# Physics: CBSE Marking Scheme with Solution (2019-20) Notes | Study Physics Class 12 - NEET

## NEET: Physics: CBSE Marking Scheme with Solution (2019-20) Notes | Study Physics Class 12 - NEET

The document Physics: CBSE Marking Scheme with Solution (2019-20) Notes | Study Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
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``` Page 1

1

Class -XII
PHYSICS
SQP Marking Scheme
2019-20

Section – A
1. a , ?=  (for one face) 1

2. b , Conductor  1

3. a , 1?. 1

4. c ,12.0kJ 1

5. a , speed 1

6. d, virtual and inverted 1

7. a, straight line 1

8. d, 60
O
1

9. b, work function 1

10. b, third orbit 1

11. 45
O
or vertical 1

12. 2 H 1

13. double 1

14. 1.227 A
o
1

15. 60° 1

16. Difference in initial mass energy and energy associated with mass of products
Or
Total Kinetic energy gained in the process
1

17. Increases   1

18. N o/8 1

19. 0.79 eV 1

20. Diodes with band gap energy in the visible spectrum range can function as LED 1
Page 2

1

Class -XII
PHYSICS
SQP Marking Scheme
2019-20

Section – A
1. a , ?=  (for one face) 1

2. b , Conductor  1

3. a , 1?. 1

4. c ,12.0kJ 1

5. a , speed 1

6. d, virtual and inverted 1

7. a, straight line 1

8. d, 60
O
1

9. b, work function 1

10. b, third orbit 1

11. 45
O
or vertical 1

12. 2 H 1

13. double 1

14. 1.227 A
o
1

15. 60° 1

16. Difference in initial mass energy and energy associated with mass of products
Or
Total Kinetic energy gained in the process
1

17. Increases   1

18. N o/8 1

19. 0.79 eV 1

20. Diodes with band gap energy in the visible spectrum range can function as LED 1
2

OR
Any one use
Section – B

21. When electric field E is applied on conductor force acting on free electrons
= -e
m = -e
=
Average thermal velocity of electron in conductor is zero
(u t) av= 0
Average velocity of electron in conductors in t (relaxation time) = v d (drift velocity)
v d = (u t) av + a t
v d = 0 +
=

1

1

22.

C 2 and C 3 are in series
=  +  = 1
= 1µf
& C 4 are in ?
C” = 1 + 1 = 2µf
C” & c 5 are in series
=  +  ?  = 1µf
& c 1 are in ?
C eq = 1 + 1 = 2µf
Energy stored
U =  cv
2
=  ×2×10
-6
×6
2

= 36×10
-6
J

1

1

Page 3

1

Class -XII
PHYSICS
SQP Marking Scheme
2019-20

Section – A
1. a , ?=  (for one face) 1

2. b , Conductor  1

3. a , 1?. 1

4. c ,12.0kJ 1

5. a , speed 1

6. d, virtual and inverted 1

7. a, straight line 1

8. d, 60
O
1

9. b, work function 1

10. b, third orbit 1

11. 45
O
or vertical 1

12. 2 H 1

13. double 1

14. 1.227 A
o
1

15. 60° 1

16. Difference in initial mass energy and energy associated with mass of products
Or
Total Kinetic energy gained in the process
1

17. Increases   1

18. N o/8 1

19. 0.79 eV 1

20. Diodes with band gap energy in the visible spectrum range can function as LED 1
2

OR
Any one use
Section – B

21. When electric field E is applied on conductor force acting on free electrons
= -e
m = -e
=
Average thermal velocity of electron in conductor is zero
(u t) av= 0
Average velocity of electron in conductors in t (relaxation time) = v d (drift velocity)
v d = (u t) av + a t
v d = 0 +
=

1

1

22.

C 2 and C 3 are in series
=  +  = 1
= 1µf
& C 4 are in ?
C” = 1 + 1 = 2µf
C” & c 5 are in series
=  +  ?  = 1µf
& c 1 are in ?
C eq = 1 + 1 = 2µf
Energy stored
U =  cv
2
=  ×2×10
-6
×6
2

= 36×10
-6
J

1

1

3

23. Gain in KE of particle = Qv

= K P = q pV p ----------(i)V p = V ? =V
= K ? = q ?V ? -----------(ii)

(ii)/(i)
=   =
=  =  =
v ? : v p = 1:

1

1

24. “The angle of incidence at which the reflected light is completely plane
polarized, is called as Brewster’s angle (i B)

At i = i B, reflected beam 1 to refracted beam
? i B + r = 90 ? r = 90-i B
Using snell’s law
= µ
= µ ?  = µ
µ = tan

1

1

25. wave function
? = 2.14eV
(a) Threshold frequency  ? = h? 0
? 0 =  =

1

Page 4

1

Class -XII
PHYSICS
SQP Marking Scheme
2019-20

Section – A
1. a , ?=  (for one face) 1

2. b , Conductor  1

3. a , 1?. 1

4. c ,12.0kJ 1

5. a , speed 1

6. d, virtual and inverted 1

7. a, straight line 1

8. d, 60
O
1

9. b, work function 1

10. b, third orbit 1

11. 45
O
or vertical 1

12. 2 H 1

13. double 1

14. 1.227 A
o
1

15. 60° 1

16. Difference in initial mass energy and energy associated with mass of products
Or
Total Kinetic energy gained in the process
1

17. Increases   1

18. N o/8 1

19. 0.79 eV 1

20. Diodes with band gap energy in the visible spectrum range can function as LED 1
2

OR
Any one use
Section – B

21. When electric field E is applied on conductor force acting on free electrons
= -e
m = -e
=
Average thermal velocity of electron in conductor is zero
(u t) av= 0
Average velocity of electron in conductors in t (relaxation time) = v d (drift velocity)
v d = (u t) av + a t
v d = 0 +
=

1

1

22.

