Kerala Plus Two Physics Question Paper March 2024 with Answers

Reviewing Kerala Syllabus Plus Two Physics Previous Year Question Papers and Answers Pdf March 2024 helps in understanding answer patterns.

Kerala Plus Two Physics Previous Year Question Paper March 2024

Time: 2 Hours
Total Scores : 60

I. Answer any 5 questions from 1 to 7. Each carries 1 score. (5 × 1 = 5)

Question 1.
The SI unit of electric flux
(a) NC^-1
(b) NmC
(c) Nm²C²
(d) Nm²C
Answer:
(c) Nm²C²

Question 2.
The net electric field inside a conductor when placed in an external electric field is
(a) Zero
(b) Half
(c) Two times
(d) Four times
Answer:
(a) Zero

Question 3.
The SI unit of power of lens
(a) N
(b) J
(c) W
(d) D
Answer:
(d) D

Question 4.
‘The locus of points which have the same phase is called a wave front” the statement is True/Faise.
Answer:
True.

Question 5.
The expression for de Broglie wavelength associated with a particle is ………….. .
Answer:
\(\frac{h}{mV}\) = \(\frac{h}{P}\)

Question 6.
Which element in the periodic table shows maximum binding energy per nucleon?
Answer:
Fe⁵⁶ (Iron)

Question 7.
What is an intrinsic semiconductor?
Answer:
A semiconductor in its pure form is called in-trinsic semiconductor.

Answer any 5 questions from 8 to 14. Each carries 2 scores. (5 × 2 = 10)

Question 8.
What is an equipotential surface? Give an example.
Answer:
An equipotential surface is a surface in a region of space where every point on that surface has the same electric potential. In other . words, it’s a surface where the electric potential is constant.
Example: All points equidistant from a point charge.

Question 9.
Define drift velocity, give its equation.
Answer:
Drift velocity is defined as the average velocity acquired by and electron under the applied electric field.
V_d = \(\frac{-eEτ}{m}\)

Question 10.
State Gauss’s law in maghetism.
Answer:
The net magnetic flux through ant closed surface is zero.
ϕ_B = \(\sum_{\text {all }} \mathrm{B} \cdot \Delta \mathrm{~S}=0\)

Question 11.
What is magnetic flux and how is it measured?
Answer:
Magnetic flux is defined as the number of magnetic field lines passing through a given area.
The number of magnetic field lines passing through the surface gives the measure of magnetic flux.
ϕ = \(\sum_{\text {all }} \overrightarrow{\mathrm{B}} . \Delta \overrightarrow{\mathrm{S}}\)

Question 12.
The household line voltage of ac measured is 220 V, calculate its peak voltage.
Answer:
V_m = V_rms × \(\sqrt{2}\)
= 1.414 × 220
= 311 V

Question 13.
What is stopping potential?
Answer:
This is minimum negative potential given to collector plate for which the photo electric current becomes zero.
Stoping potential is denoted as V₀
\(\frac{1}{2}\) mV² max = eV₀

Question 14.
What is nuclear fission? Give one example.
Answer:
Nuclear fusion is the splitting of a heavy nucleus into two lighter nuclei.
Splitting of the uranium-235 nucleus when it is bombarded with neutrons.

Answer any 6 questions from 15 to 21. Each carries 3 scores. (6 × 3=18)

Question 15.
State and explain the force between electric charges.
Answer:

The force of attraction or repulsion between two point charges is directly proportional to the product of the charge and inversely proportional to the square of the distance between them. The force between the charges is given by,
F ∝ q₁q₂
F ∝ \(\frac{1}{r^2}\)
F ∝ \(\frac{q_1 q_2}{r^2}\)
F = \(K \frac{q_1 q_2}{\mathbf{r}^2}\)
when the medium between the charges is air or vacuum.
\(F_{\text {vecuum }}\) = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}\)

Question 16.
Figure shows the two current carrying conductors. Derive the expression for force between the conductors.

