Kerala Plus Two Physics Question Paper March 2021 with Answers

Reviewing Kerala Syllabus Plus Two Physics Previous Year Question Papers and Answers Pdf March 2021 helps in understanding answer patterns.

Kerala Plus Two Physics Previous Year Question Paper March 2021

Answer the following questions from 1 to 45 upto a maximum score of 60.

I. Questions 1 to 8 carry 1 score each. (8 × 1 = 8)

Question 1.
Fill in the blanks:
” The force between two point charges is directly proportional to the product ……… of and inversely proportional to the ………. of the distance between them”.
Answer:
charges, square

Question 2.
The expression ΣB.AS = 0 is
(i) Gauss Law in Electrostatics
(ii) Gauss Law in Magnetism
(iii) Ampere’s circuital law ‘
(iv) Lenz’s law
Answer:
(ii) Gauss Law in Magnetism

Question 3.
The electromagnetic waves used in LASIK eye surgery is
(i) microwaves
(ii) ultraviolet rays
(iii) infra-red waves
(iv) gamma rays
Answer:
(ii) ultraviolet rays

Question 4.
Write Lens maker’s formula:
Answer:
\(\frac{1}{f}\) = (n-1) [\(\frac{1}{R_1}\) – \(\frac{1}{R_2}\)]

Question 5.
Name the property of light that proves its transverse nature.
Answer:
Polarisation

Question 6.
Write the equation for the wavelength of de Broglie wave associated with a moving particle.
Answer:
Wave length λ = \(\frac{h}{mV}\)

Question 7.
Energy of electron in the n^th orbit of hydrogen atom is E_n =\(\frac{13.6}{n^2}\) eV. What is the energy required to make electron free from first orbit of hydrogen atom?
Answer:
+ 13.6 eV

Question 8.
If radius of first electron orbit of hydrogen is a₀, radius of second electron orbit of hydrogen is ……… .
Answer:
a = 4a₀ [a = n²a₀; n = 2]

Questions from 9 to 22 carry 2 scores eqch. (14 × 2 = 28)

Question 9.
Calculate the electric potential at a point 9.0 cm away from a point charge of 4 × 10^-7 C.
Answer:
V = \(\frac{1}{4 \pi \varepsilon_0}\) \(\frac{q}{r}\)
= 9 × 10⁺⁹ × \(\frac{4 \times 10^{-7}}{9 \times 10^{-2}}\) = 4 × 10⁴ V

Question 10.
State Biot – Savart law and express it mathematically.
Answer:
dB = \(\frac{\mu_0}{4 \pi}\) \(\frac{\mathrm{Id} \ell \sin \theta}{\mathrm{r}^2}\)

Question 11.
Draw Wheatstone’s bridge and write its balancing condition.
Answer:

When bridge is balanced, \(\frac{P}{Q}\) = \(\frac{R}{S}\)

Question 12.
Determine the value of resistance R in the figure, assuming that the current through the galvanometer (G) is zero.

Answer:
When galvanometer current is zero,
We can write \(\frac{P}{Q}\) = \(\frac{R}{S}\)
\(\frac{R}{3}\) = \(\frac{40}{60}\)
R = 3 × \(\frac{40}{60}\) = 2 Ω

Question 13.
Write any two properties of nuclear force.
Answer:
(i) N uclear force is indepentant of charge
(ii) It is a short range force

Question 14.
Define half life of a radioactive sample. Write the equation that connects half life with disintegration constant.
Answer:
Half life is the time taken to reduce half of initial value of sample.
T_1/2 = \(\frac{0.693}{λ}\)

Question 15.
An air cored solenoid has 1000 turns per metre and carries a current of 2A. Calculate the magnetic intensity (H).
Answer:
n = 1000, I = 2 A
H = 1000 × 2 = 2000 Alm³

Question 16.
The behaviour of magnetic filed lines near two magnetic substances P and Q are shown below.

(a) From the figure identify paramagnetic substance.
Answer:
Substance Q

(b) Susceptibility of substance P is ………….. (positive /negative).
Answer:
Negative

Question 17.
Current in a curcuit fall from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced. Calculate the self inductance of the curcuit.
Answer:
ε = L \(\frac{dI}{dt}\) dI = 5 – 0
200 = L \(\frac{5 – 0}{0.1}\) dt = 0.1 ; ε = 200
L = \(\frac{200 \times 0.1}{5}\) = 4 H

Question 18.
Using a suitable ray diagram prove that the radius of curvature of a spherical mirror is twice its focal length.
Answer:

