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

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

``` Page 1

Page 1 of 13 Final Draft         11/03/16 03:00 p.m
.
MARKING SCHEME
SET 55/1/S
Q. No. Expected Answer / Value Points Marks Total
Marks
Section A
Set1,Q1
Set2,Q3
Set3,Q2
(i) Manganin
(ii)  As ? increases A also increases
Alternatively,
½
½ 1
Set1,Q2
Set2,Q2
Set3,Q5
Phase angle = 60°
[Note : If the student  only writes, [ ] , give ½ mark]
1
1
Set1,Q3
Set2,Q1
Set3,Q4
Between plates of capacitor during charging / discharging
Alternatively,
In the region of time varying electric field
1
1
Set1,Q4
Set2,Q5
Set3,Q1
(i) P = NOT gate
(ii) Q = OR gate
½
½ 1
Set1,Q5
Set2,Q4
Set3,Q3
Def: The average time, between successive collisions of electrons, (in a
conductor) is known as relaxation time
1
1
Section B
Set1,Q6
Set2,Q6
Set3,Q10
The field inside a conductor is zero.
Sensitive instruments are shielded from outside electrical influences by
enclosing them in a hollow conductor .
Potential inside the cavity is not zero/ potential is constant.
½
1
½ 2
Set1,Q7
Set2,Q7
Set3,Q8
Any two properties of electromagnetic waves
Such as (a) transverse nature (b) does not get deflected by electric fields or
magnetic fields (c) same speed in vacuum for all waves (d) no material
medium required for propagation (e) they get refracted, diffracted and
polarised / (any two properties)
Electric charges present on a plane, kept normal to the direction of
propagation of an e.m. wave can be set and sustained in motion by the electric
and magnetic field of the electromagnetic wave.  The charges thus acquire
energy and momentum from the waves.
½ + ½
1
Electrostatic Shielding  ½
Using this property in actual practice 1
Potential in a cavity  ½
Two properties of electromagnetic waves ½ +½
Showing e m waves have momentum  1
Page 2

Page 1 of 13 Final Draft         11/03/16 03:00 p.m
.
MARKING SCHEME
SET 55/1/S
Q. No. Expected Answer / Value Points Marks Total
Marks
Section A
Set1,Q1
Set2,Q3
Set3,Q2
(i) Manganin
(ii)  As ? increases A also increases
Alternatively,
½
½ 1
Set1,Q2
Set2,Q2
Set3,Q5
Phase angle = 60°
[Note : If the student  only writes, [ ] , give ½ mark]
1
1
Set1,Q3
Set2,Q1
Set3,Q4
Between plates of capacitor during charging / discharging
Alternatively,
In the region of time varying electric field
1
1
Set1,Q4
Set2,Q5
Set3,Q1
(i) P = NOT gate
(ii) Q = OR gate
½
½ 1
Set1,Q5
Set2,Q4
Set3,Q3
Def: The average time, between successive collisions of electrons, (in a
conductor) is known as relaxation time
1
1
Section B
Set1,Q6
Set2,Q6
Set3,Q10
The field inside a conductor is zero.
Sensitive instruments are shielded from outside electrical influences by
enclosing them in a hollow conductor .
Potential inside the cavity is not zero/ potential is constant.
½
1
½ 2
Set1,Q7
Set2,Q7
Set3,Q8
Any two properties of electromagnetic waves
Such as (a) transverse nature (b) does not get deflected by electric fields or
magnetic fields (c) same speed in vacuum for all waves (d) no material
medium required for propagation (e) they get refracted, diffracted and
polarised / (any two properties)
Electric charges present on a plane, kept normal to the direction of
propagation of an e.m. wave can be set and sustained in motion by the electric
and magnetic field of the electromagnetic wave.  The charges thus acquire
energy and momentum from the waves.
½ + ½
1
Electrostatic Shielding  ½
Using this property in actual practice 1
Potential in a cavity  ½
Two properties of electromagnetic waves ½ +½
Showing e m waves have momentum  1
Page 2 of 13 Final Draft         11/03/16 03:00 p.m
.
Alternatively
they carry momentum. 2
Set1,Q8
Set2,Q8
Set3,Q9
Diffraction effects are observed for beams of electrons scattered by the
crystals
1.227nm
V
? ?
1.227
120
nm
? ?
Value ?= 0.112nm
Alternatively
=
=0.112nm
½
½
½
½
½
½
½ 2
Set1,Q9
Set2,Q10
Set3,Q7
(i) Transducer: The device which converts one form of energy into
another
(ii) Repeater:  A repeater picks up signal, amplifies and retransmits them
1
1
2
Set1,Q10
Set2,Q9
Set3,Q6
(i)
2
0
r r n ?
