Kerala Plus Two Physics Board Model Paper 2022 with Answers

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

Kerala Plus Two Physics Board Model Paper 2022

PART – I

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

Question 1.
Si unit of electric charge is ………….. .
Answer:
Coulumb

Question 2.
When alternating current is flowing through an inductor, the phase difference between potential and current is ……….
(a) 0
(b) 45°
(c) 90°
(d) 180°
Answer:
(c) 90°

Question 3.
Momentum of a photon with wavelength λ is,
(a) hλ
(b) \(\frac{\mathrm{h}}{\lambda}\)
(c) \(\frac{\lambda}{\mathrm{h}}\)
(d) λ + h
Answer:
(b) \(\frac{\mathrm{h}}{\lambda}\)

Question 4.
According to Bohr, the electron can revolve around the nilcleus only in those orbits where angular momentum is an integral multiple of ………….
Answer:
\(\frac{\mathrm{h}}{2 π}\)

Question 5.
Write the number of neutrons and protons in the nucleus of \({ }_Z^A X\).
Answer:
No. of protons = Z
No. of nuetrons = A – Z

Question 6.
A stationary charge can produce electromagnetic waves. State true or false.
Answer:
False.

Question 7.
What is the value of magnetic field at the centre of a circular coil having n turns carrying current?
Answer:
B = \(\frac{\mu_0 n I}{2 R}\)

Question 8.
When bulk pieces of conductors are subjected to changing magnetic flux, currents are induced in it. Name the current.
Answer:
Eddy current

Question 9.
The redistribution of energy due to the superposition of two or more light waves is called ………..
Answer:
Interference

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

Question 10.
When a dielectric is placed between the plates of a capacitor, its capacity,
(a) remains the same
(b) increases
(c) decreases
Answer:
(b) increases

Question 11.
Which of the following material is used to make standard resistance ?
(a) Carbon
(b) Copper
(c) Germanium
(d) Manganin
Answer:
(d) Manganin

Question 12.
The susceptibility of a diamagnetic material is. (negative/positive/zero)
Answer:
Negative

Question 13.
The blue colour of the sky is due to …………. .
Answer:
Scattering

PART -II

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

Question 14.
State Gauss’s law in magnetism.
Answer:
The net magnetic fiux through any closed surface is zero.

Question 15.
Draw the phasor diagram for an A.C. voltage applied to a series LCR circuit.
Answer:

Question 16.
A gate Is a digital circuit that follows a logical relationship between the input and output.
(a) Which are the universal gates ?
Answer:
NAAD gate and NOR

(b) Why are they called so ?
Answer:
They are so called because all other basic gates can be realised using them.

Question 17.
What are coherent sources ?
Answer:
Two source are said to be coherent if phase difference between the dispalcements produced by each of the waves does not change with the time.

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

Question 18.
A charged particle is moving perpendicular to a uniform magnetic field.
(a) What is the shape of the path followed by the particle ?
Answer:
Circular

(b) If the field is non-uniform, what is its new trajectory ?
Answer:
Spiral

Question 19.
Write any two uses of X-rays. (2)
Answer:
• Used in investigation of structure of solids.
• Diagnostic tool in medicine and in treatment of certain form of cancer.

Question 20.
(a) What is the equation for torque acting on a dipole placed in a uniform electric field ?
Answer:
\(\vec{\tau}\) = \(\overrightarrow{\mathbf{P}} \times \overrightarrow{\mathbf{E}}\)

(b) What is the condition for maximum torque ?
Answer:
θ = 90° (The angle between the direction of electric dipole moment and electric field should be 90°).

PART-III

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

Question 21.
The resistance is the opposition offered by a material to the flow of current.
(a) The unit of resistance is
Answer:
Ω(Ohm)

(b) Find the equivalent resistance of two resistors R₁ and R₂ connected in parallel.
Answer:
Consider two resistors R₁ and R₂ connected in parallel across a potential of V volt as shown.

