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 . 
(any other relevant answer.) 
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 . 
(any other relevant answer.) 
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  
Radiation Pressure – Electromagnetic waves exert radiation pressure. Hence, 
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 
to receiver  
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 . 
(any other relevant answer.) 
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  
Radiation Pressure – Electromagnetic waves exert radiation pressure. Hence, 
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 
to receiver  
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 . 
(any other relevant answer.) 
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  
Radiation Pressure – Electromagnetic waves exert radiation pressure. Hence, 
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 
to receiver  
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 Einstein’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 . 
(any other relevant answer.) 
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  
Radiation Pressure – Electromagnetic waves exert radiation pressure. Hence, 
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 
to receiver  
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 Einstein’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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