Kerala Plus Two Physics Board Model Paper 2021 with Answers

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

Kerala Plus Two Physics Board Model Paper 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.
The name of the wave associated with matter is called …………… .
Answer:
Matter wave

Question 2.
The vertical’ plane passing through the axis of rotation of earth is called …………… .
Answer:
Geographical Meridian

Question 3.
What happens to the ray of light when it travels from rarerto denser medium?
a) Bends towards the normal
b) Bends awayfrom the normal
c) No change
Answer:
Bends towards the normal

Question 4.
Which physical quantify is quantised in Bohr’s second postulate?
Answer:
Angular momentum

Question 5.
Infrared spectrum lies between
a) Radio and microwave
b) Visible and UV
c) Microwave and visible
d) UV and X rays
Answer:
Microwave and visible

Question 6.
When a ray of light enters a glass slab from air:
a) Its wavelength decreases
b) Its wavelength increases
c) Its frequency increases
d) Its frequency decrases
Answer:
Its wavelength decreases

Question 7.
How many electrons constitute 1 coulomb of charge (e = 1.6 × 10^-19 C)?
Answer:
Q = ne
1 = n × 1.6 × 10^-19
n = \(\frac{1}{1.6 \times 10^{-19}}\) = 0.625 × 10¹⁹
No. of electrons, n = 6.25 × 10¹⁸

Question 8.
Name the series of hydrogen spectrum which has least wavelength.
Answer:
Lyman series

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

Question 9.
a) Define electrid potential. (1)
Answer:
Electric potential at a point is the workdone required to bring a unit positive charge from infinity to that point without acceleration.

b) Give the relation between electric intensity and electric potential. srii (1)
Answer:
E = \(\frac{-dV}{dx}\)

Question 10.
a) What is the principle of Potentiometer? (1)
Answer:
Potential difference between two points of a current carrying conductor is directly proportional to length between two points.

b) Write one practical application of Wheatstone’s bridge (1)
Answer:
Wheatstone’s bridge can be used to find unknown resistance.

Question 11.
A wire has a resistance of 16 Ohms. It is bent in the form of a circle. Find the effective resistance between two points on any diameter of the circle. (2)
Answer:

Each half will have 8Ω resistance and these can be considered to be connected in parallel as shown.
\(\frac{1}{2}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\)
Effective resistance, R = \(\frac{R_1 R_2}{R_1+R_2}\) = \(\frac{8 \times 8}{8+8}\)
= \(\frac{64}{16}\) = 4Ω

Question 12.
a) A stationary charge can produce magnetic field. (True/False) (1)
Answer:
False

b) Write down the equation for magnetic Lorentz force. (1)
Answer:
\(\overline{\mathrm{F}}=\mathrm{q}(\bar{v} \times \overline{\mathrm{B}})\)

Question 13.
a) What is the intensity of magnetisation for magnetic materials? (1)
Answer:
Magnetic moment

b) Give the relation between B and H (1)
Answer:
B = μ₀ H

Question 14.
State Faraday’s laws of electromagnetic induction.(2)
Answer:
(i) Whenever a flux linked with a coil is changed, an emf is induced in the coil.
(ii) The magnitude of induced emf is directly proportional to rate of change of flux.
ε = \(\frac{d \phi}{d t}\)

Question 15.
Draw the ray diagram for a convex lens producing virtual image. (2)
Answer:

Question 16.
State any two postulates of Bohr atom model: (2)
Answer:
(i) Electrons move around the positively charged nucleus in circular orbjts.
(ii) The electron which remains in a privileged path cannot radiate its energy.

