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 ‘Q’ 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 ‘Q’ 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 Bohr’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 Bohr’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 ‘Q’ 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 Bohr’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 Bohr’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 ‘Q’ 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 Bohr’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 Bohr’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 ‘Q’ 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 Bohr’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 Bohr’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 
‘R’ 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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