21.2x10
-11
= 5.3x10
-11
n
2
implies n = 2
(ii)
2
13.6eV
E
n
?
?
=  =-3.4eV
[Award ½ mark if the student just writes E= E
1
/4]
OR
(i) Energy of photon = = =4.5
(ii) The corresponding transition is B
½
½
½
½
½ + ½
+ ½
½
2
2
Principle ½
Calculation of ? 1 ½
Function of Transducer 1
Function of Repeater  1
Finding the principal quantum number 1
Finding the total energy 1
Calculation of energy of photon 1½
Identification of transistion  ½
Page 3

Page 1 of 13 Final Draft         11/03/16 03:00 p.m
.
MARKING SCHEME
SET 55/1/S
Q. No. Expected Answer / Value Points Marks Total
Marks
Section A
Set1,Q1
Set2,Q3
Set3,Q2
(i) Manganin
(ii)  As ? increases A also increases
Alternatively,
½
½ 1
Set1,Q2
Set2,Q2
Set3,Q5
Phase angle = 60°
[Note : If the student  only writes, [ ] , give ½ mark]
1
1
Set1,Q3
Set2,Q1
Set3,Q4
Between plates of capacitor during charging / discharging
Alternatively,
In the region of time varying electric field
1
1
Set1,Q4
Set2,Q5
Set3,Q1
(i) P = NOT gate
(ii) Q = OR gate
½
½ 1
Set1,Q5
Set2,Q4
Set3,Q3
Def: The average time, between successive collisions of electrons, (in a
conductor) is known as relaxation time
1
1
Section B
Set1,Q6
Set2,Q6
Set3,Q10
The field inside a conductor is zero.
Sensitive instruments are shielded from outside electrical influences by
enclosing them in a hollow conductor .
Potential inside the cavity is not zero/ potential is constant.
½
1
½ 2
Set1,Q7
Set2,Q7
Set3,Q8
Any two properties of electromagnetic waves
Such as (a) transverse nature (b) does not get deflected by electric fields or
magnetic fields (c) same speed in vacuum for all waves (d) no material
medium required for propagation (e) they get refracted, diffracted and
polarised / (any two properties)
Electric charges present on a plane, kept normal to the direction of
propagation of an e.m. wave can be set and sustained in motion by the electric
and magnetic field of the electromagnetic wave.  The charges thus acquire
energy and momentum from the waves.
½ + ½
1
Electrostatic Shielding  ½
Using this property in actual practice 1
Potential in a cavity  ½
Two properties of electromagnetic waves ½ +½
Showing e m waves have momentum  1
Page 2 of 13 Final Draft         11/03/16 03:00 p.m
.
Alternatively
they carry momentum. 2
Set1,Q8
Set2,Q8
Set3,Q9
Diffraction effects are observed for beams of electrons scattered by the
crystals
1.227nm
V
? ?
1.227
120
nm
? ?
Value ?= 0.112nm
Alternatively
=
=0.112nm
½
½
½
½
½
½
½ 2
Set1,Q9
Set2,Q10
Set3,Q7
(i) Transducer: The device which converts one form of energy into
another
(ii) Repeater:  A repeater picks up signal, amplifies and retransmits them
1
1
2
Set1,Q10
Set2,Q9
Set3,Q6
(i)
2
0
r r n ?
21.2x10
-11
= 5.3x10
-11
n
2
implies n = 2
(ii)
2
13.6eV
E
n
?
?
=  =-3.4eV
[Award ½ mark if the student just writes E= E
1
/4]
OR
(i) Energy of photon = = =4.5
(ii) The corresponding transition is B
½
½
½
½
½ + ½
+ ½
½
2
2
Principle ½
Calculation of ? 1 ½
Function of Transducer 1
Function of Repeater  1
Finding the principal quantum number 1
Finding the total energy 1
Calculation of energy of photon 1½
Identification of transistion  ½
Page 3 of 13 Final Draft         11/03/16 03:00 p.m
.
Section C
Set1,Q11
Set2,Q20
Set3,Q22
E
+q
=      Kq / (r
2
+a
2
)   and E
-q
=   Kq / (r
2
+a
2
)
The two Electric fields have equal magnitudes and their directions are as
shown in diagram
Components along dipole axis get added up while normal components cancel
each other.
E=-[ E
-q
+E
+q
] cos? r so E = -
= ( =
Direction of electric field is opposite to that of dipole moment.
