Kerala Plus Two Physics Question Paper March 2022 with Answers

Reviewing Kerala Syllabus Plus Two Physics Previous Year Question Papers and Answers Pdf March 2022 helps in understanding answer patterns.

Kerala Plus Two Physics Previous Year Question Paper March 2022

Time : 2 1/2 Hours
Maximum : 80 scores

PART – I

A. Answer any 5 questions from 1 to 9. Each carries 1 score. (5 × 1 = 5)

Question 1.
SI unit of electric field
(a) NC^-1
(b)V_m
(c) C_m
(d) N_m
Answer:
(a) NC^-1

Question 2.
Name the force experienced by a charge q moving through a uniform magnetic field with a velocity V.
Answer:
Lorentz force or \(\vec{f}\) = q(\(\vec{V}\) × \(\vec{B}\))

Question 3.
Changing magnetic fields can set up current loops in nearby metal bodies. They dissipate electrical energy as heat. Such currents are ………
Answer:
Eddy current

Question 4.
In purely inductive or capacitive circuit, power factor (cosϕ) is .
(a) 0
(b) 1
(c) -1
Answer:
(a) 0

Question 5.
Relation between velocity of light (c), permeability of free space (μ₀), permittivity of free space (ε₀) is ………..
(a) C = \(\frac{1}{\mu_0 \varepsilon_0}\)
(b) C = \(\mu 0 \varepsilon_0\)
(c) C = \(\frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_0}}\)
(d) C = \(\sqrt{\mu_{\mathrm{o}} \varepsilon_0}\)
Answer:
(c) C = \(\frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_0}}\)

Question 6.
Light waves are …………. in nature (transverse, longitudinal)
Answer:
transverse

Question 7.
Photons are electrically ………….
(a) neutral
(b) positive
(c) negative
(d) unpredictable
Answer:
(a) neutral

Question 8.
The minimum energy required to free an electron from the ground state of hydrogen atom is …………..
(a)+13.6 J
(b) -13.6 J
(c) +13.6 eV
(d) -13.6 eV
Answer:
(c) +13.6 eV

Question 9.
Complete the general equation of α – decay.

\({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}\) → …………. + \({ }_2^4 \mathrm{He}\)
(a) \({ }_{\mathrm{Z-4}}^{\mathrm{A-2}} \mathrm{Y}\)
(b) \({ }_{\mathrm{Z-2}}^{\mathrm{A-4}} \mathrm{Y}\)
(c) \({ }_{\mathrm{Z-2}}^{\mathrm{A-2}} \mathrm{Y}\)
(d) \({ }_{\mathrm{Z+1}}^{\mathrm{A}} \mathrm{Y}\)
Answer:
(b) \({ }_{\mathrm{Z-2}}^{\mathrm{A-4}} \mathrm{Y}\)

B. Answer all questions from 10 to 13. Each carries 1 score. (4 × 1=4)

Question 10.
Electrostatic field at the surface of a charged conductor must be normal to the surface at every point. Is the statement true or false?
Answer:
True

Question 11.
Magnitude of the drift velocity per unit electric field is ………… .
Answer:
Mobility [μ = \(=\frac{V_d}{E}\)]

Question 12.
The temperature at which a ferromagnetic material become paramagnetic is …………
(a) Cut-off temperature
(b) Absolute temperature
(c) Curie temperature
Answer:
(c) Curie temperature

Question 13.
Optical fibres make use the phenomenon of ……….
Answer:
Total internal reflection

Part – II

A. Answer any 2 questions from 14 to 17. Each carries 2 scores. (2 × 2 = 4)

Question 14.
Draw the input and output waveform of a half-wave rectifier.
Answer:

Question 15.
State Malus’ law.
Answer:
This law states that when a beam of plane polarized light is incident on an analyzer, the intensity (I) of ‘the emergent light is directly proportional to the square ‘x erf the cosine of the ang le (0) between the polarizing directions of the polarizer and the analyzer.
ie, I = I_m cos² θ
I = I_m cos² θ
Where I_m is the maximum intensity.

Question 16.
What is angle of dip?
Answer:
Dip is the angle between earth’s magnetic field and the horizontal component of earth’s magnetic field at a place.

