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# Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12 Class 12 Notes | EduRev

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## Class 12 : Physics Past Year Paper Outside Delhi Anskey (Central Zone All Set) - 2016, Class 12 Class 12 Notes | EduRev

``` Page 1

1     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
MARKING SCHEME  55/1/C
Q. No. Expected Answer / Value Points Marks Total
Marks
SECTION-A
SET1,Q1
SET2,Q4
SET3,Q5
No work is done /
W = qV
AB
= q x 0 = 0
1 1
SET1,Q2
SET2,Q1
SET3,Q3
A diamagnetic specimen would move towards the weaker region of the field
while a paramagnetic specimen would move towards the stronger region./
A diamagnetic specimen is repelled by a magnet while a paramagnetic
specimen moves towards the magnet./
The paramagnetic get aligned along B and the diagrammatic perpendicular
to the field.
1
1
SET1,Q3
SET2,Q5
SET3,Q2
Transmitter, Medium or Channel and Receiver. 1 1
SET1,Q4
SET2,Q3
SET3,Q1 .
It is due to least scattering of red light as it has the longest wavelength/
As per Rayleighâ€™s scattering, the amount of light scattered
1 1
SET1,Q5
SET2,Q2
SET3,Q4
E = 2V ½
½ 1
SECTION B
SET1,Q6
SET2,Q9
SET3,Q8.
Modulation index is the ratio of the amplitude of modulating signal to that of carrier
wave
Alternatively
c
m
A
A
Reason-
To avoid distortion.
Role-
A bandpass filter rejects low and high frequencies and allows a band of frequencies
to pass through.
1
½
½ 2
Definition-      1
Reason-      ½
Role of bandpass filter-  ½
Page 2

1     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
MARKING SCHEME  55/1/C
Q. No. Expected Answer / Value Points Marks Total
Marks
SECTION-A
SET1,Q1
SET2,Q4
SET3,Q5
No work is done /
W = qV
AB
= q x 0 = 0
1 1
SET1,Q2
SET2,Q1
SET3,Q3
A diamagnetic specimen would move towards the weaker region of the field
while a paramagnetic specimen would move towards the stronger region./
A diamagnetic specimen is repelled by a magnet while a paramagnetic
specimen moves towards the magnet./
The paramagnetic get aligned along B and the diagrammatic perpendicular
to the field.
1
1
SET1,Q3
SET2,Q5
SET3,Q2
Transmitter, Medium or Channel and Receiver. 1 1
SET1,Q4
SET2,Q3
SET3,Q1 .
It is due to least scattering of red light as it has the longest wavelength/
As per Rayleighâ€™s scattering, the amount of light scattered
1 1
SET1,Q5
SET2,Q2
SET3,Q4
E = 2V ½
½ 1
SECTION B
SET1,Q6
SET2,Q9
SET3,Q8.
Modulation index is the ratio of the amplitude of modulating signal to that of carrier
wave
Alternatively
c
m
A
A
Reason-
To avoid distortion.
Role-
A bandpass filter rejects low and high frequencies and allows a band of frequencies
to pass through.
1
½
½ 2
Definition-      1
Reason-      ½
Role of bandpass filter-  ½
2     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
SET1,Q7
SET2,Q10
SET3,Q6
Face-AC
Here )
=
on face AC is 30º which is less than .Hence the ray get replaced here.
1
½
½
2
SET1,Q8
SET2,Q6
SET3,Q7
Kinetic Energy for the second state-
E
k
=  = =3.4X1.6X10
-19
J
De Broglies wavelength ?=
= 0.067nm
½
½
½
½ 2
SET1,Q9
SET2,Q8
SET3,Q10
The minimum energy, required to free the electron from the ground state of
the hydrogen atom, is known as Ionization Energy.
