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

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

``` Page 1

Page | 1
East Final Draft-11 March 2016 2:00 p.m.
MARKING SCHEME
SET 55/1/E
Q. No. Expected Answer / Value Points Marks Total
Marks
(SECTION A)
Set1,Q1
Set2,Q5
Set3,Q2
Potentiometer â€˜Qâ€™ will be preferred
Reason:- ?????????????????????? ?
1
?????????????????? ???????????????? (?? )
Since potential gradient is less, sensitivity is more.
[Note: Also accept if the student just writs that potential gradient is less for
potentiometer Q]
½
½ 1
Set1,Q2
Set2,Q3
Set3,Q1
Graph of V
Graph of I
½
½ 1
Set1,Q3
Set2,Q2
Set3,Q4
[Note: If students write truth table correctly then award ½ mark.]
1
1
Set1,Q4
Set2,Q4
Set3,Q5
For a.c. source, circuit is complete due to the presence of displacement
current in the capacitor. For steady dc, there is no displacement current,
therefore, circuit is not complete.
[Alternatively, Capacitive reactance ?? ?? =
1
2?????? =
1
????
So, capacitor allows easy path for a.c. source.
For d.c, f= 0, so X
c
= infinity,
So capacitor blocks d.c]
½+½
½+½ 1
Set1,Q5
Set2,Q1
Set3,Q3
Conductivity of a conductor is the current flowing per unit area per unit
electric field applied.
[Alternatively, conductivity ?? =
?? ?? ]
½
Page 2

Page | 1
East Final Draft-11 March 2016 2:00 p.m.
MARKING SCHEME
SET 55/1/E
Q. No. Expected Answer / Value Points Marks Total
Marks
(SECTION A)
Set1,Q1
Set2,Q5
Set3,Q2
Potentiometer â€˜Qâ€™ will be preferred
Reason:- ?????????????????????? ?
1
?????????????????? ???????????????? (?? )
Since potential gradient is less, sensitivity is more.
[Note: Also accept if the student just writs that potential gradient is less for
potentiometer Q]
½
½ 1
Set1,Q2
Set2,Q3
Set3,Q1
Graph of V
Graph of I
½
½ 1
Set1,Q3
Set2,Q2
Set3,Q4
[Note: If students write truth table correctly then award ½ mark.]
1
1
Set1,Q4
Set2,Q4
Set3,Q5
For a.c. source, circuit is complete due to the presence of displacement
current in the capacitor. For steady dc, there is no displacement current,
therefore, circuit is not complete.
[Alternatively, Capacitive reactance ?? ?? =
1
2?????? =
1
????
So, capacitor allows easy path for a.c. source.
For d.c, f= 0, so X
c
= infinity,
So capacitor blocks d.c]
½+½
½+½ 1
Set1,Q5
Set2,Q1
Set3,Q3
Conductivity of a conductor is the current flowing per unit area per unit
electric field applied.
[Alternatively, conductivity ?? =
?? ?? ]
½
Page | 2
East Final Draft-11 March 2016 2:00 p.m.
Depends upon number density i.e. nature of material, and relaxation time i.e.
temperature.
½ 1
(SECTION B)
Set1,Q6
Set2,Q8
Set3,Q7
???? =???? sin?? ????
Work done against the restoring torque
???? =t d?
?,?? = pE sin?? ???? ?? 1
?? 0

=  pE cos?? 0
- cos?? 1
½
½
½
½ 2
Set1,Q7
Set2,Q9
Set3,Q6
? 2???? =
?? ?? ???? de Broglie wavelength, ? =
?? ???? For electron moving in the n
th
orbit, 2???? =???? ? mvr =
?? ?? 2?? = L (orbital angular momentum)
This is Bohrâ€™s Postulate of quantization of orbital angular momentum.
½
½
½
½ 2
Set1,Q8
Set2,Q10
Set3,Q9
Concept of mobile telephony is to divide the service area into a suitable
number of cells centred on an office MTSO (Mobile Telephone Switching
Office) / Mobile telephony means that you can talk to any person from
anywhere.