Answer:

The magnitude of the magnetic induction B-p at the right side conductor Q is,
\(B_{P}\) = \(\frac{\mu_0 i_1}{2 \pi d}\)
The force on a length L of the ‘Q’ conductor is Fg/> = Bpi2L
\(F_{QP}\) = \({B_P i_2 L}\)
= \(\frac{\mu_0 i_1}{2 \pi d} i_2 L\)
= \(\frac{\mu_0 i_1 i_2}{2 \pi d} L\)
Force per unit length can be written as
\(F_{QP}\) = \(\frac{\mu_0 i_1 i_2}{2 \pi d}\)

Question 17.
Compare dia, para and ferromagnetic substances with suitable examples.
Answer:
Diamagnetic substance:

• These substances when placed in a magnetic field, acquire feeble magnetism opposite to the direction of the magnetic field.
• They are feebly repelled by a magnet.
• The relative permeability of these substances are slightly less than 1 .
• The susceptibility has a small and negative value.
• When they are placed in a magnetic field, the magnetic lines of force are repelled. They do not pass through the material.
• Examples: Copper, Silver, Bismuth.
  

Paramagnetic substance:

• These substances when placed in a magnetic field, acquire feeble magnetism in the direction of the magnetic field.
• They are feebly attracted by a magnet.
• The relative permeability of these substances are slightly greater than 1.
• The susceptibility value is small and positive.
• When they are placed in a magnetic field, the magnetic lines offeree pass through them.
• Examples: Aluminium, Sodium, Calcium, Oxygen (at STP)
  

Ferromagnetic substance :

• These substances when placed in a magnetic field are strongly magnetised in the direction of the magnetic field.
• They are strongly attracted by a magnet.
• The relative permeability of these substances is very high.
• The susceptibility is large and positive.
• When they are placed in a magnetic field, most of the magnetic lines of force pass through them.
• Examples : Iron, Nickel, Cobalt, natural magnet.

Question 18.
What is self-induction and define the expression for self-inductance of a solenoid.
Answer:
When current in a coil changes the magnetic flux linked with the coil also changes and a back emf is induced in the coil. This phenomenon is known as self induction.
The magnetic flux through one turn of the solenoid is given by,
ϕ = BA = (μ₀nI)A
The magnetic flux linkage in the solenoid with a total number of n/turns is given by,
ϕ = μ₀n²AIl
But ϕ = LI
LI = μ₀n²AIl
L = μ₀n²Al

Question 19.
Briefly explain the electromagnetic spectrum.
Answer:
Electromagnetic spectrum is the arrangement of electro magnetic waves in the increasing order of wavelength. Thee.m waves have been arranged in terms of increasing wavelength y-ray, X – ray, Ultraviolet rays,Visible light, Infrared rays, Microwaves, Radio waves.

1. Radio Waves: These have the lowest frequencies and longest wavelengths in the spectrum. They are used for communication, broadcasting, and radar.
2. Microwaves: With slightly higher frequencies and shorter wavelengths than radio waves, microwaves are used in microwave ovens, satellite communication, and certain types of radar.
3. Infrared Radiation: Just beyond the visible spectrum, infrared radiation has frequencies slightly higher and wavelength slightly lower than microwaves.
4. Visible Light: This is the only part of the electromagnetic spectrum that humans can directly perceive with their eyes.
5. Ultraviolet (UV) Radiation: Beyond the visible spectrum, UV radiation has shorter wavelengths and higher frequencies than visible light.
6. X-rays: With even shorter wavelengths and higher frequencies than UV radiation, X-rays are used in medical imaging (X-ray radiography), security scanning, and industrial applications.
7. Gamma Rays: These have the shortest wavelengths and highest frequencies in the spectrum.