1
Path of emergent ray  1
Naming the face      ½
Justification      ½
Formulae of Kinetic energy and deBrogliea wavelength  ½ +½
Calculation and Result      ½+½
Definition      1
Formula     ½
Calculation and Result ½
Page 3

1     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
MARKING SCHEME  55/1/C
Q. No. Expected Answer / Value Points Marks Total
Marks
SECTION-A
SET1,Q1
SET2,Q4
SET3,Q5
No work is done /
W = qV
AB
= q x 0 = 0
1 1
SET1,Q2
SET2,Q1
SET3,Q3
A diamagnetic specimen would move towards the weaker region of the field
while a paramagnetic specimen would move towards the stronger region./
A diamagnetic specimen is repelled by a magnet while a paramagnetic
specimen moves towards the magnet./
The paramagnetic get aligned along B and the diagrammatic perpendicular
to the field.
1
1
SET1,Q3
SET2,Q5
SET3,Q2
Transmitter, Medium or Channel and Receiver. 1 1
SET1,Q4
SET2,Q3
SET3,Q1 .
It is due to least scattering of red light as it has the longest wavelength/
As per Rayleighâ€™s scattering, the amount of light scattered
1 1
SET1,Q5
SET2,Q2
SET3,Q4
E = 2V ½
½ 1
SECTION B
SET1,Q6
SET2,Q9
SET3,Q8.
Modulation index is the ratio of the amplitude of modulating signal to that of carrier
wave
Alternatively
c
m
A
A
Reason-
To avoid distortion.
Role-
A bandpass filter rejects low and high frequencies and allows a band of frequencies
to pass through.
1
½
½ 2
Definition-      1
Reason-      ½
Role of bandpass filter-  ½
2     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
SET1,Q7
SET2,Q10
SET3,Q6
Face-AC
Here )
=
on face AC is 30º which is less than .Hence the ray get replaced here.
1
½
½
2
SET1,Q8
SET2,Q6
SET3,Q7
Kinetic Energy for the second state-
E
k
=  = =3.4X1.6X10
-19
J
De Broglies wavelength ?=
= 0.067nm
½
½
½
½ 2
SET1,Q9
SET2,Q8
SET3,Q10
The minimum energy, required to free the electron from the ground state of
the hydrogen atom, is known as Ionization Energy.
1
Path of emergent ray  1
Naming the face      ½
Justification      ½
Formulae of Kinetic energy and deBrogliea wavelength  ½ +½
Calculation and Result      ½+½
Definition      1
Formula     ½
Calculation and Result ½
3     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
m E e i
h
me
E
o
o
o
,. .
8
2 2
4
Therefore, Ionization Energy will become 200 times
OR
2 2
1
2
1 1
R
For shortest wavelength, n = a
Therefore, = => ?= =4x10
-7
m
½
½
1
½
½
2
2
SET1,Q10
SET2,Q7
SET3,Q9
a) Effective resistance of the circuit R
E
= 6?
Terminal potential difference across the cell, V=E-ir
Also p.d. across 4? resistor =4X2V= 8V
Hence the volmeter gives the same reading in the two cases.
b) In series -current same
In parallel â€“ potential same
½
½
½
½ 2
SECTION C
SET1,Q11
SET2,Q15
SET3,Q22
Surface with a constant value of potential at all points on the surface. ½
Formula      1
Calculation and Result  ½+½
½
a) Relation for terminal potential ½
b) Justification ½
c) Explanation (parallel and series) ½ + ½
Definition-  ½
i.Diagram of Equipotential Surface ½
ii.Diagram and reason ½ +½
Page 4

1     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
MARKING SCHEME  55/1/C
Q. No. Expected Answer / Value Points Marks Total
Marks
SECTION-A
SET1,Q1
SET2,Q4
SET3,Q5
No work is done /
W = qV
AB
= q x 0 = 0
1 1
SET1,Q2
SET2,Q1
SET3,Q3
A diamagnetic specimen would move towards the weaker region of the field
while a paramagnetic specimen would move towards the stronger region./
A diamagnetic specimen is repelled by a magnet while a paramagnetic
specimen moves towards the magnet./
The paramagnetic get aligned along B and the diagrammatic perpendicular
to the field.