Question 17.
a) State the law of radioactive decay. (1)
Answer:
Rate of disintegration of radio active nuclei is directly proportional to number of atoms
i.e. \(\frac{dN}{dt}\) ∝ N

b) What are the number of protons and neutrons in a nucleus ₉₂U²³⁸. (1)
Answer:
Number of protons = 92
number of neutrons = 238 – 92 = 146

Question 18.
In the magnetic meridian of certain place, the horizontal component of earth’s magnetic field is 0.26G and the dip angle is 60°. What is the magnetic field of the earth at this location? (2)
Answer:
Horizontal component of magnetic filed, B_H = 0.2669
dip angle, 0 = 60°
Earths magnetic filed B_H = Bcos60
0. 26 = B cos 60°
B = \(\frac{0.26}{\cos 60}\)
= 0.52 G

Question 19.
a) Give the principle of a transformer. (1)
Answer:
Mutual induction

b) Give the two energy losses in transformer. (1)
Answer:
(i) Eddy current loss
(ii) Flux leakage loss

Question 20.
a) Draw the Phasor diagram with V and I for an inductive circuit. (1)
Answer:
V_L lags I by π/2.

b) What is the phase difference between V and I in an inductive circuit. (1)
Answer:
π/2

Question 21.
Give two differences between nuclear fission and fusion. (2)
Answer:

Fission	Fusion
Spliting of heavy nuclei in to two lighter nucleus	Fusion of two light nuclei in to single nuclei
Alpha and beta particles are emitted	Alpha and Beta particles are not emitted

Question 22.
a) What is meant by for hidden energy gap? (1)
Answer:
The energy gap between valence band and conduction band is called forbidden energy gap.

b) Write any one use of zener diode. (1)
Answer:
Zener diode is used as voltage regulator.

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

Question 23.
a) State Guass’s theorem. (1)
Answer:
Total flux over a closed surface is \(\frac{1}{\varepsilon_0}\) times charge enclosed by the surface.

b) Give the equation forelectric flux through the given surface when the angle between electric field and area is 45°. , (1)

Answer:
flux dϕ = E. ∆ S
= E ∆S cos θ
= E ∆Scos45°
= \(\frac{1}{\sqrt{2}}\) E ∆S

c) What is the flux through the surface if the surface is parallel to the lines offeree? (1)
Answer:
If the surface is parallel to the electric field line,
θ = 90°
∴ Flux ∆ϕ = ∆S cos 90 = θ

Question 24.
Find the effective capacitance when three capacitors are connected in parallel. (3)
Answer:
Capacitors in parallel

Let three capacitors C₁, C₂ and C₃ be connected in parallel to p.d of V. Let q₁, q₂ and q₃ be the charges on C₁, C₂ and C₃.
If ‘q’ is the total charge .then’q’can be written as
q = q₁ + q₂ + q₃
But q₁ = C₁V, q₂ = C₂V and q₃ = C₃V
Hence eq (2) can be written as
CV = C₁V + C₂V + C₃V
C = C₁ + C₂ + C₃
Effective capacitance increases in parallel connection.

Question 25.
A solendid of length of 0.5 m has radius 1 cm and is made up 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid? (3)
Answer:
l = 0.5 m, r = 1 × 10_-2 m, N = 500, I = 5 A
Magnetic field = μ₀ nl
When n = \(\frac{N}{l}\) = \(\frac{500}{0.5}\)
n = 1000
B = μ₀ × 1000 × 5
= 4π x 10^-7 × 1000 × 5
= 6.2 × 10^-3 T

Question 26.
a) Name the angle between horizontal component of earth’s magnetic filed and earth’s magnetic field. (1)
Answer:
dip

b) Define twd magnetic elements of the earth. (2)
Answer:
The magnetic elements of the earth are dip, declination and horizontal intensity.

Declination :
Declination at a place is the angle between the geographic meridian and magnetic meridian at that place.

Horizontal intensity:
The horizontal intensity at a place is the horizontal components of the earths field.

Question 27.
a) Name the principle of AC Generator. (1)
Answer:
Electro magnetic induction

b) Derive the equation for instantaneous e.m.f. in an AC Generator. (2)
Answer:

Question 28.
a) Give two properties of electromagnetic waves.(2)
Answer:
(i) No medium is required for propagation of em wave.
(ii) The ratio of magnitudes of electric and magnetic field vectors in free space is constant.

b) Give one use of radio waves. (1)
Answer:
Radio waves are used in radio and television communication systems.