1
½
½
½
½ 3
Set1,Q12
Set2,Q15
Set3,Q16
a) Zero
b) We have  C
series
=  = 1 µF
Also, C
parallel
= (3+3+3)=9µF
Energy stored =
Energy in series combination =
Energy in parallel combination =
Ratio = 1:9
½
½
½
½
½
½
3
Diagram 1
Deriving expression for E
eq
1 ½
Direction of E
eq
½
a) To find charge accumulated in capacitor  C
2
½
b) To find the ratio of energy stored 2  ½
Page 4

Page 1 of 13 Final Draft         11/03/16 03:00 p.m
.
MARKING SCHEME
SET 55/1/S
Q. No. Expected Answer / Value Points Marks Total
Marks
Section A
Set1,Q1
Set2,Q3
Set3,Q2
(i) Manganin
(ii)  As ? increases A also increases
Alternatively,
½
½ 1
Set1,Q2
Set2,Q2
Set3,Q5
Phase angle = 60°
[Note : If the student  only writes, [ ] , give ½ mark]
1
1
Set1,Q3
Set2,Q1
Set3,Q4
Between plates of capacitor during charging / discharging
Alternatively,
In the region of time varying electric field
1
1
Set1,Q4
Set2,Q5
Set3,Q1
(i) P = NOT gate
(ii) Q = OR gate
½
½ 1
Set1,Q5
Set2,Q4
Set3,Q3
Def: The average time, between successive collisions of electrons, (in a
conductor) is known as relaxation time
1
1
Section B
Set1,Q6
Set2,Q6
Set3,Q10
The field inside a conductor is zero.
Sensitive instruments are shielded from outside electrical influences by
enclosing them in a hollow conductor .
Potential inside the cavity is not zero/ potential is constant.
½
1
½ 2
Set1,Q7
Set2,Q7
Set3,Q8
Any two properties of electromagnetic waves
Such as (a) transverse nature (b) does not get deflected by electric fields or
magnetic fields (c) same speed in vacuum for all waves (d) no material
medium required for propagation (e) they get refracted, diffracted and
polarised / (any two properties)
Electric charges present on a plane, kept normal to the direction of
propagation of an e.m. wave can be set and sustained in motion by the electric
and magnetic field of the electromagnetic wave.  The charges thus acquire
energy and momentum from the waves.
½ + ½
1
Electrostatic Shielding  ½
Using this property in actual practice 1
Potential in a cavity  ½
Two properties of electromagnetic waves ½ +½
Showing e m waves have momentum  1
Page 2 of 13 Final Draft         11/03/16 03:00 p.m
.
Alternatively
they carry momentum. 2
Set1,Q8
Set2,Q8
Set3,Q9
Diffraction effects are observed for beams of electrons scattered by the
crystals
1.227nm
V
? ?
1.227
120
nm
? ?
Value ?= 0.112nm
Alternatively
=
=0.112nm
½
½
½
½
½
½
½ 2
Set1,Q9
Set2,Q10
Set3,Q7
(i) Transducer: The device which converts one form of energy into
another
(ii) Repeater:  A repeater picks up signal, amplifies and retransmits them
1
1
2
Set1,Q10
Set2,Q9
Set3,Q6
(i)
2
0
r r n ?
21.2x10
-11
= 5.3x10
-11
n
2
implies n = 2
(ii)
2
13.6eV
E
n
?
?
=  =-3.4eV
[Award ½ mark if the student just writes E= E
1
/4]
OR
(i) Energy of photon = = =4.5
(ii) The corresponding transition is B
½
½
½
½
½ + ½
+ ½
½
2
2
Principle ½
Calculation of ? 1 ½
Function of Transducer 1
Function of Repeater  1
Finding the principal quantum number 1
Finding the total energy 1
Calculation of energy of photon 1½
Identification of transistion  ½
Page 3 of 13 Final Draft         11/03/16 03:00 p.m
.
Section C
Set1,Q11
Set2,Q20
Set3,Q22
E
+q
=      Kq / (r
2
+a
2
)   and E
-q
=   Kq / (r
2
+a
2
)
The two Electric fields have equal magnitudes and their directions are as
shown in diagram
Components along dipole axis get added up while normal components cancel
each other.
E=-[ E
-q
+E
+q
] cos? r so E = -
= ( =
Direction of electric field is opposite to that of dipole moment.