Explanation:
1. Entire service area is divided into smaller parts called cells.
2. Each cell has a base station to receive and send signals to mobiles.
3. Each base station is linked to MTSO. MTSO co-ordinates between
½
½
½
½ 2
Derivation of expression for work done     2
de-Broglie wavelength          ½
Condition of stationary orbits          ½
Obtaining Bohrâ€™s Postulate of quantization of orbital angular momentum.1
Explanation of the concept of Mobile Telephony  ½
Explanation of working  1½
Page 3

Page | 1
East Final Draft-11 March 2016 2:00 p.m.
MARKING SCHEME
SET 55/1/E
Q. No. Expected Answer / Value Points Marks Total
Marks
(SECTION A)
Set1,Q1
Set2,Q5
Set3,Q2
Potentiometer â€˜Qâ€™ will be preferred
Reason:- ?????????????????????? ?
1
?????????????????? ???????????????? (?? )
Since potential gradient is less, sensitivity is more.
[Note: Also accept if the student just writs that potential gradient is less for
potentiometer Q]
½
½ 1
Set1,Q2
Set2,Q3
Set3,Q1
Graph of V
Graph of I
½
½ 1
Set1,Q3
Set2,Q2
Set3,Q4
[Note: If students write truth table correctly then award ½ mark.]
1
1
Set1,Q4
Set2,Q4
Set3,Q5
For a.c. source, circuit is complete due to the presence of displacement
current in the capacitor. For steady dc, there is no displacement current,
therefore, circuit is not complete.
[Alternatively, Capacitive reactance ?? ?? =
1
2?????? =
1
????
So, capacitor allows easy path for a.c. source.
For d.c, f= 0, so X
c
= infinity,
So capacitor blocks d.c]
½+½
½+½ 1
Set1,Q5
Set2,Q1
Set3,Q3
Conductivity of a conductor is the current flowing per unit area per unit
electric field applied.
[Alternatively, conductivity ?? =
?? ?? ]
½
Page | 2
East Final Draft-11 March 2016 2:00 p.m.
Depends upon number density i.e. nature of material, and relaxation time i.e.
temperature.
½ 1
(SECTION B)
Set1,Q6
Set2,Q8
Set3,Q7
???? =???? sin?? ????
Work done against the restoring torque
???? =t d?
?,?? = pE sin?? ???? ?? 1
?? 0

=  pE cos?? 0
- cos?? 1
½
½
½
½ 2
Set1,Q7
Set2,Q9
Set3,Q6
? 2???? =
?? ?? ???? de Broglie wavelength, ? =
?? ???? For electron moving in the n
th
orbit, 2???? =???? ? mvr =
?? ?? 2?? = L (orbital angular momentum)
This is Bohrâ€™s Postulate of quantization of orbital angular momentum.
½
½
½
½ 2
Set1,Q8
Set2,Q10
Set3,Q9
Concept of mobile telephony is to divide the service area into a suitable
number of cells centred on an office MTSO (Mobile Telephone Switching
Office) / Mobile telephony means that you can talk to any person from
anywhere.
Explanation:
1. Entire service area is divided into smaller parts called cells.
2. Each cell has a base station to receive and send signals to mobiles.
3. Each base station is linked to MTSO. MTSO co-ordinates between
½
½
½
½ 2
Derivation of expression for work done     2
de-Broglie wavelength          ½
Condition of stationary orbits          ½
Obtaining Bohrâ€™s Postulate of quantization of orbital angular momentum.1
Explanation of the concept of Mobile Telephony  ½
Explanation of working  1½
Page | 3
East Final Draft-11 March 2016 2:00 p.m.
base station and TCO (Telephone Control Office)
Set1,Q9
Set2,Q7
Set3,Q10
1
?? ?????? =?? 1
?? ?? 2
-
1
?? ?? 2
1
?? ?????? =
1
2²
-
1
3²
??
?? ?????? =
36
5?? The energy of the incident photon = 12.5 eV
Energy of ground state = -13.6eV
?, Energy after absorption of photon can be -1.1eV
This means that electron can go to the excited state ?? ?? = 3. It emits photons
of maximum wavelength on going to ?? ?? = 2 i.e.
=
36
5×1.1×10
7
= 6.555 × 10
-7
m = 6555 ?? °
It belongs to Balmer Series.
[Note:-
(1) If student just writes the formula
1
?? ?????? =?? 1
?? ?? 2
-
1
?? ?? 2
for the wavelength of different levels in the Hydrogen spectrum and
calculates ?? ??????  for any series, award full 3 marks.
