Kerala Plus Two Physics Question Paper March 2020 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 2020

Time : 2 1/2 Hours
Maximum : 80 scores

I. Answer all questions from 1 to 6. Each carries 1 Score. (6 × 1 = 6)

Question 1.
How capacitance changes if the distance between the plates of a parallel plate capacitor is halved?
a) Does not change
b) Becomes half
c) Doubled
d) Becomes one fourth
Answer:
c) Doubled

Initial capaciatatance C₀ = \(\frac{A \varepsilon_0}{d}\)
New capaciatatance C₁ = \(=\frac{A \varepsilon_0}{d / 2}\) = \(\frac{2 A \varepsilon_0}{d}\)
C₁ = 2 C₀

Question 2.
The path of a charged particle entering parallel to uniform magnetic field will be
a) Circular
b) helical
c) Straight line
d) None of these
Answer
c) Straight line

Force acting on the charge F = q(\(\vec{υ} \times \vec{B}\))
= qυB
in this case θ = 0, Hence F = 0
No force acting on the partical, hence partical continues its path.

Question 3.
Coefficient of mutual inductance of two coils is 1H. Current in one of the coils is increased from 4 to 5 A in 1ms. What average emf will bw induced in the other coil?
a) 100V
b) 2000V
c) 100V
d) 200V
Answer:
a) 100V

Induced emf ε = M \(\frac{\mathrm{dI}}{\mathrm{dt}}\)
M = 1 H,dI = 5 – 4, dt = 1 × 10^-3
ε = \(=1 \times \frac{(5-4)}{1 \times 10^{-3}}\) = \(\frac{1}{10^-3}\)
= 10³
= 1000 V

Question 4.
Total internal reflection may be observed if
a) Light ray is travelling from denser medium to rarer medim
b) Light ray is travelling from rarer medium to denser medium
c) light ray is travelling from any medium to another medium
Answer:
a) Light ray is travelling from denser medium to rarer medim

Question 5.
Optical fibres make use of the phenomenon of ……….
a) refraction
b) total internal reflection
c) interference
d) diffraction
Answer:
b) total internal reflection

Question 6.
The size of the atom in Thomson’s model is ……… the atomic size in Rutherford’s model.
a) much greater than
b) not different from
c) much less than
Answer:
a) much greater than

Question 7.
A Permanent electric dipole of dipole moment P is placed in a uniform external electric field E, as shown in Figure.

(a) Redraw the figure and show the magnitude and direction of force actng on the charges.
Answer:

(b) Write an expresson of the torque acting on this dipole in vector form.
Answer:
Torque acting on the dipole τ = \(\overrightarrow{\mathbf{P}} \times \overrightarrow{\mathbf{E}}\)

Question 8.
Ampere’s theorem helps to find the magnetic field in a region around a current carrying conductor.
a) Write the expression of Ampere’s theorem.
Answer:
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{~d} \ell}=\gamma_0 \mathrm{I}\)

b) Draw a graph showing the variation of intensity of magnetic field with the distance from the axis of a current carrying conductor.
Answer:

Question 9.
A magnetised needle in uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however experiences a force of attraction in addition to a torque. Why?
Answer:
A bar magnet produces non – uniform field. Hence iron nail experiences a force of attraction in addition to a torque.

Question 10.
What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5.0 cm at a distance of 50cm from its mid-point? The magnetic moment of the bar magnet is 0.40 Am².
Answer:
If magnet is short,
Magnetic field along the equatorial line
B_E = \(\frac{\mu_0}{4 \pi} \frac{M}{r^3}\)
\(\frac{\mu_0}{4 K} \) = 10^-7, M = 0.40 Am², r = 50 cm = 0.5 cm
∴ B_E = 10^-7 × \(\frac{0.40}{(0.5)^3}\)
B_E = 3.2 × 10^-7 T
Magnetic field along the axial line
B_axial = \(\frac{\mu_0}{4 \pi} \frac{2 M}{2 r^3}\)
B_axial = 2 B_E
= 2 × 3.2 × 10^-7 T
= 6.4 × 10^-7 T

Question 11.
A magician during a show makes a glass lens with n = 1.47 disappear in a liquid.
a) What is the refractive index of the liquid?
Answer:
1.47

b) Could the liquid be water?
Answer:
No, the reflactive index of the water is 1.33

Question 12.
Explain why the bluish colour predominates in a clear sky.
Answer:
Blue colour undergoes for more scattering.

Question 13.
Match the following:
Tablee 1
Answer:
Tablee 2

Question 14.
Diodes are cone of the building elements of electronic circuits. Some type of diodes are shown in the figure.

a) Identify rectifier diode from the figure.
Answer:
(iii)

b) Draw the circuit diagram of a forward biased rectifier diodes are shown in the figure.
Answer:

Question 15.
The given figure shows the various propagation modes of e.m. waves in communication.

a) Write the names of propagation modes in A, B, C.
Answer:
A – ground wave
B – sky wave
C – space wave

b) Why transmission of TV signals via sky wave is not possible?
Answer:
The frequencies below 40 MHz is reflected by ionosphere. Tv signals have frequences above 54 MHz. Hence sky wave propagation;is not possible in Tv transmission.

