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

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

``` Page 1

Compartment Page No. 1 20
th
July, 2014  final
MARKING SCHEME
SET 55/1/1 (Compartment)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. 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
2. Random motion of free electrons gets directed towards the point at a higher
potential.
Alternatively:
Random motion becomes  a (partially) directed motion.
1
1
3. Diamagnetic material
= 1 +
½
½ 1
4. Due to the heating effect of eddy currents set up in the metallic piece. 1 1
5.
Effective power
(Alternatively:  Effective power radiated decreases with an increase in
wavelength.)
1
1
6.
Alternatively: Also accept if the student gives only the following diagram:
1
1
7. Two monochromatic sources, which produce light waves, having a constant phase
difference, are known as coherent sources.
1 1
8. 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
9
Charge on inner surface  :  - Q
Charge on outer surface  :  + Q
Electric field at point
E   =
½
½
1
2
Charges on the inner and outer surfaces ½ + ½
Expression for electric field  1
Page 2

Compartment Page No. 1 20
th
July, 2014  final
MARKING SCHEME
SET 55/1/1 (Compartment)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. 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
2. Random motion of free electrons gets directed towards the point at a higher
potential.
Alternatively:
Random motion becomes  a (partially) directed motion.
1
1
3. Diamagnetic material
= 1 +
½
½ 1
4. Due to the heating effect of eddy currents set up in the metallic piece. 1 1
5.
Effective power
(Alternatively:  Effective power radiated decreases with an increase in
wavelength.)
1
1
6.
Alternatively: Also accept if the student gives only the following diagram:
1
1
7. Two monochromatic sources, which produce light waves, having a constant phase
difference, are known as coherent sources.
1 1
8. 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
9
Charge on inner surface  :  - Q
Charge on outer surface  :  + Q
Electric field at point
E   =
½
½
1
2
Charges on the inner and outer surfaces ½ + ½
Expression for electric field  1
Compartment Page No. 2 20
th
July, 2014  final
10.
Alternatively:
Applying Ampere circuital law for the rectangular loop abcd
=
Bh = (nh)
B =
½
½
½
½ 2
11.
=
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 3

Compartment Page No. 1 20
th
July, 2014  final
MARKING SCHEME
SET 55/1/1 (Compartment)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. 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
2. Random motion of free electrons gets directed towards the point at a higher
potential.
Alternatively:
Random motion becomes  a (partially) directed motion.
1
1
3. Diamagnetic material
= 1 +
½
½ 1
4. Due to the heating effect of eddy currents set up in the metallic piece. 1 1
5.
Effective power
(Alternatively:  Effective power radiated decreases with an increase in
wavelength.)
1
1
6.
Alternatively: Also accept if the student gives only the following diagram:
1
1
7. Two monochromatic sources, which produce light waves, having a constant phase
difference, are known as coherent sources.
1 1
8. 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
9
Charge on inner surface  :  - Q
Charge on outer surface  :  + Q
Electric field at point
E   =
½
½
1
2
Charges on the inner and outer surfaces ½ + ½
Expression for electric field  1
Compartment Page No. 2 20
th
July, 2014  final
10.
Alternatively:
Applying Ampere circuital law for the rectangular loop abcd
=
Bh = (nh)
B =
½
½
½
½ 2
11.
=
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. 3 20
th
July, 2014  final
½
2
12.
(Any two of the following)
i. No chromatic aberration.
ii. Less spherical aberration.
iii. Large magnifying power.
iv. Large resolving power
v. [Note : or any other two advantages.]
1
½ + ½
2
13.
Input Output
A A' B B' Y
0 1 0 1 0
0 1 1 0 0
1 0 0 1 0
1 0 1 0 1
Identification :  AND Gate
½
1
½ 2
Schematic diagram of reflecting telescope 1
Drawing the logic circuit of the combination ½
Truth table  1
Identification ½
Page 4

Compartment Page No. 1 20
th
July, 2014  final
MARKING SCHEME
SET 55/1/1 (Compartment)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. 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
2. Random motion of free electrons gets directed towards the point at a higher
potential.
Alternatively:
Random motion becomes  a (partially) directed motion.
1
1
3. Diamagnetic material
= 1 +
½
½ 1
4. Due to the heating effect of eddy currents set up in the metallic piece. 1 1
5.
Effective power
(Alternatively:  Effective power radiated decreases with an increase in
wavelength.)
1
1
6.
Alternatively: Also accept if the student gives only the following diagram:
1
1
7. Two monochromatic sources, which produce light waves, having a constant phase
difference, are known as coherent sources.
1 1
8. 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
9
Charge on inner surface  :  - Q
Charge on outer surface  :  + Q
Electric field at point
E   =
½
½
1
2
Charges on the inner and outer surfaces ½ + ½
Expression for electric field  1
Compartment Page No. 2 20
th
July, 2014  final
10.
Alternatively:
Applying Ampere circuital law for the rectangular loop abcd
=
Bh = (nh)
B =
½
½
½
½ 2
11.
=
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. 3 20
th
July, 2014  final
½
2
12.
(Any two of the following)
i. No chromatic aberration.
ii. Less spherical aberration.
iii. Large magnifying power.
iv. Large resolving power
v. [Note : or any other two advantages.]
1
½ + ½
2
13.
Input Output
A A' B B' Y
0 1 0 1 0
0 1 1 0 0
1 0 0 1 0
1 0 1 0 1
Identification :  AND Gate
½
1
½ 2
Schematic diagram of reflecting telescope 1
Drawing the logic circuit of the combination ½
Truth table  1
Identification ½
Compartment Page No. 4 20
th
July, 2014  final
14.

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
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.

i. emf induced
e =
= 0.1 x 10 x 10
-2
x 20 V
= 0.2 volt
ii. Current in the loop
i =
=  A  = 0.1 A
½
½
½
½ 2
Obtaining the expression for the torque 2
Drawing the two plots  ½ + ½
Explanation of Behaviour ½ + ½
Calculation of
i. emf induced in the arm PQ 1
ii. Current induced in the loop 1
Page 5

Compartment Page No. 1 20
th
July, 2014  final
MARKING SCHEME
SET 55/1/1 (Compartment)
Q.No. Expected Answer/Value Points Marks Total
Marks
1. 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
2. Random motion of free electrons gets directed towards the point at a higher
potential.
Alternatively:
Random motion becomes  a (partially) directed motion.
1
1
3. Diamagnetic material
= 1 +
½
½ 1
4. Due to the heating effect of eddy currents set up in the metallic piece. 1 1
5.
Effective power
(Alternatively:  Effective power radiated decreases with an increase in
wavelength.)
1
1
6.
Alternatively: Also accept if the student gives only the following diagram:
1
1
7. Two monochromatic sources, which produce light waves, having a constant phase
difference, are known as coherent sources.
1 1
8. 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
9
Charge on inner surface  :  - Q
Charge on outer surface  :  + Q
Electric field at point
E   =
½
½
1
2
Charges on the inner and outer surfaces ½ + ½
Expression for electric field  1
Compartment Page No. 2 20
th
July, 2014  final
10.
Alternatively:
Applying Ampere circuital law for the rectangular loop abcd
=
Bh = (nh)
B =
½
½
½
½ 2
11.
=
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. 3 20
th
July, 2014  final
½
2
12.
(Any two of the following)
i. No chromatic aberration.
ii. Less spherical aberration.
iii. Large magnifying power.
iv. Large resolving power
v. [Note : or any other two advantages.]
1
½ + ½
2
13.
Input Output
A A' B B' Y
0 1 0 1 0
0 1 1 0 0
1 0 0 1 0
1 0 1 0 1
Identification :  AND Gate
½
1
½ 2
Schematic diagram of reflecting telescope 1
Drawing the logic circuit of the combination ½
Truth table  1
Identification ½
Compartment Page No. 4 20
th
July, 2014  final
14.

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
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.

i. emf induced
e =
= 0.1 x 10 x 10
-2
x 20 V
= 0.2 volt
ii. Current in the loop
i =
=  A  = 0.1 A
½
½
½
½ 2
Obtaining the expression for the torque 2
Drawing the two plots  ½ + ½
Explanation of Behaviour ½ + ½
Calculation of
i. emf induced in the arm PQ 1
ii. Current induced in the loop 1
Compartment Page No. 5 20
th
July, 2014  final
17.

(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
18.

Work done in bringing the charge  from infinity to position
Hence, total work done in assembling the  two charges
OR

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
½
½
½
½
2
2
Explanation of parts (i) and (ii) 1+1
Obtaining the expression for total work done  2
Derivation  of relation between Electric field and potential gradient  1
Two important conclusions  ½ + ½
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