Question 20.
Write the postulates of Bohr’s atom model.
Answer:

1. The nucleus of an atom is a small positively charged core where whole of its mass is supposed to be concentrated.
2. The electrons revolve round the nucleus in fixed orbits of definite radii. They do not radiate any energy as long as the electron is in certain orbit.
3. The electrons can revolve only in those orbits, in which its angular momentum is an integral multiple of h/27i. L = mvr = nh/2. where n is known as quantum number of the orbit and h is Planck’s constant.
4. Electron jumps from higher energy orbit to lower energy orbit,by emitting the energy equivalent to the energy gap, in the form of radiations.
   hv = E₂ – E₁

Question 21.
What is a rectifier ? Draw the circuit diagram and input, output wave forms of a full wave rectifier.
Answer:
A rectifier is an electrical device that converts alternating current (AC) into direct current (DC). The process of converting AC to DC is known as rectification. Rectifiers are essential components in various electronic circuits and power supply systems, as many electronic devices required DC power to operate. Rectifiers can be classified into two main types:
Half-Wave Rectifier: This type of rectifier allows only one half of the AC input waveform to pass through to the output, while blocking the other half. It typically consists of a single diode connected in series with the load.

Full-Wave Rectifier: Full-wave rectifiers allow both halves of the AC input waveform to be utilized, resulting in a more continuous DC output.

Answer any 3 questions from 22 to 25. Each carries 4 scores. (3 × 4 = 12)

Question 22.
(a) Complete the diagram with proper marking of direction. (1)

Answer:

(b) Derive the expression for electric field intensity at a point from an infinitely long straight conductor carrying charge. (3)
Answer:

Consider a thin infinitely long straight rod conductor having charge density λ. (λ = \(\frac{q}{I}\))
To find the electric field at P ,we imagine a Gaussian surface passing through P.
Then according to Gauss’s law we can write,
\(\int \overrightarrow{\mathrm{E}} \cdot \mathrm{~d} \overrightarrow{\mathrm{~s}}\) = \(\frac{1}{\varepsilon_0} q\)
\(\int E d s \cos \theta\) = \(\frac{1}{\varepsilon_0} q\) (θ = 0°)
\(E \int d s\) = \(\frac{\lambda I}{\varepsilon_0}\) (Since q = λI)
Integrating over the Gaussian surface, we get (we need not integrate the upper and lower surface because, electric lines do not pass through these surfaces.)
E 2πrI = \(\frac{\lambda I}{\varepsilon_0}\) (L.S.A. of cyhnder = 2πrI)
E = \(\frac{1}{2 \pi r I} \frac{\lambda I}{\varepsilon_0}\)
E = \(\frac{1}{2 \pi \varepsilon_0} \frac{\lambda}{r}\)

Question 23.
(a) State Ohm’s, law. (1)
Answer:
Ohm’s Law states that the current through a conductor between two “points is directly proportional to the potential difference across its ends. Mathematically, we can represent it as V = IR, where V is the potential difference, I is the current, and R is the resistance of the conductor.

(b) Derive Wheatstone’s network principle.
Answer:
Four resistances P,Q,R and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I₁, I₂,I₃ and I₄ be the four currents passing through P, R,Q and S respectively.

Working
The voltage across R.
When key is closed, current flows in different branches as shown in figure. Under this situation
The voltage across P, V_AB = I₁P
The voltage across Q, V_BC = I₃Q ………(1)
The voltage across R, V_AD = I₂R
The voltage across S, V_DC = I₄S
The value of R is adjusted to get zero deflection in galvanometer. Under this condition,
I₁ = I₃ and I₂ = I₄ ………(2)
Using Kirchoffs second law in loopABDA and BCDB, we get
V_AB = V_AD …………(3)
and V_BC = V_DC ……….. (4)
Substituting the values from eq(1) in to (3) and (4), we get
I₁P = I₂R …………..(5)
and I₃Q = I₄S ………….(6)
Dividing Eq(5)byEq(6)
\(\frac{I_1 P}{I_3 Q}\) = \(\frac{\mathrm{I}_2 \mathrm{R}}{\mathrm{I}_4 \mathrm{~S}}\)
\(\frac{P}{Q}\) = \(\frac{R}{S}\) [since I₁ = I₃ and I₂ = I₄]
This is called Wheatstone condition.