(2) Also award full 3 marks if the student writes that the energy of the excited
state cannot be 12.5eV]
OR
½
½
½
½
Formula ½
Calculation ½
Longest Wavelength  ½
Identification of Series ½
Formula        1
Calculation   1
Page 4

Page | 1
East Final Draft-11 March 2016 2:00 p.m.
MARKING SCHEME
SET 55/1/E
Q. No. Expected Answer / Value Points Marks Total
Marks
(SECTION A)
Set1,Q1
Set2,Q5
Set3,Q2
Potentiometer â€˜Qâ€™ will be preferred
Reason:- ?????????????????????? ?
1
?????????????????? ???????????????? (?? )
Since potential gradient is less, sensitivity is more.
[Note: Also accept if the student just writs that potential gradient is less for
potentiometer Q]
½
½ 1
Set1,Q2
Set2,Q3
Set3,Q1
Graph of V
Graph of I
½
½ 1
Set1,Q3
Set2,Q2
Set3,Q4
[Note: If students write truth table correctly then award ½ mark.]
1
1
Set1,Q4
Set2,Q4
Set3,Q5
For a.c. source, circuit is complete due to the presence of displacement
current in the capacitor. For steady dc, there is no displacement current,
therefore, circuit is not complete.
[Alternatively, Capacitive reactance ?? ?? =
1
2?????? =
1
????
So, capacitor allows easy path for a.c. source.
For d.c, f= 0, so X
c
= infinity,
So capacitor blocks d.c]
½+½
½+½ 1
Set1,Q5
Set2,Q1
Set3,Q3
Conductivity of a conductor is the current flowing per unit area per unit
electric field applied.
[Alternatively, conductivity ?? =
?? ?? ]
½
Page | 2
East Final Draft-11 March 2016 2:00 p.m.
Depends upon number density i.e. nature of material, and relaxation time i.e.
temperature.
½ 1
(SECTION B)
Set1,Q6
Set2,Q8
Set3,Q7
???? =???? sin?? ????
Work done against the restoring torque
???? =t d?
?,?? = pE sin?? ???? ?? 1
?? 0

=  pE cos?? 0
- cos?? 1
½
½
½
½ 2
Set1,Q7
Set2,Q9
Set3,Q6
? 2???? =
?? ?? ???? de Broglie wavelength, ? =
?? ???? For electron moving in the n
th
orbit, 2???? =???? ? mvr =
?? ?? 2?? = L (orbital angular momentum)
This is Bohrâ€™s Postulate of quantization of orbital angular momentum.
½
½
½
½ 2
Set1,Q8
Set2,Q10
Set3,Q9
Concept of mobile telephony is to divide the service area into a suitable
number of cells centred on an office MTSO (Mobile Telephone Switching
Office) / Mobile telephony means that you can talk to any person from
anywhere.
Explanation:
1. Entire service area is divided into smaller parts called cells.
2. Each cell has a base station to receive and send signals to mobiles.
3. Each base station is linked to MTSO. MTSO co-ordinates between
½
½
½
½ 2
Derivation of expression for work done     2
de-Broglie wavelength          ½
Condition of stationary orbits          ½
Obtaining Bohrâ€™s Postulate of quantization of orbital angular momentum.1
Explanation of the concept of Mobile Telephony  ½
Explanation of working  1½
Page | 3
East Final Draft-11 March 2016 2:00 p.m.
base station and TCO (Telephone Control Office)
Set1,Q9
Set2,Q7
Set3,Q10
1
?? ?????? =?? 1
?? ?? 2
-
1
?? ?? 2
1
?? ?????? =
1
2²
-
1
3²
??
?? ?????? =
36
5?? The energy of the incident photon = 12.5 eV
Energy of ground state = -13.6eV
?, Energy after absorption of photon can be -1.1eV
This means that electron can go to the excited state ?? ?? = 3. It emits photons
of maximum wavelength on going to ?? ?? = 2 i.e.
=
36
5×1.1×10
7
= 6.555 × 10
-7
m = 6555 ?? °
It belongs to Balmer Series.
[Note:-
(1) If student just writes the formula
1
?? ?????? =?? 1
?? ?? 2
-
1
?? ?? 2
for the wavelength of different levels in the Hydrogen spectrum and
calculates ?? ??????  for any series, award full 3 marks.