Question 29.
State Brewster’s law. A glass plate of refractive index. 1.60 is used as a pplarizer. Find the polarising angle. (3)
Answer:
Brewster’s law states that the tangent of the polaring angle is equal to the refractive index of the material of reflector.

n = 1.6, θ = ?
n = tan θ
1. 6 = tan θ
θ = tan^-1 (1.6)
θ = 58°

Question 30.
Calculate the work function in electron volt for a metal, given that the photoelectric threshold wavelength is 6800 Å. (3)
Answer:
Threshold wavelength A,θ = 6800 × 10^-10 m
Work function E = \(\frac{\mathrm{hc}}{\lambda_0}\)
= \(\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6800 \times 10^{-10}}\)
= 2.9 × 10^-19 J

Question 31.
Derive the equation for the electric field intensity due to an infinite thin sheet of charge using Gauss’s law. (3)
Answer:

Consider an infinite thin plane sheet of charge of. density σ.
To find electric field at a point P (at a distance ‘r’ from sheet), imagine a Gaussian surface in the form of cylinder having area of cross section ‘ds’.
According to Gauss’s law we can write,
\(\int \overrightarrow{\mathrm{E}} \cdot \mathrm{~d} \overrightarrow{\mathrm{~s}}\) = \(\frac{1}{\varepsilon_0} \mathrm{q}\)
\(\mathrm{E} \int \mathrm{ds}\) = \(=\frac{\sigma d s}{\varepsilon_0}\) (Since q = σds)
But electric field passes only through end surfaces, so we get \(\int \mathrm{ds}\) = ‘2ds
ie., E 2ds = \(\frac{\sigma d s}{\varepsilon_0}\)
E = \(\frac{\sigma \mathrm{ds}}{2 \mathrm{ds} \varepsilon_0}\), E = \(\frac{\sigma}{2 \varepsilon_0}\)

Question 32.
Derive the equation for the capacitance of a parallel plate capacitor. (3)
Answer:

Potential difference between the two plates is,
V = Ed
= \(\frac{σ}{\varepsilon_0 A} d\)
V = \(\frac{Q}{A \varepsilon_0} d\) …….(1)

Capacitance C of the parallel plate capacitor,
C = \(\frac{Q}{V}\) ……..(2)
Sub. eq(1) in eq (2)
C = \(\frac{Q}{\frac{Q}{A \varepsilon_0} d}\)
C = \(\frac{{A \varepsilon_0}}{d}\)

Question 33.
Find the value of current in the circuit shown in the figure.

Answer:
Effective resistance = \(\frac{1}{R}\) = \(\frac{1}{30}\) + \(\frac{1}{60}\)
= 20Ω
Current I = \(\frac{V}{R}\) = \(\frac{2}{20}\) = 0.1 A

Question 34.
a) State Ampere’s circuital law.
Answer:
According to ampere’s law the line integral of magnetic field along any closed path is equal to μ₀ times the current passing through the surface.

b) Find the magnetic field along the axis of a solenoid, at its centre, carrying current. (3)
Answer:

Consider a solenoid having radius Y. 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 so-lenoid ) consider an Amperian loop PQRS. Let ‘l ‘ be the length and ‘b’ the breadth Applying Amperes law, we can write

Substituting the above values in eq (1 ),we get
Bl = μ₀ I_enc ……….(2)
But I_enc = nl I
where ‘n l ’ is the total number of turns that carries current I (inside the loop PQRS)
eq (2) can be written as
B l = μ₀nI l
B = μ₀nI
If core of solenoid is filled with a medium of relative permittivity μ_r then
B = μ₀ μ_r n l.

Questions from 35 to 41 carry 4 scores each. (7 × 4 = 28)

Question 35.
a) Give the SI unit of capacitance. (1)
Answer:
Farad

b) Two capacitors of capacitance 2pFand 4pF connected in series to potential difference of 100 Volt. Calculate the potential difference across each capacitor. (3)
Answer:
Effective capacitance C = \(\frac{C_1 C_2}{C_1+C_2}\)
CV = \(\frac{2 \times 4}{2+4}\) = \(\frac{4}{3} \mu \mathrm{~F}\)

Charge Q = CV
= \(\frac{4}{3}\) × 10^-6 × 100
= 133.3 pC

Capacitors are connected in series. Hence charge on each is same.
ie., Q = C₁V₁
133 × 10^-6 = 2 × 10^-6 × V₁