Consider a ray AB parallel to principal axis incident on a cpncave mirror at point B and is reflected along BF. The line CB is normal to the mirror as shown in the figure.
Let θ be angle of incidence and reflection.
Draw BD ⊥ CP,
In right angled ∆ BCD,
Tanθ = \(\frac{BD}{CD}\) ……….(1)
In right angled ∆ BFD,
Tan2θ = \(\frac{BD}{FD}\) ……….(2)
Dividing (1) and(2)
\(\frac{Tanθ}{Tan2θ}\) = \(\frac{CD}{FD}\) ………(3)
If θ is very small, then tanθ ≈ θ and tan2θ ≈ 2θ
The pointB lies very close to P. Hence CD ≈ CP and FD ≈ FP From (3) we get
\(\frac{2θ}{θ}\) = \(\frac{PC}{PF}\) = \(\frac{R}{f}\)
R = 2f

Question 19.
A light bulb of resistance 484 Ω is connected with 200 V ac supply. Find peak value of current through the bulb.
Answer:
R = 484 Ω
V = 220 V
I_rms = \(\frac{V}{R}\) = \(\frac{220}{484}\) = 0.45 A
I_rms = \(\frac{I_0}{\sqrt{2}}\)
I₀ = \(\sqrt{2} \times I_{m s}\)
= √2 × 0.45 = 0.64 A

Question 20.
Write any two postulates of Bohr model of hydrogen atom.
Answer:
1. Electrons revolve round the positively charged nucleus in circular orbits.
2. The electron which remains in a privileged path cannot radiate its energy.
3. The orbital angular momentum of the electron is an integral multiple of h/2π.
4. Emission or Absorption of energy takes place when an electron jumps from one orbit to another.

Question 21.
The symbol of a logic gate is given below. Identify the gate and write its truth table.

Answer:
OR gate
Tablee

Question 22.
When bulk pieces of conductors are subjected to changing magnetic flux, currents are induced in them.
(a) Write the name of this induced current.
Answer:
Eddy current

(b) Write any two practical applications of this current.
Answer:
(i) Magnetic breaking
(ii) Induction furnance

Questions from 23 to 34 carry 3 scores each. (12 × 3 = 36)

Question 23.
(a) Define electric dipole moment. (1)
Answer:
Electric dipole moment is product of magnitude of charge of dipole and its length.

(b) A system has two charges 2.5 × 10^-7 C and 2.5 × 10^-7 C located at points (0, 0, -15 cm) and (0,0, +15 cm), respectively.
Determine the magnitude and direction of electric dipole moment of the system. (2)
Answer:
q = 2.5 × 10^-7 C
2a = 2 × 15 = 30 cm
P = q × 2a = 2.5 × 10^-7 × 30 × 10^-2
= 75 × 10^-9 Cm
direction → -q to +q

Question 24.
(a) Write any two properties of electric field lines. (2)
Answer:
(i) Electric field lines never intercept each other.
(ii) Electric field lines are originated from positive charge and terminated at negative charge.

(b) Observe the figure and write the signs of the charges q₁ and q₂.

Answer:
q₁ is positive and q₂ is negative.

Question 25.
Derive an expression for the energy stored in a ca-pacitor in terms of capacitance and potential differ-ence across the capacitor.
Answer:
Energy of a capacitor is the work done in charging it. Consider a capacitor of capacitance ‘C’. Let ‘q’ be the charge at any instant and ‘V’ be the potential. If we supply a charge ‘dq’ to the capacitor, then work done can be written as,
dw = Vdq
dw = \(\frac{q}{C}\)dq (Since V = \(\frac{q}{C}\))
∴ total work done to change the capacitor (from 0 to Q) is
W = \(\int_0^Q \frac{q}{C} d q\)
W = \(\frac{1\left[q^2\right]_0^Q}{\mathrm{C} 2}\)
W = \(\frac{1}{C} \frac{Q^2}{2}\)
but Q = CV
W = \(\frac{1}{2}\) CV²
This work done is stored in the capacitor as electric potential energy.
Energy stored in the capacitor is,
U = \(\frac{1}{2}\) CV²

Question 26.
Write any one difference between polar and non-polar molecule: Give one example each for polar and non-polar molecule.
Answer:
In polar molecule, positive centre and negative centre does not coincide each other. But in non polar molecule, positive centre and negative centre coin-cide each other.
Example:
Polar molecule – H₂O, HCl
Non polar molecule – H₂, O₂

Question 27.
(a) Define angle of dip. (1)
Answer:
The angle between earths magnetic field with its horizontal is called dip.