Question 17.
A light bulb is rated at 100 W. for a 220 V supply. Find the resistance of the bulb.
Answer:
R = \(\frac{V^2}{P}\)
V = 220 V, P = 100 W
R = \(\frac{(220)^2}{100}\) = 484

B. Answer any 2 questions from 18 to 20. Each carries 2 scores. (2 × 2 = 4)

Question 18.
Infra-red waves are also referred to as heat waves. Why?
Answer:
Infrared rays produces heat and hence it heat up surroundings. Infrared rays are responsible for green house effect. Hence infrared rays are called heat waves

Question 19.
(i) What is a solenoid? (1)
Answer:
Solenoid is a long wire wound in the form of a helix.

(ii) Write down the equation for magnetic field inside a solenoid. (1)
Answer:
B = \(\frac{\mu_0 n I}{L}\)
n = no.of turns
L = Length of coil
I = Current in coil

Question 20.
Explain earthing. (2)
Answer:
When a charged body bring in contact with earth, all the excess charge pass to the earth through the connecting conductor. This process of sharing the charges with the earth is called grounding or earthing. Earthing provides protection to electrical circuits and appliances.

Part-III

A. Answer any 3 questions from 21 to 24. Each carries 3 scores. (3 × 3 = 9)

Question 21.
Write down any three properties of an equipotential surface.
Answer:
1) Direction of electric field is perpendicular to the equipotential surface.
2) No work is done to move a charge from one point to another along the equipotential surface.
3) Two equipotential surface can never interesect.

Question 22.
(i) SI unit of resistance is ………… (1)
Answer:
Ohm(Ω)

(ii) Obtain the equivalent value of resistance when two resistors R1 and R2 are connected in series. (2)
Answer:

In series combination of resistors, current is same and voltage is different
V = V¹ + V² ……….(1)
V¹ = IR¹ & V² = IR²
If R is the effective resistance of combination,
then V = IR
Substituting in equation (1), we get
IR = IR¹ = IR²
R = R¹ + R²

Question 23.
(i) The a1ngle between magnetic meridian and geographic meridian is ……….. (1)
Answer:
Angle of declination

(ii) The declination is ………. (higher/smaller) at higher lattitudes and ………. (higher/smaller) near the equator. (2)
Answer:
Higher,smaller

Question 24.
(i) If f = 0.5 m, for a glass lens, what is the power of the lens? (1)
Answer:
f = 0.5 m
P = \(\frac{1}{f}\) = \(\frac{1}{0.5}\) = 2D

(ii) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass? (2)
Answer:
f = +12 cm
R¹ = +10 cm
R² = -15 cm
\(\frac{1}{f}\) = (n-1) [\(\frac{1}{R_1}\) – \(\frac{1}{R_2}\)]
\(\frac{1}{12}\) = (n-1) [\(\frac{1}{10}\) – \(\frac{1}{-15}\)]
\(\frac{1}{12}\) = (n-1) × \(\frac{1}{6}\)
n – 1 = \(\frac{1}{2}\)
n = 1 + \(\frac{1}{2}\) = 1.5

B. Answer any 2 questions from 25 to 27r.Each carries 3 scores. (2 x 3 = 6)

Question 25.
(i) Drawthe energy level diagram for hydrogen atom and mark the transition corresponding to Balmer series. (2)
Answer:

(ii) Name the spectral series which lies in the ultraviolet region of the spectrum. (1)
Answer:
Lyman series

Question 26.
(i) What is meant by the threshold frequency of a photosensitive metal?
Answer:
It is the minimum frequency required for incident light to emit electrons from the surface of photosensitive metal.

(ii) Draw the graph showing the variation of stopping potential with frequency. (1)
Answer:

Question 27.
Nuclear reactor is a device used to initiate and control a nuclear chain reaction. Explain the major parts of a nuclear reactor.
Answer:
i) Fissionable material or fuel: The fissionable material is \(\left({ }_{92}^{235} \mathrm{U}\right)\). It is placed inside the core where the fission takes place.
ii) Moderator: It is used to slow down fast moving neutron. Commonly used moderators are water, heavy water (D²O) and graphite.
iii) Reflector: The core is surrounded by reflector to prevent the leakage.
iv) Control rods: Its purpose is to absorb neutron and hence to control reaction rate. It is made up of neutron-absorbing material like Cadmium.
v) Coolant:- The energy released in the form of heat is continuously removed by coolant. It transfers heat to the working fluid.