(2) Also award full 3 marks if the student writes that the energy of the excited
state cannot be 12.5eV]
OR
½
½
½
½
Formula ½
Calculation ½
Longest Wavelength  ½
Identification of Series ½
Formula        1
Calculation   1
Page | 4
East Final Draft-11 March 2016 2:00 p.m.
?? =
?? ?? 2?????? And ?? =
1
??
?? 2
?? 2
4?? 2
?? ?? 2
So, ?? =??
2?? ?? 2
?? ??    In first excited state
n = 2
So velocity ?? 2
=
2?? ???? 2
2?? = 1.09 × 10
6
ms
-1
OR
Velocity of electron, ?? ?? =
1
137
?? ??
In first excited state n =2
So velocity in first excited state (?? 2
)
=
1
137
?? 2
= 1.09 × 10
6
ms
-1
½
½
½
½
1
½
½ 2
Set1,Q10
Set2,Q6
Set3,Q8
(i) Infrared waves are produced by hot bodies and molecules.
Important use( Any one)
To treat muscular strains/ To reveal the secret writings on the ancient
walls/ For producing dehydrated fruits/ Solar heater/ Solar cooker
Ozone layer protects us from harmful U-V rays
½
½
1 2
(SECTION C)
Set1,Q11
Set2,Q15
Set3,Q12

(i) Electric flux through a Gaussian surface,
?? =
?????????? ???????????????? ?? ?? ???????? ?
0
½
(i) How are infrared waves produced    ½
One important use                             ½
(ii) Reason (any one)      1
(i) Electric Flux through the shell 1
(ii) Statement of Law        1
(iii) Force on charge at C
1
2

Force on charge at A
1
2

Page 5

Page | 1
East Final Draft-11 March 2016 2:00 p.m.
MARKING SCHEME
SET 55/1/E
Q. No. Expected Answer / Value Points Marks Total
Marks
(SECTION A)
Set1,Q1
Set2,Q5
Set3,Q2
Potentiometer â€˜Qâ€™ will be preferred
Reason:- ?????????????????????? ?
1
?????????????????? ???????????????? (?? )
Since potential gradient is less, sensitivity is more.
[Note: Also accept if the student just writs that potential gradient is less for
potentiometer Q]
½
½ 1
Set1,Q2
Set2,Q3
Set3,Q1
Graph of V
Graph of I
½
½ 1
Set1,Q3
Set2,Q2
Set3,Q4
[Note: If students write truth table correctly then award ½ mark.]
1
1
Set1,Q4
Set2,Q4
Set3,Q5
For a.c. source, circuit is complete due to the presence of displacement
current in the capacitor. For steady dc, there is no displacement current,
therefore, circuit is not complete.
[Alternatively, Capacitive reactance ?? ?? =
1
2?????? =
1
????
So, capacitor allows easy path for a.c. source.
For d.c, f= 0, so X
c
= infinity,
So capacitor blocks d.c]
½+½
½+½ 1
Set1,Q5
Set2,Q1
Set3,Q3
Conductivity of a conductor is the current flowing per unit area per unit
electric field applied.
[Alternatively, conductivity ?? =
?? ?? ]
½
Page | 2
East Final Draft-11 March 2016 2:00 p.m.
Depends upon number density i.e. nature of material, and relaxation time i.e.
temperature.
½ 1
(SECTION B)
Set1,Q6
Set2,Q8
Set3,Q7
???? =???? sin?? ????
Work done against the restoring torque
???? =t d?
?,?? = pE sin?? ???? ?? 1
?? 0

=  pE cos?? 0
- cos?? 1
½
½
½
½ 2
Set1,Q7
Set2,Q9
Set3,Q6
? 2???? =
?? ?? ???? de Broglie wavelength, ? =
?? ???? For electron moving in the n
th
orbit, 2???? =???? ? mvr =
?? ?? 2?? = L (orbital angular momentum)
This is Bohrâ€™s Postulate of quantization of orbital angular momentum.
½
½
½
½ 2
Set1,Q8
Set2,Q10
Set3,Q9
Concept of mobile telephony is to divide the service area into a suitable
number of cells centred on an office MTSO (Mobile Telephone Switching
Office) / Mobile telephony means that you can talk to any person from
anywhere.