(b) At a particular place the horizontal and vertical components of earth’s magnetic field are found to be equal. What is the value of dip at this place? (2)
Answer:
B_H = B_V
B cosθ = B sinθ
cosθ = sinθ
∴ θ = 45°

Question 28.
In the figure shown below
(a) Which are the resistors Connected in parallel? (1)
Answer:
6Ω and 3Ω

(b) Calculate the current drawn from the cell. (2)

Answer:
6Ω and 3Ω resistors are’connected in parallel. Hence effecting resistance
R = \(\frac{6 \times 3}{6+3}\) = \(\frac{18}{9}\) = 2Ω
This 2Ω connected 8Ω in series.
.-. Total resistance = 8 + 2 = 10Ω
Current I = \(\frac{\text { Total voltage }}{\text { Total resistance }}\)
I = \(\frac{24}{10}\) = 2.4 A

Question 29.
Using Ampere’s circuital law show that the intensity of magnetic field at an axial point near the centre of a current carrying solenoid is B = μ₀n I.
Answer:

Consider a solenoid having radius ‘r’. Let ‘n’ be the number of turns per unit length and I be the current flowing through it.
In order to find the magnetic field (inside the solenoid ) consider an Amperian loop PQRS. Let ‘e ‘ be the length and ‘b’ the breadth.
Applying Amperes law, we can write

(since RS is completely out side the solenoid, for which B = 0)
Substituting the above values in eq (1 ),we get
Bl = μ₀ I_ene …….(2)
But I_ene = n l I
where ‘nl’ is the total number of turns that carries current I (inside the loop PQRS)
∴ eq (2) can be written as
Bl = μ₀ nI l
B = μ₀ nI
If core of solenoid is filled with a medium of relative permittivity nr then
B = μ₀μ_rnl

Question 30.
Write a circuit diagram explain how a moving coil galvanometer can be converted to an ammeter.
Answer:
A galvanometer can be converted into an ammeter by a low resistance (shunt) connected parallel to it.
Theory: Let G be the resistance of the galvanometer, giving full deflection for a current I_g.
To convert it into an ammeter, a suitable shunt resistance ‘S’ is connected in parallel. In thisoafirange- ment Ig current flows through Galvanometer and remaining (I-Ig) current flows through shunt resistance.
Since G and S are parallel
P.d Across G = p.d across S
Ig × G = (I-Ig)S

Question 31.
Prove that when an alternating voltage is applied to an inductor, the current thrcjugh it lags behind voltage by an angle \(\frac{\phi}{2}\).
Answer:

Consider a circuit containing an inductor of inductance ‘L’ connected to an alternating voltage.

Let the applied voltage be
V = V₀ sin ωt ………(1)
Due to the flow of alternating current through coil, an emf, L\(\frac{dI}{dt}\) is produced in the coil. This induced emf is equaland opposite to the applied emf (in the case of ideal inductor)
ie. L\(\frac{dI}{dt}\) = V₀ sin ωt
dI = \(\frac{dI}{dt}\) sin ωt dt
Integrating, we get
I = \(\frac{V_0}{L \omega}\) cos ωt
I = \(\frac{V_0}{L \omega}\) Sin (ωt – \(\frac{\phi}{2}\))
I = I₀ Sin (ωt – \(\frac{\phi}{2}\)) ………(2)
Where I₀ = \(\frac{V_0}{L \omega}\)
The term Lω is called inductive reactance. Comparing eq(1) and eq(2), we can understand that, the current lags behind the voltage by an angle 90°.

Question 32.
(a) The current due to time varying electric field is called ……… (1)
Answer:
displacement current

(b) An electromagneticjwave travels in free space with a velocity of 3 × 10⁸ m/s. At a particular point in space and timte, magnitude of intensity of electric field is 6.3 V/m. What is magnitude of magnetic field at this point? (2)
Answer:
C = \(\frac{E}{B}\)
3 × 10⁸ = \(\frac{63}{B}\)
Magnetic field, B = \(\frac{63}{3 × 10^8}\) = 2.1 × 10^-8 T

Question 33.
Using Huygens wave theory prove that angle of incidence is equal tojangle of reflection.
Answer:
Reflection of plane wave by a plane surface.

AB is the incident wavefront and CD is the reflected wavefront, ‘i’ is the angle of incidence and ‘r’ is the angle of reflection. Let c₁ be the velocity of light in the medium. Let PO be the incident ray and OQ be the reflected ray.
The time taken for the ray to travel from P to Q is
t = \(\frac{\mathrm{PO}}{\mathrm{C}_1}+\frac{\mathrm{OQ}}{\mathrm{C}_1}\) …….(1)
t = \(\frac{A O \sin i}{C_1}+\frac{O D \sin r}{C_1}\)

O is an arbitrary point. Hence AO is a variable. But the time to travel for a wave front from AB to CD is a constant. So eq.(2) should be independent of AO. i.e., the term containing AO in eq.(2) should be zero.

\(\frac{A O}{C_1}\) (sin i – sin r) = 0
sin i – sin r = 0
sin i = sinr i = r

Question 34.
(a) Write Einstein’s photoelectric equation.
Answer:
hυ = hυ₀ + \(\frac{1}{2}\)mv²

(b) Using this equation show that, “photoelectric emission is not possible if the frequency of incident radiation is less than threshold – frequency”. (2)
Answer:
hυ = hυ₀ + \(\frac{1}{2}\)mv²
If incident frequency is less than threshold frequence (υ < υ₀), kinetic energy of photoelectron becomes negative, which means that no photo emission takes place.