Question 24.
(a) State Snell’s law of refraction. (1)
Answer:
The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.
\(\frac{\sin i}{\sin r}\) = n₂₁
where n₂₁ is a constant, called the refractive index of medium 2 with respect to medium 1.

(b) Explain critical angle and total internal reflection.
Answer:
The angle of incidence for which angle of refraction becomes 90° is called critical angle and total internal reflection is the phenomenon where light travels from denser to rarer medium and the angle of incidence when becomes greater than critical angle, there is no refracted light and all the light is reflected back to the denser medium. This phenomenon is known as total internal reflection.

Question 25.
(a) What are coherent sources? (1)
Answer:
Coherent sources of light are those which have constant their difference between them.

(b) In Young’s double slit experiment, interference pattern is observed at 5 cm from the slits with a fringe width of 1 mm. Calculate the separation between the slits. (k = 500 A) (3)
Answer:
λ = 5000 A
n = 3
D = 1.5 m
wkt,
β = \(\frac{nλD}{d}\)
d = \(\frac{nλD}{ß}\)
= \(\frac{3 \times 5000 \times 10^{-10} \times 1.5}{10^{-2}}\)
= 225 × 10^-6 m
d = 0.0225 cm

Answer any 3 questions from 26 to 29. Each carries 5 scores. (3×5 = 15)

Question 26.
(a) What is the principle of a capacitor? (1)
Answer:
Whenever two neutral conductors are placed nearby, and a potential difference is applied to them, then equal and opposite charges are induced on them.
Therefore, due to these charges, Energy is stored in the form of Electric Field in the gap between them.
A capacitor is device used to store Energy.
The charge appearing on the conductors is directly proportional to the Potential difference applied to them. i.e.
Q ∝ V
⇒ Q = CV
Where C is the constant of proportionality, called capacitance of the system, it depends on physical dimensions of the conductors.

(b) Derive the expression for capacitance of a parallel plate capacitor. (2)
Answer:

Electric field between the plate is given by,
E = \(\frac{\sigma}{\varepsilon_0}\) = \(\frac{Q}{\varepsilon_0 \mathrm{~A}}\) [σ = \(\frac{Q}{A}\)]

Potential difference between the two plates is,
V = Ed = \(\frac{Qd}{\varepsilon_0 A}\)]

Capacitance of capacitor,
C = \(\frac{Q}{V}\) = \(\frac{Q}{\frac{Q d}{\varepsilon_0 A}}\) = \(\frac{{\varepsilon_0 A}}{d}\)

(c) A 12 pF capacitor i connected to 50 V battery. How much electrostatic energy is stored in the capacitor? (2)
Answer:
U = \(\frac{1}{2}\) CV²
= \(\frac{1}{2}\) × 12 × 10^-12 ×50²
= 1.5 × 10^-8 J

Question 27.
(a) The direction of magnetic field around a current carrying conductor is given by . (1)
Answer:
Eight hand thumb rule or right hand grip rule or Maxwell’s Cork screw rule(write name of any one rule).

(b) State Biot-Savart law. (1)
Answer:

The strength of the magnetic field dB is
(i) directly proportional to the current
I, dB ∝ I
(ii) directly proportional to the length
dl, dB ∝ dl
(iii) directly proportional to the sine of the angle (θ) between the element and the line joining the midpoint of the element to the point, dB ∝ sin θ.
(iv) inversely proportional to the square of the distance between the element and the point,
Combining these we get
dB ∝\(\frac{1}{r^2}\)
dB ∝\(\frac{I d l \sin \theta}{r^2}\) = \(\frac{\mu_0}{4 \pi} \frac{I d l \sin \theta}{r^2}\)

(c) Derive the expression for magnetic field on the axis of a circular coil carrying current.(3)
Answer:

