1
CLASS- XII MODEL QUESTION PAPER - 13 (2024-25)
SUBJECT: PHYSICS (THEORY)
(MARKING SCHEME & KEY)
Q.No. Option/Ans/Key point weightage Marks
SECTION:A
1. C 1 1
2. D 1 1
3. C 1 1
4. C 1 1
5. D 1 1
6. D 1 1
7. D 1 1
8. B 1 1
9. C 1 1
10. C 1 1
11. C 1 1
12. B 1 1
13. A 1 1
14. C 1 1
15. D 1 1
16. C 1 1
SECTION: B
17.
½
2
D < C < B < A. ½
½
2
18. A diamond cutter uses a large angle of incidence to ensure that 1+1 2
the light entering the diamond is totally reflected from its face.
19. ½
½ 2
½+½
R = 0.15 m
(OR)
Magnification when image formed at infinity
½
½
½ 2
[ ]
½
[ ]
20. ½
½
½ 2
As sodium and Potassium are having work function less than ½
energy of photon. These two metals exhibit photoelectric effect.
21. Differences two points ½+½ 2
Diagrams ½+½
SECTION - C
22. (i) ½
½
½
(ii) ½ 3
½
23. Derivation 2
Charge flowing through the given cross-section is equal to area
under the curve of current (I) versus time (t). ½ 3
( )
½
q = 12.5 + 25 = 37.5 C.
3
24. (a) Principle. 1
(b) Two reasons. ½+½ 3
(c) Definitions of voltage sensitivity and currentsensitivity. ½+½
(OR)
1
1 3
1
25. Let ON be at some point x. 1
The emf induced in the loop
1
Current in the arm,
I= 1
A
26. (i) Microwaves are suitable for RADAR systems that are used in 1
aircraft navigation. These rays are produced by special vacuum
tubes, namely klystrons and magnetrons diodes.
(ii) Infrared rays are used to treat muscular strain. These rays are 1 3
produced by hot bodies and molecules.
(iii) X-rays are used as a diagnostic tool in medicine. These rays
1
are produced, when high energy electrons are stopped suddenly
on a metal of high atomic number.
(OR)
(i) ϒ-rays are used for the treatment of certain forms of cancer. 1
Its frequency range is 3 X 1019 Hz to 5 X 1022 Hz.
(ii) The thin ozone layer on top of stratosphere absorbs most of
the harmful ultraviolet rays coming from the sun towards the 1
earth. They include UVA, UVB and UVC radiations, which can
destroy the life system on the earth. Hence, this layer is
3
crucial for human survival.
(iii) An electromagnetic wave transports linear momentum as it
travels through space. If an electromagnetic wave transfers a 1
total energy U to a totally absorbing surface in time t, then
total linear momentum delivered to the at surface.
This means, the momentum range of EM waves is 10-19 to 10- 41.
Thus, the amount of momentum transferred by the EM waves
incident on the surface is very small.
27. Energy difference = energy emitted by photon 1
= -1.51 – (-3.4) = 1.89 eV
4
= 1.89 × 1.6 × 10-19 J
1 3
=
= 6548A0. 1
This wavelength belongs to Balmer series of hydrogen spectrum.
28. Using the given data
Energy released 1
⟦ ⟧ 1
3
SECTION - D
29. (i) b 1
(ii) d 1
(iii) d 1 4
(iv) a 1
Or (v) d
30. (i) c 1
(ii) c 1 4
(iii) d 1
(iv) c 1
Or (v) b
SECTION - E
31. (a) W=q×dV=2× ×1 1
=3.2×10−19 J
(b) Zero .Work done in moving a charge in a closed path is zero.
(c) (i) Since the battery remains connected, the potential difference 1
5
remains constant, hence E also remain unchanged 1
(ii)Capacitance becomes K times 1
(iii).Charge becomes K times since capacitance becomes K times. 1
(OR)
(a) (i) Ф1= /𝜀0and Ф2= 3 /𝜀0 So, Ф1:Ф2 =1:3
1
(ii) Ф1=∫ . 𝑆= /𝜀0.
On introducing medium of dielectric constant L inside the sphere S1,
5
the electric field becomes K times. 2
Now the new flux Ф1′= /𝐾𝜀0 On solving K=5 .
So new flux Ф1′= /5𝜀0
(b) Derivation of electric field intensity 2
32. (i) EP= 2200 V, nP= 3000, nS= ?, ES= 220 V
2
ES/EP = nS/nP
So nS= 3000 ×1/10 = 300
5
(ii) A step up transformer converts a low voltage into high voltage, 1 5
it does not violate principle of conservation of energy as the
increase in voltage is at the cost of current. When voltage
increases the current decreases.
(iii) Energy loss in a transformer:
(a)Eddy current loss: Alternating magnetic flux induces eddy 1
currents in the iron core, which leads to energy loss in the
form of heat. It can be minimized by using laminated core.
(b) Hysteresis loss: AC carries the
core to the process of magnetization and demagnetization. 1
Work is done in each of these cycles resulting into loss of energy.
(OR)
(i) Consider a coil consisting of N turns of insulated copper wire
rotated in a uniform magnetic field B. Let the angle between 1
magnetic field and area vector at any
point of time be θ. The coil is rotated
with angular velocity ω. 1
ϕ= NBA cos θ
θ= ωt
So, ϕ= NBA cos ωt 5
E= -dϕ/ dt 1
= -NBAω (- sin ωt)
= ANBω sin ωt
E=0 when ωt=0
E= max when ωt=𝝅/2
Emax= NBAω =E0
Ein = E0 sin ωt
(ii) A= 200 cm2= 200 ×10-4m2, N=20, ω= 50 rad/s , B= 3×10-2T
E0= NBA ω = 20× 3×10-2× 200 ×10-4× 50 = 0.6 V 2
33. Huygen’s principle 1
Definition 1
Ray diagram 1½
derivation 1½
(OR)
5
Two points 2
Ray diagram 1½
derivation 1½