Explanation:
1. Entire service area is divided into smaller parts called cells.
2. Each cell has a base station to receive and send signals to mobiles.
3. Each base station is linked to MTSO. MTSO co-ordinates between
½
½
½
½ 2
Derivation of expression for work done     2
de-Broglie wavelength          ½
Condition of stationary orbits          ½
Obtaining Bohrâ€™s Postulate of quantization of orbital angular momentum.1
Explanation of the concept of Mobile Telephony  ½
Explanation of working  1½
Page | 3
East Final Draft-11 March 2016 2:00 p.m.
base station and TCO (Telephone Control Office)
Set1,Q9
Set2,Q7
Set3,Q10
1
?? ?????? =?? 1
?? ?? 2
-
1
?? ?? 2
1
?? ?????? =
1
2²
-
1
3²
??
?? ?????? =
36
5?? The energy of the incident photon = 12.5 eV
Energy of ground state = -13.6eV
?, Energy after absorption of photon can be -1.1eV
This means that electron can go to the excited state ?? ?? = 3. It emits photons
of maximum wavelength on going to ?? ?? = 2 i.e.
=
36
5×1.1×10
7
= 6.555 × 10
-7
m = 6555 ?? °
It belongs to Balmer Series.
[Note:-
(1) If student just writes the formula
1
?? ?????? =?? 1
?? ?? 2
-
1
?? ?? 2
for the wavelength of different levels in the Hydrogen spectrum and
calculates ?? ??????  for any series, award full 3 marks.
(2) Also award full 3 marks if the student writes that the energy of the excited
state cannot be 12.5eV]
OR
½
½
½
½
Formula ½
Calculation ½
Longest Wavelength  ½
Identification of Series ½
Formula        1
Calculation   1
Page | 4
East Final Draft-11 March 2016 2:00 p.m.
?? =
?? ?? 2?????? And ?? =
1
??
?? 2
?? 2
4?? 2
?? ?? 2
So, ?? =??
2?? ?? 2
?? ??    In first excited state
n = 2
So velocity ?? 2
=
2?? ???? 2
2?? = 1.09 × 10
6
ms
-1
OR
Velocity of electron, ?? ?? =
1
137
?? ??
In first excited state n =2
So velocity in first excited state (?? 2
)
=
1
137
?? 2
= 1.09 × 10
6
ms
-1
½
½
½
½
1
½
½ 2
Set1,Q10
Set2,Q6
Set3,Q8
(i) Infrared waves are produced by hot bodies and molecules.
Important use( Any one)
To treat muscular strains/ To reveal the secret writings on the ancient
walls/ For producing dehydrated fruits/ Solar heater/ Solar cooker
Ozone layer protects us from harmful U-V rays
½
½
1 2
(SECTION C)
Set1,Q11
Set2,Q15
Set3,Q12

(i) Electric flux through a Gaussian surface,
?? =
?????????? ???????????????? ?? ?? ???????? ?
0
½
(i) How are infrared waves produced    ½
One important use                             ½
(ii) Reason (any one)      1
(i) Electric Flux through the shell 1
(ii) Statement of Law        1
(iii) Force on charge at C
1
2

Force on charge at A
1
2

Page | 5
East Final Draft-11 March 2016 2:00 p.m.
Net charge enclosed inside the shell q=0
? Electric flux through the shell
?? ?
?? =0
Award ½ mark even when the student writes - Electric flux through the shell
is zero as electric field inside the shell is zero.
(ii) Gauss Law- Electric flux through a Gaussian surface is
1
?
0

times the net charge enclosed with in it.
Alternatively,  ??
.????
=
?? ?? 0
(iii) Force on the charge at the centre i.e. Charget
?? 2
= 0
?? ?? =
1
4?? ?? 0
2?? × ?? +
?? 2

?? 2
=
1
4?? ?? 0
3?? 2
?? 2
½
1
½
½ 3
Set1,Q12
Set2,Q13
Set3,Q21
A galvanometer is converted into a voltmeter by connecting a high resistance
â€˜Râ€™ in series with it.
A galvanometer is converted into an ammeter by connecting a small
resistance (called shunt) in parallel with it.
½
½
½
½
How galvanometer is converted in to a voltmeter and an Ammeter   ½ + ½
Diagram for conversion of galvanometer into a voltmeter and an      ½ + ½
Ammeter.
Resistance of each arrangement         ½ + ½
```
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