Answer any 6 questions from 16 to 23. Each carries 3 scores. (6 × 3 = 18)

Question 16.
An infinitely long straight wire with uniform linear charge density is shown in figure.

a) Draw a Gaussian surface in order to calculate the electric field at P and mark direction of electric field at this point.
Answer:

b) Derive an expression to calculate electric field at this point P.
Answer:

Consider a thin infinitely long straight rod
conductor having charge density λ. [ λ = \(\frac{q}{l}\)]
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°)
\(E \int d s\)= \(\frac{\lambda I}{\varepsilon_0}\) q (since q = λ.I)
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.)

Question 17.
Three resistors R₁, R₂, R₃ are to be combined as shown in the figures.

a) Identify the series and parallel combinations,
Answer:
Fig 1 – parallel combination
Fig 2 – series combination

b) Which combination has lowest effective resistance?
Answer:
Parallel combination

c) Arrive at the expression for the effective resistance of parallel combination.
Answer:
Consider three resistors R₁, R₂ and R₃ connected in parallel across a pd of V volt. Since all the resistors are connected across same terminals, pd across all the resistors are equal.
Img 18
As the value of resistors are different current will be different in each resistor and is given by Ohm’s law
Current through the first resistor
I₁ = \(\frac{V}{R_1}\)
Current through the second resistor
I₂ = \(\frac{V}{R_2}\)
Current through the third resistor
I₃ = \(\frac{V}{R_3}\)

Total current through the combination is I = \(\frac{V}{R}\), where R is the effective resistance of parallel combination.
Total current through the combination = the sum of current through each resistor
I = I₁ + I₂ + I₃
Substituting the values of current we get
\(\frac{V}{R}\) = \(\frac{V}{R_1}\) + \(\frac{V}{R_2}\) + \(\frac{V}{R_3}\)
Eliminating V from all terms on both sides of the equations, we get
\(\frac{1}{R}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\)

Question 18.
a) State Faraday’s law of electomagnetic induction.
Answer:
Faraday’s Law of induction
Faraday’s law of electromagnetic induction states that the magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.
Mathematically, the induced emf is given by
\(\epsilon=\frac{d \phi}{d t}\)
If the coil contain N turns, the total induced emf is given by
\(\epsilon=N \frac{d \phi}{d t}\)

b) How does the magnetic energy stored in an inductor and electrostatic energy stored in a capacitor related to their respective field strengths?
Answer:
In capacitor energy is stored in electronic field
E = \(\frac{1}{2}\) cv²
In inductor energy is stored in magnetic field
E = \(\frac{1}{2}\)LI²

Question 19.
A typical plane electromagnetic wave propagating along the Z direction is shown in figure.

a) Write the equation for electric and magnetic fields.
Answer:
This em wave is travelling in z direction and direction of oscillation of electric field is along x. Hence equation for electric field
E_x = E₀ Sin(Kz-wt)
The direction of oscillation of magnetic field is along the y direction, hence
B_y = B₀ Sin (Kz-wt)

b) Write the methods of production of radio waves and microwaves. Write any one use of these waves. (1+2)
Answer:
Radio wave: Rapid accelartion and deacceleration of electrons in aerials. These are used in radio and televisiion communication systems.

Microwaves: Microwaves are produced by special vaccum tubes. Microwaves ovens are domestic applica¬tion of those waves.

Question 20.
The figure shows the image formation of an object in simple microscope.

a) Find out the object distance and image distance from the figure.
Answer:
OQ is the object distance.
OQ₁ is the image distance.

b) Derive an equation for magnifying power of the simple microscope.
Answer:
A simple microscope is a converging lens of small focal length.

Working: The object to be magnified is placed very close to the lens and the eye is positioned close to the lens on the other side. Depending upon the position of object, the position of image is changed.
If the image is formed at ‘D’, (D = 25 cm) we can take u = -D. Hence the lens formula can be written as
\(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)
1 – \(\frac{v}{u}\) = \(\frac{v}{f}\)
\(\frac{v}{u}\) = 1 – \(\frac{v}{f}\)
m = 1 – \(\frac{v}{f}\)
The image is formed at D, ie. v = -D
m = 1 – \(\frac{-D}{f}\)
m = 1 + \(\frac{D}{f}\)

Question 21.
The atomic line spectra of hydrogen atom is shown in figure.

Answer:
A – Lyman series
B – Balmor series
C – Paschen series

Question 22.
Spontaneous and continuous disinflation of a nucleus of a heavy element with the emission of certain types of radiation is known as radioactivity,
a) The radioactive isotope ‘D’ decays according to the sequence.

If the mass number and atomic number of D₂ are 172 and 71 respectively, what are the (i) Mass number, (ii) atomic number of D.
Answer:

When an α (₂He⁴) emitted, mass number 4 and atomic number 2 is reduced from the parent nucleus. Hence pamet nucleus,
D₁ = ₇₃Y¹⁷⁶
When is emitted, atomic number of daughter nuclei increases by 1 unit and mass numbers constant
Hence D = ₇₂Z¹⁷⁶

atomic number of D is 72, and mass number is 176.

b) State radioactive decay law.
Answer:
The number of nuclei undergoing decay per unit time is proportional to number of nuclei in the sample at that time.

c) Write he relation connecting half-life and mean life of radioactive element.
Answer:
T_1/2 = 0.693 τ
T_1/2 is the half life
τ is the mean life

Question 23.
In the broadcast of communication, modulation is necesary.
a) What do you mean by modulation?
Answer:
Modulation is the process of superprosing a low ferquency signal on a high frequency carrier wave.

b) Explain any two reason why modulation is necessary?
Answer:
i) