively:  Effective power radiated decreases with an increase in
wavelength.)
1
1
4.
1
1
5. i. The two point charges( and ) should be of opposite nature.
ii. Magnitude of charge must be greater than that of charge
½
½ 1
6. Two monochromatic sources, which produce light waves, having a constant phase
difference, are known as coherent sources.
1
1
7. Ferromagnetic material 1 1
8. Random motion of free electrons gets directed towards the point at a higher
potential.
Alternatively:
Random motion becomes  a (partially) directed motion.
1
1
9

Work done in bringing the charge  from infinity to position
Hence, total work done in assembling the  two charges
OR
½
½ +½
½ 2
Obtaining the expression for total work done  2
Compartment Page No. 2 20
th
July, 2014 final

Work done in moving a unit  positive charge along distance
=
= V – (V +
= -
E= -
(i) Electric field is in the direction in which the potential decreases steepest.
(ii) Magnitude of Electric field is given by the change in the magnitude of
potential per unit displacement, normal to the equipotential surface at the
point.
½
½
½
½
2
10.
Charge on inner surface  :  - Q
Charge on outer surface  :  + Q
Electric field at point
E   =
½
½
1
2
11.

Equivalent magnetic moment of the coil
=IA
=I b
( = unit vector  to the plane of the coil)
Torque =
=  I b
= 0
(as  are parallel or antiparallel, to each other)
[Note: Also give credit, when student obtains the relation
, and substitutes  and writes
½
½
½
½
2
12.
i. emf induced
e =
=  0.2 x 20 x 10
-2
x 15
= 0.6 volt
ii. Current in the loop
i =
=    = 0.12 A
½
½
½
½ 2
Calculation of
i. emf induced in the arm PQ 1
ii. Current induced in the loop 1
Derivation  of relation between Electric field and potential gradient  1
Two important conclusions  ½ + ½
Charges on the inner and outer surfaces ½ + ½
Expression for electric field  1
Obtaining the expression for the torque 2
Compartment Page No. 3 20
th
July, 2014 final
13.
Alternatively:
Applying Ampere circuital law for the rectangular loop abcd
=
Bh = (nh)
B =
½
½
½
½ 2
14.
=
As  = 0
? =
? =  =
?
wavelength ( )of emitted electrons,
½
½
½
Drawing of magnetic field lines ½
Obtaining the expression for magnetic field 1 ½
Finding the relation 1 ½
Drawing the graph ½
Page 4

Compartment Page No. 1 20
th
July, 2014 final
MARKING SCHEME
SET 55/1/2 (Compartment)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. When a constant current flows through a wire, the Potential difference, between
any two points on the wire of uniform cross section, is directly proportional to the
length of the wire between these points.
Alternatively:
V or  = constant
1
1
2. Due to the heating effect of eddy currents set up in the metallic piece. 1 1
3.
Effective power
(Alternatively:  Effective power radiated decreases with an increase in
wavelength.)
1
1
4.
1
1
5. i. The two point charges( and ) should be of opposite nature.
ii. Magnitude of charge must be greater than that of charge
½
½ 1
6. Two monochromatic sources, which produce light waves, having a constant phase
difference, are known as coherent sources.
1
1
7. Ferromagnetic material 1 1
8. Random motion of free electrons gets directed towards the point at a higher
potential.
Alternatively:
Random motion becomes  a (partially) directed motion.
1
1
9

Work done in bringing the charge  from infinity to position
Hence, total work done in assembling the  two charges
OR
½
½ +½
½ 2
Obtaining the expression for total work done  2
Compartment Page No. 2 20
th
July, 2014 final

Work done in moving a unit  positive charge along distance
=
= V – (V +
= -
E= -
(i) Electric field is in the direction in which the potential decreases steepest.
(ii) Magnitude of Electric field is given by the change in the magnitude of
potential per unit displacement, normal to the equipotential surface at the
point.
½
½
½
½
2
10.
Charge on inner surface  :  - Q
Charge on outer surface  :  + Q
Electric field at point
E   =
½
½
1
2
11.

Equivalent magnetic moment of the coil
=IA
=I b
( = unit vector  to the plane of the coil)
Torque =
=  I b
= 0
(as  are parallel or antiparallel, to each other)
[Note: Also give credit, when student obtains the relation
, and substitutes  and writes
½
½
½
½
2
12.
i. emf induced
e =
=  0.2 x 20 x 10
-2
x 15
= 0.6 volt
ii. Current in the loop
i =
=    = 0.12 A
½
½
½
½ 2
Calculation of
i. emf induced in the arm PQ 1
ii. Current induced in the loop 1
Derivation  of relation between Electric field and potential gradient  1
Two important conclusions  ½ + ½
Charges on the inner and outer surfaces ½ + ½
Expression for electric field  1
Obtaining the expression for the torque 2
Compartment Page No. 3 20
th
July, 2014 final
13.
Alternatively:
Applying Ampere circuital law for the rectangular loop abcd
=
Bh = (nh)
B =
½
½
½
½ 2
14.
=
As  = 0
? =
? =  =
?
wavelength ( )of emitted electrons,
½
½
½
Drawing of magnetic field lines ½
Obtaining the expression for magnetic field 1 ½
Finding the relation 1 ½
Drawing the graph ½
Compartment Page No. 4 20
th
July, 2014 final
½
2
15.