1
½
½
½
½ 3
Set1,Q12
Set2,Q15
Set3,Q16
a) Zero
b) We have  C
series
=  = 1 µF
Also, C
parallel
= (3+3+3)=9µF
Energy stored =
Energy in series combination =
Energy in parallel combination =
Ratio = 1:9
½
½
½
½
½
½
3
Diagram 1
Deriving expression for E
eq
1 ½
Direction of E
eq
½
a) To find charge accumulated in capacitor  C
2
½
b) To find the ratio of energy stored 2  ½
Page 4 of 13 Final Draft         11/03/16 03:00 p.m
.
Set1,Q13
Set2,Q16
Set3,Q19
a) Intensity of radiation  equals the energy of all the  Photons incident
normally  per unit area per unit time.
Alternatively, The intensity of radiation is proportional to the number
of photons emitted per unit area per unit time.
b)
c) As per Einsteinâ€™s equation,
(i) The stopping potential is same for I
1
and I
2
as they have the
same frequency.
(ii) The saturation currents are as shown , because  I
1
> I
2
> I
3

1
1
½
½ 3
Set1,Q14
Set2,Q14
Set3,Q12
(i) During Forward bias of LED, electrons move from n side to p side and
holes move from p side to n side. During recombination, energy is
released in the form of photons having energy  of the order of band
gap.
(ii) GaAs/ GaAsP (any one)
Band gap should be 1.8 eV  to 3 eV These materials have band gap which
is suitable to produce desired visible light wavelengths.
(iii)Low operational voltage, fast action , no warm up time required,  nearly
monochromatic, long life ,ruggedness, fast on and off switching capacity.
(any two points)
1
½
½
½ + ½ 3
Set1,Q15
Set2,Q13
Set3,Q14

Capacitance = C =
= F
½
½
a) Definition of intensity      1
b) Required  graph      1
c) Explanation of nature of the curves      1
(i) To explain the process of emission 1
(ii)  Material preferred to make LED and reason ½ + ½
(iii)Two advantages of using LED  ½ + ½
Calculation of capacitance  1
Calculation of Impedence  1
Calculation of Power dissipitated 1
Page 5

Page 1 of 13 Final Draft         11/03/16 03:00 p.m
.
MARKING SCHEME
SET 55/1/S
Q. No. Expected Answer / Value Points Marks Total
Marks
Section A
Set1,Q1
Set2,Q3
Set3,Q2
(i) Manganin
(ii)  As ? increases A also increases
Alternatively,
½
½ 1
Set1,Q2
Set2,Q2
Set3,Q5
Phase angle = 60°
[Note : If the student  only writes, [ ] , give ½ mark]
1
1
Set1,Q3
Set2,Q1
Set3,Q4
Between plates of capacitor during charging / discharging
Alternatively,
In the region of time varying electric field
1
1
Set1,Q4
Set2,Q5
Set3,Q1
(i) P = NOT gate
(ii) Q = OR gate
½
½ 1
Set1,Q5
Set2,Q4
Set3,Q3
Def: The average time, between successive collisions of electrons, (in a
conductor) is known as relaxation time
1
1
Section B
Set1,Q6
Set2,Q6
Set3,Q10
The field inside a conductor is zero.
Sensitive instruments are shielded from outside electrical influences by
enclosing them in a hollow conductor .
Potential inside the cavity is not zero/ potential is constant.
½
1
½ 2
Set1,Q7
Set2,Q7
Set3,Q8
Any two properties of electromagnetic waves
Such as (a) transverse nature (b) does not get deflected by electric fields or
magnetic fields (c) same speed in vacuum for all waves (d) no material
medium required for propagation (e) they get refracted, diffracted and
polarised / (any two properties)
Electric charges present on a plane, kept normal to the direction of
propagation of an e.m. wave can be set and sustained in motion by the electric
and magnetic field of the electromagnetic wave.  The charges thus acquire
energy and momentum from the waves.
½ + ½
1
Electrostatic Shielding  ½
Using this property in actual practice 1
Potential in a cavity  ½
Two properties of electromagnetic waves ½ +½
Showing e m waves have momentum  1
Page 2 of 13 Final Draft         11/03/16 03:00 p.m
.
Alternatively
they carry momentum. 2
Set1,Q8
Set2,Q8
Set3,Q9
Diffraction effects are observed for beams of electrons scattered by the
crystals
1.227nm
V
? ?
1.227
120
nm
? ?
Value ?= 0.112nm
Alternatively
=
=0.112nm
½
½
½
½
½
½
½ 2
Set1,Q9
Set2,Q10
Set3,Q7
(i) Transducer: The device which converts one form of energy into
another
(ii) Repeater:  A repeater picks up signal, amplifies and retransmits them
1
1
2
Set1,Q10
Set2,Q9
Set3,Q6
(i)
2
0
r r n ?