C 2 and C 3 are in series
=  +  = 1
= 1µf
& C 4 are in ?
C” = 1 + 1 = 2µf
C” & c 5 are in series
=  +  ?  = 1µf
& c 1 are in ?
C eq = 1 + 1 = 2µf
Energy stored
U =  cv
2
=  ×2×10
-6
×6
2

= 36×10
-6
J

1

1

3

23. Gain in KE of particle = Qv

= K P = q pV p ----------(i)V p = V ? =V
= K ? = q ?V ? -----------(ii)

(ii)/(i)
=   =
=  =  =
v ? : v p = 1:

1

1

24. “The angle of incidence at which the reflected light is completely plane
polarized, is called as Brewster’s angle (i B)

At i = i B, reflected beam 1 to refracted beam
? i B + r = 90 ? r = 90-i B
Using snell’s law
= µ
= µ ?  = µ
µ = tan

1

1

25. wave function
? = 2.14eV
(a) Threshold frequency  ? = h? 0
? 0 =  =

1

4

= 5.17× H z
(b) As k max = eV 0 = 0.6eV
Energy of photon E = k max + ? = 0.6eV + 2.14eV
= 2.74eV
Wave length of photon ? =  =
= 4530Å

1

26.

centripetal force = electrostatic attraction
=
=  --------(i)
as = n .
=  put in (i)
m . =
=

OR

Energy of electron in n = 2 is -3.4eV
? energy in ground state = -13.6eV
kE = -TE = +13.6eV

1

1

1

E n =  ? -3.4eV =  ?
energy in ground state x = - 13.6eV.
Page 5

1

Class -XII
PHYSICS
SQP Marking Scheme
2019-20

Section – A
1. a , ?=  (for one face) 1

2. b , Conductor  1

3. a , 1?. 1

4. c ,12.0kJ 1

5. a , speed 1

6. d, virtual and inverted 1

7. a, straight line 1

8. d, 60
O
1

9. b, work function 1

10. b, third orbit 1

11. 45
O
or vertical 1

12. 2 H 1

13. double 1

14. 1.227 A
o
1

15. 60° 1

16. Difference in initial mass energy and energy associated with mass of products
Or
Total Kinetic energy gained in the process
1

17. Increases   1

18. N o/8 1

19. 0.79 eV 1

20. Diodes with band gap energy in the visible spectrum range can function as LED 1
2

OR
Any one use
Section – B

21. When electric field E is applied on conductor force acting on free electrons
= -e
m = -e
=
Average thermal velocity of electron in conductor is zero
(u t) av= 0
Average velocity of electron in conductors in t (relaxation time) = v d (drift velocity)
v d = (u t) av + a t
v d = 0 +
=

1

1

22.

C 2 and C 3 are in series
=  +  = 1
= 1µf
& C 4 are in ?
C” = 1 + 1 = 2µf
C” & c 5 are in series
=  +  ?  = 1µf
& c 1 are in ?
C eq = 1 + 1 = 2µf
Energy stored
U =  cv
2
=  ×2×10
-6
×6
2

= 36×10
-6
J

1

1

3

23. Gain in KE of particle = Qv

= K P = q pV p ----------(i)V p = V ? =V
= K ? = q ?V ? -----------(ii)

(ii)/(i)
=   =
=  =  =
v ? : v p = 1:

1

1

24. “The angle of incidence at which the reflected light is completely plane
polarized, is called as Brewster’s angle (i B)

At i = i B, reflected beam 1 to refracted beam
? i B + r = 90 ? r = 90-i B
Using snell’s law
= µ
= µ ?  = µ
µ = tan

1

1

25. wave function
? = 2.14eV
(a) Threshold frequency  ? = h? 0
? 0 =  =

1

4

= 5.17× H z
(b) As k max = eV 0 = 0.6eV
Energy of photon E = k max + ? = 0.6eV + 2.14eV
= 2.74eV
Wave length of photon ? =  =
= 4530Å

1

26.

centripetal force = electrostatic attraction
=
=  --------(i)
as = n .
=  put in (i)
m . =
=

OR

Energy of electron in n = 2 is -3.4eV
? energy in ground state = -13.6eV
kE = -TE = +13.6eV

1

1

1

E n =  ? -3.4eV =  ?
energy in ground state x = - 13.6eV.
5

PE = 2TE = -2×13.6eV = -27.2eV

1

27.

P-type semiconductor n-type semiconductor
1. Density of holes >> density of
electron

2. Formed by doping trivalent
impurity
1. density of
electron>>density of holes

2.   formed by doping pentavalent
impurity
Energy band diagram for p-type

Energy band diagram of n-type
semiconductor

OR

Energy of photon E =  = eV =2.06eV
As E<E g (2.8eV), so photodiode cannot detect this photon.

Any
2x1
=1

1

1
Section – C

28.
Principle of potentiometer, when a constant current flows through a wire of uniform
area of cross-section, the potential drop across any length of the wire is directly
proportional to the length.
Let resistance of wire AB be R 1 and its length be ‘l’ then current drawn from
driving cell –
I =  and hence
P.D. across the wire AB will be
V AB = IR 1 =  ×
Where ‘a’ is area of cross-section of wire AB
? =  = constant = k
Where R increases, current and potential difference across wire AB will be

1

1

1

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