Voltage across first V₁ = \(\frac{133}{2}\) = 66.7 V

Voltage across send capacitor V₂ = \(\frac{133}{4}\) = 33.3 V

Question 36.
Derive an equation for the magnetic field due to a circular loop carrying current, at any point on the axis using Biot-Savart’s law. (4)
Answer:

Consider a circular loop of radius ‘a’ and carrying current T. 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 = \(\frac{\mu_0 \text { Idl }}{4 \pi x^2}\) ……..(1)
The dB can be resolved into dB cosϕ (along Py) and dB sinϕ (along Px).
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 px) and dB cosϕ (along py¹)
dB cosϕ components cancel each other, because they are in opposite direction. So only dB sinϕ components are found at P, so total filed at P is
B = ∫ dB sinϕ
= \(\int \frac{\mu_0 \mathrm{Idl}}{4 \pi \mathrm{x}^2} \sin \phi\)
but from ∆AOP we get, sinϕ = a/x
We get,

Point at the centre of the loop : When the point is at the centre of the loop, (r = 0)
Then,
B = \(\frac{\mu_0 \text { NI }}{2a}\)
B = \(\frac{\mu_0 R^2 I}{2\left(x^2+R^2\right)^{3 / 2}}\)

Question 37.
a) What is motional e.m.f? (1)
Answer:
e.m.f can be induced due tot he motion of a conductor in a magnetic field. This emf is called motional emf.

b) Derive the equation for the induced emf between the ends of a straight conductor moving perpendicular to a uniform magnetic field. (3)
Answer:

Consider a rectangular frame MSRN in which the conductor PQ is free to move as shown in figure. The straight conductor PQ is moved towards the left vyith a constant velocity v perpendicular to a uniform magnetic field B. PQRS forms a closed circuit enclosing an area that change as PQ moves. Let the length RQ = x and RS = l.
The magnetic flux Φ linked with loop PQRS will be Blx.
Since x is changing with time the rate of change of flux Φ will induce an e.m.f. given by
ε = \(\frac{-\mathrm{d} \phi}{\mathrm{dt}}\) = \(\frac{d}{d t}(B \ell x)\)
= – Bl \(\frac{d x}{d t}\)
ε = Blv
THe induced e.m.f Blv is called motional e.m.f.

Question 38.
Derive the mirror formula for a concave mirror. (4)
Answer:

Let points P, F, C be pole, focus and centre of curva-ture 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.

Question 39.
a) State Huygen’s principle. (2)
Answer:
According to Huygen’s principle
1. Every point in a wavefront acts as a source of secondary wavelets.
2. The secondary wavelets travel with the same velocity as the original value.
3. The envelope of all these secondary wavelets gives a new wavefront.

b) Based on Huygen’s wave theory of light, show that angle of incidence is equal to angle of reflection. (2)
Answer:

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 ,

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.
sin i – sin r = 0
sin i = sin r
i = r

Question 40.
a) What is the use of a rectifier? (1)
Answer:
Rectifier is used to convert AC in to DC

b) With the help of a fiagram explain how diode acts as a rectifier. (3)
Answer:

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 forward biased and D₂ is reverse biased. So that current flows through D₁ and R_L.

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 RL in the same direction. Thus we get a +ve voltage across R_L for +ve and -ve input. This process is called full wave rectification.

Question 41.
a) Define the principal focus of a convex lens. (1)
Answer:
A narrow beam of parallel rays, parallel and close to the principal axis, after refraction, converges to a point on the principal axis in the case of a convex lens or appears to diverge from a point on the axis in the case of a concave lens. This fixed point is called the principal focus of the lens.

b) Write the phenomenon related to the image formation in a lens. (1)
Answer:
Refraction

c) A convex lens of focal length 10 cm is combined with a concave lens of focal length 15 cm. Find the focal length of the combination. (2)
Answer:
f₁ = 10 × 10² m, f₂ = 15 × 10² m
\(\frac{1}{F}\) = \(\frac{1}{f_1}\) + \(\frac{1}{f_2}\)
\(\frac{1}{F}\) = \(\frac{f_1 f_2}{f_1 + f_2}\) = \(\left(\frac{10 \times 15}{10+15}\right) \times 10^{-2}\)
F = 6 × 10² m