(i) Conductor (ii)  Semiconductor
=
In conductors, average relaxation time decreases with increase in temperature,
resulting in an increase in resistivity.
In semiconductors,  the increase in number density (with increase in temperature)
is more than the decrease in relaxation time; the net result is, therefore, a  decrease
in resistivity.
½ + ½
½
½ 2
16.

P: OR Gate
Q: NOT Gate
Inputs Output
A B Y
0 0 1
0 1 0
1 0 0
1 1 0
NOR Gate
½
½
½
½ 2
Naming of gates P and Q ½  + ½
Truth table & Identification ½  + ½
Drawing the two plots  ½ + ½
Explanation of Behaviour ½ + ½
Page 5

Compartment Page No. 1 20
th
July, 2014 final
MARKING SCHEME
SET 55/1/2 (Compartment)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. When a constant current flows through a wire, the Potential difference, between
any two points on the wire of uniform cross section, is directly proportional to the
length of the wire between these points.
Alternatively:
V or  = constant
1
1
2. Due to the heating effect of eddy currents set up in the metallic piece. 1 1
3.
Effective power
(Alternatively:  Effective power radiated decreases with an increase in
wavelength.)
1
1
4.
1
1
5. i. The two point charges( and ) should be of opposite nature.
ii. Magnitude of charge must be greater than that of charge
½
½ 1
6. Two monochromatic sources, which produce light waves, having a constant phase
difference, are known as coherent sources.
1
1
7. Ferromagnetic material 1 1
8. Random motion of free electrons gets directed towards the point at a higher
potential.
Alternatively:
Random motion becomes  a (partially) directed motion.
1
1
9

Work done in bringing the charge  from infinity to position
Hence, total work done in assembling the  two charges
OR
½
½ +½
½ 2
Obtaining the expression for total work done  2
Compartment Page No. 2 20
th
July, 2014 final

Work done in moving a unit  positive charge along distance
=
= V – (V +
= -
E= -
(i) Electric field is in the direction in which the potential decreases steepest.
(ii) Magnitude of Electric field is given by the change in the magnitude of
potential per unit displacement, normal to the equipotential surface at the
point.
½
½
½
½
2
10.
Charge on inner surface  :  - Q
Charge on outer surface  :  + Q
Electric field at point
E   =
½
½
1
2
11.

Equivalent magnetic moment of the coil
=IA
=I b
( = unit vector  to the plane of the coil)
Torque =
=  I b
= 0
(as  are parallel or antiparallel, to each other)
[Note: Also give credit, when student obtains the relation
, and substitutes  and writes
½
½
½
½
2
12.
i. emf induced
e =
=  0.2 x 20 x 10
-2
x 15
= 0.6 volt
ii. Current in the loop
i =
=    = 0.12 A
½
½
½
½ 2
Calculation of
i. emf induced in the arm PQ 1
ii. Current induced in the loop 1
Derivation  of relation between Electric field and potential gradient  1
Two important conclusions  ½ + ½
Charges on the inner and outer surfaces ½ + ½
Expression for electric field  1
Obtaining the expression for the torque 2
Compartment Page No. 3 20
th
July, 2014 final
13.
Alternatively:
Applying Ampere circuital law for the rectangular loop abcd
=
Bh = (nh)
B =
½
½
½
½ 2
14.
=
As  = 0
? =
? =  =
?
wavelength ( )of emitted electrons,
½
½
½
Drawing of magnetic field lines ½
Obtaining the expression for magnetic field 1 ½
Finding the relation 1 ½
Drawing the graph ½
Compartment Page No. 4 20
th
July, 2014 final
½
2
15.

(i) Conductor (ii)  Semiconductor
=
In conductors, average relaxation time decreases with increase in temperature,
resulting in an increase in resistivity.
In semiconductors,  the increase in number density (with increase in temperature)
is more than the decrease in relaxation time; the net result is, therefore, a  decrease
in resistivity.
½ + ½
½
½ 2
16.

P: OR Gate
Q: NOT Gate
Inputs Output
A B Y
0 0 1
0 1 0
1 0 0
1 1 0
NOR Gate
½
½
½
½ 2
Naming of gates P and Q ½  + ½
Truth table & Identification ½  + ½
Drawing the two plots  ½ + ½
Explanation of Behaviour ½ + ½
Compartment Page No. 5 20
th
July, 2014 final
17.

i. It suffers chromatic aberration
ii. It has spherical aberration
iii. Magnifying power small.
iv. Small resolving power
(Any two)
1
½ + ½ 2
18.

(i) Intensity of incident radiation I = nhv,
where n is number of photons incident per unit time per unit area.
For same intensity of two monochromatic radiations of frequency
and
=
As  >
?
Therefore the number of electrons emitted for monochromatic radiation of
frequency , will be more than that for radiation of frequency
[Alternatively: Also accept if the student says that, for same intensity of incident
radiation, the number of emitted electrons is same for each of the two frequencies
(ii) hv =
For given  (work function of metal)
increases with
Maximum Kinetic energy of emitted photoelectrons will be more for
monochromatic light of frequency as  > )
½
½
½
½ 2
19.
a) Force acting on the charged particle, moving with a velocity  , in a magnetic
field  :
= q( )
½
Schematic diagram 1
Two important limitations ½  + ½
Explanation of parts (i) and (ii) 1+1
a) Circular path + angular frequency expression 1 + ½
b) Trace of path;  justification ½+ 1
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;