1
1
SET1,Q3
SET2,Q5
SET3,Q2
Transmitter, Medium or Channel and Receiver. 1 1
SET1,Q4
SET2,Q3
SET3,Q1 .
It is due to least scattering of red light as it has the longest wavelength/
As per Rayleighâ€™s scattering, the amount of light scattered
1 1
SET1,Q5
SET2,Q2
SET3,Q4
E = 2V ½
½ 1
SECTION B
SET1,Q6
SET2,Q9
SET3,Q8.
Modulation index is the ratio of the amplitude of modulating signal to that of carrier
wave
Alternatively
c
m
A
A
Reason-
To avoid distortion.
Role-
A bandpass filter rejects low and high frequencies and allows a band of frequencies
to pass through.
1
½
½ 2
Definition-      1
Reason-      ½
Role of bandpass filter-  ½
2     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
SET1,Q7
SET2,Q10
SET3,Q6
Face-AC
Here )
=
on face AC is 30º which is less than .Hence the ray get replaced here.
1
½
½
2
SET1,Q8
SET2,Q6
SET3,Q7
Kinetic Energy for the second state-
E
k
=  = =3.4X1.6X10
-19
J
De Broglies wavelength ?=
= 0.067nm
½
½
½
½ 2
SET1,Q9
SET2,Q8
SET3,Q10
The minimum energy, required to free the electron from the ground state of
the hydrogen atom, is known as Ionization Energy.
1
Path of emergent ray  1
Naming the face      ½
Justification      ½
Formulae of Kinetic energy and deBrogliea wavelength  ½ +½
Calculation and Result      ½+½
Definition      1
Formula     ½
Calculation and Result ½
3     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
m E e i
h
me
E
o
o
o
,. .
8
2 2
4
Therefore, Ionization Energy will become 200 times
OR
2 2
1
2
1 1
R
For shortest wavelength, n = a
Therefore, = => ?= =4x10
-7
m
½
½
1
½
½
2
2
SET1,Q10
SET2,Q7
SET3,Q9
a) Effective resistance of the circuit R
E
= 6?
Terminal potential difference across the cell, V=E-ir
Also p.d. across 4? resistor =4X2V= 8V
Hence the volmeter gives the same reading in the two cases.
b) In series -current same
In parallel â€“ potential same
½
½
½
½ 2
SECTION C
SET1,Q11
SET2,Q15
SET3,Q22
Surface with a constant value of potential at all points on the surface. ½
Formula      1
Calculation and Result  ½+½
½
a) Relation for terminal potential ½
b) Justification ½
c) Explanation (parallel and series) ½ + ½
Definition-  ½
i.Diagram of Equipotential Surface ½
ii.Diagram and reason ½ +½
4     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
i.
ii.
V /
iii.No
If the field lines are tangential, work will be done in moving a charge on the
surface which goes against the definition of equipotential surface.
½
½
½
½
½ 3
SET1,Q12
SET2,Q14
SET3,Q12
i.When the pass axis of a  poloroid makes an angle  with the plane of
polarisation of polorised light of intensity I
o
incident on it, then the intensity
of the tramsmitted emergent light is given by I=
Note: If the student writes the formula I= and draws the
1
Statement      1
Plotting the graph      1
Calculating value of ( refractive index  1
Page 5

1     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
MARKING SCHEME  55/1/C
Q. No. Expected Answer / Value Points Marks Total
Marks
SECTION-A
SET1,Q1
SET2,Q4
SET3,Q5
No work is done /
W = qV
AB
= q x 0 = 0
1 1
SET1,Q2
SET2,Q1
SET3,Q3
A diamagnetic specimen would move towards the weaker region of the field
while a paramagnetic specimen would move towards the stronger region./
A diamagnetic specimen is repelled by a magnet while a paramagnetic
specimen moves towards the magnet./
The paramagnetic get aligned along B and the diagrammatic perpendicular
to the field.