The potential across all resistors are equal but current is different.
The current through resistor R₁,
I₁ = \(\frac{V}{R_1}\)
The current through resistor R₂,
I₂ = \(\frac{V}{R_2}\)
Total current through the combination is
I = \(\frac{V}{R}\)
R is the effective resistance of combination.
Total current through combination is equal to sum of current through each resistor.
I = I₁ + I₂
Substituting we get,
\(\frac{V}{R}\) = \(\frac{V}{R_1}\) + \(\frac{V}{R_2}\)
Eliminating V, we get
\(\frac{1}{R}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\)

Question 22.
A spherical mirror is a mirror with a curved reflecting surface.
(a) What is the relation between focal length and radius of curvature of a spherical mirror ?
Answer:
f = \(\frac{R}{2}\)

(b) Prove that for a concave mirror forming real image,
\(\frac{1}{u}\) + \(\frac{1}{v}\) = \(\frac{1}{f}\)
Answer:

Let points P, F, C be pole, focus and centre of curvature of a concave mirror. Object AB is placed on the principal axis. A ray from AB incident at E and then reflected through F. Another ray of light from B incident at pole P and then reflected. These two rays meet at M. The ray of light from point B is passed through C. Draw EN perpendicular to the principal axis.
∆IMF and ∆ENF are similar.
ie, \(\frac{\mathrm{IM}}{\mathrm{NE}}=\frac{\mathrm{IF}}{\mathrm{NF}}\) …………..(1)
but IF = PI – PF and NF = PF (since aperture is small)
hence eq. (1) can be written as
\(\frac{\mathrm{IM}}{\mathrm{NE}}=\frac{\mathrm{PI – PF}}{\mathrm{PF}}\) ………….(2)
[∵ NE=AB]
∆ABP and ∆IMP are similar
\(\frac{\mathrm{IM}}{\mathrm{AB}}=\frac{\mathrm{PI}}{\mathrm{PA}}\) ……….(3)
From eq.(2) and eq.(3),weget
\(\frac{\mathrm{PI – PF}}{\mathrm{PF}}\) = \(\frac{\mathrm{PI}}{\mathrm{PA}}\)
applying sign convention we get
PI = -v
PF = -F
PA = -u
Substituting these values in eq.(4) we get
\(\frac{-v-f}{-f}\) = \(\frac{-v}{-u}\)
\(\frac{v-f}{vf}\) = \(\frac{1}{u}\)
\(\frac{1}{f}\) – \(\frac{1}{v}\) = \(\frac{1}{u}\)

Question23.
(a) Define dip at a place.
Answer:
The angle between earth’s magnetic field and the horizontal component of earth’s magnetic field at a place is called dip.

(b) Calculate the vertical component of the earth’s magnetic field at a place where the dip is 60° and the horizontal component is 0.2 × 10^-4 wb/m². (2)
Answer:
θ = 60°
B_h = 0.2 × 10^-4 wb/m²
tan θ = \(\frac{B_V}{B_H}\)
B_v = B_H tan θ
B_v = 0.2 × 10^-4 × tan 60°
= 0.2 × 10^-4 × -√3
= 0.346 × 10^-4 Wb/m²

Question 24.
(a) Write any two characteristics of equipotential surfaces.
Answer:
No work is done to move a charge from one point to another along equipotential surface.

(b) What is the shape of equipotential surface around a point charge ?
Answer:
Direction of electric field is perpendicular the equipotential surface.

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

Question 25.
Photoelectric effect is the emission of electrons from certain metal surfaces when they are exposed to light.
(a) State any two laws of photo electric emission. (2)
Answer:
1) For a given frequency of radiation, number of photoelectrons emitted is proportional to the intensity of incident radiation.
2) The kinetic energy of photoelectrons depends on the frequency of incident light but it is independent of the light intensity.
3) Photoelectric effect does not occur if the frequency is below a certain value. The minimum frequency (v₀) required to produce photo electric effect is called the threshold frequency.
4) Photoelectric effect is an instantaneous phenomenon.

(b) What is meant by stopping potential ?
Answer:
The retarding potential at which photo current reaches zero is called stopping potential.

Question 26.
A nucleus is made up of protons and neutrons.
(a) Define mass defect. Write an equation for mass defect.
Answer:
The mass defect (∆m) is the difference in the mass of nucleus and total mass of constituent nucleons.
∆m = (ZM_p + (A-Z)m_n] – M

(b) The nuclear process used in nuclear reactor is ………..
Answer:
nuclear fission

Question 27.
(a) Define ionisation energy.
Answer:
Ionization energy is the minimum energy required to free the electron from the ground state of atom

(b) Mention any two drawbacks of Rutherford model of atom.
Answer:
i) The Bohr model is applicable to hydrogenic atoms. It cannot be extended to many elec-tron atoms such as helium.
ii) The model is unable to explain the relative intensities of the frequencies in the spectrum.
iii) Bohr model could not explain fine structure of spectral lines.
iv) Bohr theory could not give a satisfactory explanation for circular orbit.