21.2x10
-11
= 5.3x10
-11
n
2
implies n = 2
(ii)
2
13.6eV
E
n
?
?
=  =-3.4eV
[Award ½ mark if the student just writes E= E
1
/4]
OR
(i) Energy of photon = = =4.5
(ii) The corresponding transition is B
½
½
½
½
½ + ½
+ ½
½
2
2
Principle ½
Calculation of ? 1 ½
Function of Transducer 1
Function of Repeater  1
Finding the principal quantum number 1
Finding the total energy 1
Calculation of energy of photon 1½
Identification of transistion  ½
Page 3 of 13 Final Draft         11/03/16 03:00 p.m
.
Section C
Set1,Q11
Set2,Q20
Set3,Q22
E
+q
=      Kq / (r
2
+a
2
)   and E
-q
=   Kq / (r
2
+a
2
)
The two Electric fields have equal magnitudes and their directions are as
shown in diagram
Components along dipole axis get added up while normal components cancel
each other.
E=-[ E
-q
+E
+q
] cos? r so E = -
= ( =
Direction of electric field is opposite to that of dipole moment.
1
½
½
½
½ 3
Set1,Q12
Set2,Q15
Set3,Q16
a) Zero
b) We have  C
series
=  = 1 µF
Also, C
parallel
= (3+3+3)=9µF
Energy stored =
Energy in series combination =
Energy in parallel combination =
Ratio = 1:9
½
½
½
½
½
½
3
Diagram 1
Deriving expression for E
eq
1 ½
Direction of E
eq
½
a) To find charge accumulated in capacitor  C
2
½
b) To find the ratio of energy stored 2  ½
Page 4 of 13 Final Draft         11/03/16 03:00 p.m
.
Set1,Q13
Set2,Q16
Set3,Q19
a) Intensity of radiation  equals the energy of all the  Photons incident
normally  per unit area per unit time.
Alternatively, The intensity of radiation is proportional to the number
of photons emitted per unit area per unit time.
b)
c) As per Einsteinâ€™s equation,
(i) The stopping potential is same for I
1
and I
2
as they have the
same frequency.
(ii) The saturation currents are as shown , because  I
1
> I
2
> I
3

1
1
½
½ 3
Set1,Q14
Set2,Q14
Set3,Q12
(i) During Forward bias of LED, electrons move from n side to p side and
holes move from p side to n side. During recombination, energy is
released in the form of photons having energy  of the order of band
gap.
(ii) GaAs/ GaAsP (any one)
Band gap should be 1.8 eV  to 3 eV These materials have band gap which
is suitable to produce desired visible light wavelengths.
(iii)Low operational voltage, fast action , no warm up time required,  nearly
monochromatic, long life ,ruggedness, fast on and off switching capacity.
(any two points)
1
½
½
½ + ½ 3
Set1,Q15
Set2,Q13
Set3,Q14

Capacitance = C =
= F
½
½
a) Definition of intensity      1
b) Required  graph      1
c) Explanation of nature of the curves      1
(i) To explain the process of emission 1
(ii)  Material preferred to make LED and reason ½ + ½
(iii)Two advantages of using LED  ½ + ½
Calculation of capacitance  1
Calculation of Impedence  1
Calculation of Power dissipitated 1
Page 5 of 13 Final Draft         11/03/16 03:00 p.m
.
= 2.5 x 10
-5
F
Impedence = resistance( since V and I are in phase)
Impedence = 100?
Power discipated =
=  = 400 watt
½
½
½
½ 3
Set1,Q16
Set2,Q19
Set3,Q20

(i)
= 2cosA/2 =
A= 60°
(ii)  =
(iii)
Lies between  and
Hence, TIR takes place.
Alternatively,
which is less  than
angle of incidence >
TIR
½
½
½
½
1 3
Set1,Q17
Set2,Q18
Set3,Q17
Nuclear force is Saturated,  or short ranged  [ any one]
The final system is more tightly bound when heavy nucleus undergoes
nuclear fission. Hence, there is a release of energy.
The final system is more tightly bound when light nuclei  undergoes nuclear
fusion. Hence, there is a releases of energy.
1½
½
½
½
(i) To calculate angle of prism           1 ½
(ii) To trace the path of incident light inside the  prism    1 ½
To plot (BE/A) vs mass number graph 1½
To state the property of nuclear force  ½
To explain the release of energy in fission and ½ + ½
fusion using the graph
```
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