Part – IV

A. Answer any 3 qeustions from 28 to 31. Each carries 4 scores. (3×4=12)

Question 28.
(i) SI unit of capacitance is ………. (1)
Answer:
Farad

(ii) Two capacitors C¹ and C² are connected in series. Derive an expression for the capacitance of the combination. (3)
Answer:

Consider two capacitors C¹ and C² connected in series as shown in series connection charge will be same and potential will be different.
V = V¹ + v²
V¹ = \(\frac{q}{C_1}\), V² = \(\frac{q}{C_2}\)
V = \(\frac{q}{C_1}\) + \(\frac{q}{C_2}\)
If C is the effective capacitance, then V = \(\frac{q}{C}\)
∴ \(\frac{q}{C}\) = \(\frac{q}{C_1}\) + \(\frac{q}{C_2}\)
\(\frac{1}{C}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\)
or C = \(\frac{C_1 C_2}{C_1+C_2}\)

Question 29.
(i) Wliich law helps us to find the magnetic field on the axis of a circular current loop? (3)
Answer:
Bio-Savart Law

(ii) Consider a tightly wound 100 turn coil of radius 10 cm, carrying current of 1A. What is the magnitude of the magnetic field at the centre of the coil? (3)
Answer:
N = 100
R = 10 cm = 10 × 10^-2 m
I = 1 A
B = \(\frac{\mu_0 N I}{2 R}\)
B = \(\frac{4 \pi \times 10^{-7} \times 100 \times 1}{2 \times 10 \times 10^{-2}}\)
= 6.28 × 10^-4 T

Question 30.
(i) Which is the working principle of an a.c. generator? (1)
Answer:
Electromanetic Induction

(ii) With the help of a diagram explain the working of a.c. generator. (3)
Answer:

Take the area of coil as A and magnetic field produced by the magnet as B .Let the coil be rotating about an axis with an angular velocity ω.
Let θ be the angle made by the areal vector with the magnetic field B. The magnetic flux linked with the coil can be written as
ϕ = B. A
ϕ = BA cos θ
ϕ = BA cos ωt [since θ = ωt)
If there are N turns
ϕ = NBA cos ωt
∴ The induced e.m.f. in the coil,

ε = \(\frac{-\mathrm{d} \phi}{\mathrm{dt}}\)
ε = \(\frac{-\mathrm{d}}{\mathrm{dt}}\) (NBA cos ωt)
ε = NBA sin ωt ω
Let ε₀ = NBA ω
then ε = ε₀ sin ωt
When this emf is appiled ti an external circuit, alternating current is produced.

The current is any instant is given by
I = \(\frac{V}{R}\)
= \(\frac{\varepsilon_0 \sin \omega t}{R}\) (V = ε₀ sin ωt)
I = I₀ sin ωt

Question 31.
(i) Identify the logic gate.

Answer:
NAND gate

(ii) Write down the truth table of this gate. (2)
Answer:
Tablee

(iii) Why this gate is also called universal gate? (1)
Answer:
We can construct all other basic gates using NAND gate. Hence it is called Universal gate.

B. Answer any 1 question from 32 to 33. Each carries 4 scores. (1 × 4 = 4)

Question 32.
Using Huygen’s principle, explain refraction of a plane wave, with the help of a diagram.
Answer:
AB is the incident wavefront and C₁ is the velocity of the wavefront in the first medium. CD is the refracted wavefront and C₂ is the velocity of the wavefront in the second medium. AC is a plane separating the two media.

The time taken for the ray to travel from P to R is

O is an arbitrary point. Hence AO is a variable. But the time to travel a wavefront from AB to CD is constant. In order to satisfy this condition, the term containing AO in eq.(2) should be zero.

where \({ }^1 n_2\) is the refractive index of the second medium w.r.t. the first. This is the law of refraction.

Question 33.
(i) State the principle of working of a transformer.
Answer:
Mutual induction

(ii) Explain briefly any three energy losses in a transformer.
Answer:

1. Joule loss or Copper loss : When current passes through a coil heat is produced. This energy loss is called Joule loss. It can be minimized by using thick wires.
2. Eddy current loss : This can be minimized by using laminated cores. Laminated core increases the resistance of the coil. Thus eddy current decreases.
3. Hysteresis loss : When the iron core undergoes cycles of magnetization, energy is lost. This loss is called hysteresis loss. This is minimized by using soft iron core.
4. Magnetic flux loss: The total flux linked with the coil may not pass through secondary coil.
   This loss is called magnetic flux loss. This loss can be minimized by closely winding the wires.

Part-V

Answer any 2 questions from 34 to 36. Each Carries 6 scores. (2 × 6 = 12)

Question 34.
(i) Write down the Wheatstone bridge principle.
Answer:
\(\frac{P}{Q}\) = \(\frac{R}{S}\)

(ii) A meter bridge circuit is given below.

Using this circuit, derive an expression for finding the unknown resistance.(3)
Answer:
Meter Bridge is used to measure unknown resistance.
Principle : It works on the principle of Wheatstone bridge condition (P/Q=R/S).

Circuit details : Unknown resistance ‘X’ is connected in between A and B. Known resistance (box) is connected in between B and C. Voltage is applied between A and C. A 100cm wire is connected between A and C. Let r be’the resistance per unit length. Jockey is connected to ‘B’ through galvanometer.

Working : A suitable resistance R is taken in the box. The position of jockey is adjusted to get zero deflection.
If ‘l’ is the balancing length from A, using Wheatstone’s condition,
\(\frac{X}{R}\) = \(\frac{lr}{(100-l)r}\)
Or X = \(\frac{Rl}{100-l}\)
knowing R and l, we can find X (resistance of wire)

(iii) In the above circuit, the balance point is found to be at 40 cm from the end A, when the resistance S,is of 12Ω. Determine the resistance R.
Answer:
S = 12 Ω
l = 40 cm
R = S × \(\frac{l}{100-l}\)
R = 12 × \(\frac{40}{60}\) = 8 Ω

(iv) Would the galvanometer show any current if the galvanometer and cell are interchanged?
Answer:
No change in the galvenometer reading

Question 35.
(i) State Gauss’s law. (2)
Answer:
Gauss’s Law: Gauss’s theorem states that the total electric flux over a closed surface is 1/ε₀ times the total charge enclosed by the surface. Gauss’s theorem may be expressed as
\(\int \overrightarrow{\mathrm{E}} . \mathrm{d} \overrightarrow{\mathrm{~s}}\) = \(\frac{1}{\varepsilon_0}\) q or ϕ = \(\frac{1}{\varepsilon_0}\) q

(ii) What is meant by a Gaussian surface? (1)
Answer:
Guassian surface is a closed, symmetrical surface which may or may not enclose any charge.

(iii) Using Gauss’s law, finfl the electric field due to a uniformly charged thin spherical shell at a point outside the shell.
Answer:
Consider a uniformly charged hollow spherical conductor of radius R. Let ‘q’ be the total charge on the surface.

To find the electric field at P (at a distance r from the centre), we imagine a Gaussian spherical surface having radius Y.
Then, according to Gauss’s theorem we can write,
\(\int \overrightarrow{\mathrm{E}} . \mathrm{d} \overrightarrow{\mathrm{~s}}\) = \(\frac{1}{\varepsilon_0}\) q
The electric field is constant ,at a distance ‘r’. So we can write,
E \(\int ds\) = \(\frac{1}{\varepsilon_0}\) q
E 4πr² = \(\frac{1}{\varepsilon_0}\) q
E = \(\frac{1}{4 \pi \varepsilon_0}\) q \(\frac{q}{r^2}\)

Question 36.
(i) State Laws of refraction. (2)
Answer:
Laws of refraction :

First law : The incident ray, the refracted ray and the normal at the point of incidence are all in the same plane.
Second law (Snell’s law) : The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a giyen pair of media and for the given colour of light used.

This constant is known as the refractive index of second medium w.r.t. the first medium.

(ii) Obtain a relation for the total deviation produced for a ray incident on a prism with the help of a ray diagram.
Answer:

ABC is a section Of a prism. AB and AC are the refracting faces, BC is the base of the prism. ∠A is the angle of prism.
A ray PQ incidents on the face AB at an angle i₁. QR is the refracted ray inside the prism, which makes two angles r₁ and r₂ (inside the prism).
RS is the emergent ray at angle i₂.
The angle between the emergent ray and incident ray is the deviation ‘d’.
In the quadrilateral AQMR,
∠Q + ∠R = 180° [since and NM are normal]
ie, ∠A + ∠M = 180° ………..(1)
In the A QMR,
∴ r₁ + r₂ + ∠M = 180° ……..(2)
Comparing eq (1) and eq (2)
r₁ + r₂ = ZA …………(3)
From the ∆QRT,
(i₁ – r₁) + (i₂ – r₂) = d
[since exterior angle equal sum of the Spposite interior angles]
(i₁ + i₂) – (r₁ + r₂) = d
but, r₁ + r₂ = A
∴ (i₁ + i₂) – A = d