PART – IV

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

Question 28.
A capacitor is a system of two conductors separated by an insulator.
(a) Obtain an expression for the equivalent capacitance of two capacitors connected in Series.
Answer:
Let two capacitors C₁ and C₂ be connected in series to a potential of V. V₁ and V₂ are the potential across C₁ and C₂.

The applied voltage can be written as
V = V₁ + V₂
Charge ‘q’ is same in all capacitors.
V₁ = \(\frac{q}{C_1}\)
V₂ = \(\frac{q}{C_2}\)
Substituting values in (1), we get
V = \(\frac{q}{C_1}\) + \(\frac{q}{C_2}\)
If C is the effective capacitance of combination.
V = \(\frac{q}{C}\)
Equation (2) can be written as
\(\frac{q}{C}\) = \(\frac{q}{C_1}\) + \(\frac{q}{C_2}\)
Eliminating q, we get
\(\frac{1}{C}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\)

(b) The potential across a 900 pF capacitor is 100 volts. How much energy is stored by the capacitor? (2)
Answer:
V = 100 V
C = 900μF = 900 × 10^-6 F
Energy stored in a capacitor,
E = \(\frac{1}{2}\) CV²
E = \(\frac{1}{2}\) 900 × 10^-6 × (100)² = 4.5 J

Question 29.
A moving coil galvanometer is used as a current detector in a circuit.
(a) What is the working principle of a moving coil galvanometer?
Answer:
A current carrying loop placed in magnetic field experience a torque . T=NIABsin0

(b) How can you convert a moving coil galvanometer into
(i) an ammeter
(ii) a voltmeter
Answer:
(i) A galvanometer can be converted in to ammeter by connecting a small resistance (shunt resistance) parallel to it.

(ii) A galvanometer can be converted in to voltmeter by connecting a high resistance in series to it.

Question 30.
An emf is induced in a coil due to the change of flux in the same coil.
(a) This phenomenon is called ………….
Answer:
Self induction

(b) Derive an expression for the energy stored in an inductor.
Answer:
When the current in the coil is switched on, a back emf (ε = -Ldt/dt) is produced. This back emf opposes the growth of current. Hence work should be done, against this e.m.f.
Let the current at any instant be ‘I’ and induced emf
E = \(\frac{-d \phi}{d t}\)
i.e., vyork done, dw = Eldt dl
= L \(\frac{d I}{d t}\)Idt
dw = LIDI
Hence the total work done (when the current grows from 0 to I₀ is)
w = \(L \int_0^{I_0} I d I\)
w = \(\frac{L}{2}\left[I^2\right]_0^{t_0},\)
w = \(\frac{1}{2} L I_0^2\)

This work is stored as potential energy.
V = \(\frac{1}{2} L I^2\)

Question 31.
A diode is used to rectify alternating voltages.
(1) Which property of diode is used here?
Answer:
A p-n junction diode conducts current when it is 4 forward biased, and does not conduct when it is reverse biased.

(2) Explain a full wave rectifier using necessary circuit diagram.
Answer:
Circuit

Full wave rectifier consists of transformer, two diodes and a load resistance R_L. Input a.c signal is ^ applied across the primary of the transformer. Secondary of the transformer is connected to D₁ and D₂. The output is taken across R_L.

Working : During the +ve half cycle of the a.c signal at secondary, the diode D₁ is fprward biased and D₂ is reverse biased. So that current flows through D, and Ru.

During the negative half cycle of the a.c signal at secondary, the diode D₁ is reverse biased and D₂ is forward biased. So that current flows through D₂ and R_L. Thus during both the half cycles, the current flows through R_L in the same direction. Thus we get a +ve voltage across RL for +ve and -ve input. This process is called full wave rectification.

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

Question 32.
A transfer is used to change alternating voltage from one to another of greater or smaller value.
(a) Which principle is used in transformer ?
Answer:
Mutual Induction

(b) A power transmission line feeds input power at 2300 volts to a stepdown transformer with primary coil having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V ?
Answer:
N_P = 4000
V_P = 2300 V
V_S = 230 V
\(\frac{N_S}{N_P}\) = \(\frac{V_S}{V_p}\)
N_S = \(\frac{V_S}{V_p}\) × N_P
= \(\frac{230}{2300}\) × 4000 = 400