Consider a circular loop of radius ‘a’ and carrying current I. Let P be a point on the axis of the coil, at distance x from A and r from ‘O’. Consider a small length dl at A.
The magnetic field at ‘p’ due to this small element dI,
dB = \(\frac{\mu_0 \mathrm{Idl} \sin 90}{4 \pi \mathrm{x}^2}\)
dB = \(\frac{\mu_0 \mathrm{Idl}}{4 \pi \quad \mathrm{x}^2}\) …………..(1)
The dB can be resolved into dB cosϕ (along P_y) and dB sinϕ) (along P_x).
Similarly consider a small element at B, which produces a magnetic field ‘dB’ at P. If we resolve this magnetic field we get.
dB sinϕ (along P_x) and dB cosϕ (along P_y1)
dB cosϕ components cancel each other, because they are in opposite direction. So only dB sinϕ components are found at P, so total field at P is

Question 28.
(a) Write the expression for instantaneous emfofa.c. (1)
Answer:
V = V_m sin t or E = E₀

(b) Identify A, B and C in figure. (1)

Answer:
A – Resistance
B – Inductor
C – Capacitor

(c) Draw the phasor diagram of the above circuit and write the expression for imped-ance in the circuit, then mention the terms. (3)
Answer:

Impendence of the circuit Z is given by
Z = \(\sqrt{\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2}\)
R = Resistance, XL = Lω; – inductive reactance, x_c = \(\frac{1}{\mathrm{C} \omega}\) :- Capacitive reactance.

Question 29.
(a) Derive lens maker’s formula. (3)
Answer:
Consider a thin lens of refractive index n₂ formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n₁. Let an object ‘O’ is placed in the medium of refractive index n₁. Hence the incident ray OM is in the medium of refractive index n1 and the refracted ray MN is in the medium of refractive index n₂.

The spherical surface ABC (radius of curvature R₁) forms the image at I₁. Let ‘u’ be the object distance and V₁ be the image distance. Then we can write,
\(\frac{n_2}{v_1}\) – \(\frac{n_1}{u}\) = \(\frac{n_2-n_1}{R_1}\) ……….(1)
This image I₁ will act as the virtual object for the surface ADC and forms the image at v. Then we can write,
\(\frac{n_1}{v}\) – \(\frac{n_2}{v_1}\) = \(\frac{n_1-n_2}{R_2}\) ……….(2)
Adding eq (1) and eq (2) we get
\(\frac{n_2}{v_1}\) – \(\frac{n_1}{u}\) + \(\frac{n_1}{v}\) – \(\frac{n_2}{v_1}\) = \(\frac{n_2-n_1}{R_1}\) + \(\frac{n_1-n_2}{R_2}\)
\(\frac{n_1}{v}\) – \(\frac{n_1}{u}\) = (n₂ – n₁) (\(\frac{1}{R_1}\) – \(\frac{1}{R2}\))
Dividing throughout by n,, we get
\(\frac{1}{v}\) – \(\frac{1}{u}\) = (\(\frac{n_2}{n_1}\) – 1) (\(\frac{1}{R_1}\) – \(\frac{1}{R2}\))
if the lens is kept in air, \(\frac{n_2}{n_1}\) = n
So the above equation can be written as,
\(\frac{1}{v}\) – \(\frac{1}{u}\) = (n – 1) (\(\frac{1}{R_1}\) – \(\frac{1}{R2}\))
From the definition of the lens, we can take,
when u = ∞, f = v
Substituting these values in the eq (3), we get
\(\frac{1}{f}\) – \(\frac{1}{∞}\) = (n – 1) (\(\frac{1}{R_1}\) – \(\frac{1}{R2}\))
This is lens maker’s formula
\(\frac{1}{f}\) = (n – 1) (\(\frac{1}{R_1}\) – \(\frac{1}{R2}\))

(b) Draw the image formation in a simple microscope. (1)
Answer:

(c) Write the value of least distance of distinct vision. (1)
Answer:
25 cm is the value of least distance of distinct vision.