1
1
SET1,Q3
SET2,Q5
SET3,Q2
Transmitter, Medium or Channel and Receiver. 1 1
SET1,Q4
SET2,Q3
SET3,Q1 .
It is due to least scattering of red light as it has the longest wavelength/
As per Rayleighâ€™s scattering, the amount of light scattered
1 1
SET1,Q5
SET2,Q2
SET3,Q4
E = 2V ½
½ 1
SECTION B
SET1,Q6
SET2,Q9
SET3,Q8.
Modulation index is the ratio of the amplitude of modulating signal to that of carrier
wave
Alternatively
c
m
A
A
Reason-
To avoid distortion.
Role-
A bandpass filter rejects low and high frequencies and allows a band of frequencies
to pass through.
1
½
½ 2
Definition-      1
Reason-      ½
Role of bandpass filter-  ½
2     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
SET1,Q7
SET2,Q10
SET3,Q6
Face-AC
Here )
=
on face AC is 30º which is less than .Hence the ray get replaced here.
1
½
½
2
SET1,Q8
SET2,Q6
SET3,Q7
Kinetic Energy for the second state-
E
k
=  = =3.4X1.6X10
-19
J
De Broglies wavelength ?=
= 0.067nm
½
½
½
½ 2
SET1,Q9
SET2,Q8
SET3,Q10
The minimum energy, required to free the electron from the ground state of
the hydrogen atom, is known as Ionization Energy.
1
Path of emergent ray  1
Naming the face      ½
Justification      ½
Formulae of Kinetic energy and deBrogliea wavelength  ½ +½
Calculation and Result      ½+½
Definition      1
Formula     ½
Calculation and Result ½
3     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
m E e i
h
me
E
o
o
o
,. .
8
2 2
4
Therefore, Ionization Energy will become 200 times
OR
2 2
1
2
1 1
R
For shortest wavelength, n = a
Therefore, = => ?= =4x10
-7
m
½
½
1
½
½
2
2
SET1,Q10
SET2,Q7
SET3,Q9
a) Effective resistance of the circuit R
E
= 6?
Terminal potential difference across the cell, V=E-ir
Also p.d. across 4? resistor =4X2V= 8V
Hence the volmeter gives the same reading in the two cases.
b) In series -current same
In parallel â€“ potential same
½
½
½
½ 2
SECTION C
SET1,Q11
SET2,Q15
SET3,Q22
Surface with a constant value of potential at all points on the surface. ½
Formula      1
Calculation and Result  ½+½
½
a) Relation for terminal potential ½
b) Justification ½
c) Explanation (parallel and series) ½ + ½
Definition-  ½
i.Diagram of Equipotential Surface ½
ii.Diagram and reason ½ +½
4     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
i.
ii.
V /
iii.No
If the field lines are tangential, work will be done in moving a charge on the
surface which goes against the definition of equipotential surface.
½
½
½
½
½ 3
SET1,Q12
SET2,Q14
SET3,Q12
i.When the pass axis of a  poloroid makes an angle  with the plane of
polarisation of polorised light of intensity I
o
incident on it, then the intensity
of the tramsmitted emergent light is given by I=
Note: If the student writes the formula I= and draws the
1
Statement      1
Plotting the graph      1
Calculating value of ( refractive index  1
5     FOURTH  DRAFT     1O
th
March/2016 4:00 PM
diagram give 1mark.
i.
iii.
=  =
1
½
½ 3
SET1,Q13
SET2,Q13
SET3,Q14
(i) For material B
From the graph for the same value of â€˜ â€˜, stopping potential is more for material â€˜Bâ€™/
is higher for lower value of ]
(ii) No
As slope is given by  which is constant.
1
½
½
½
½ 3
Sketch of the Graph      1
(i) Stopping Potential and Reason  ½+ ½
(ii) Dependence of Slope and Explanation   ½+ ½
```
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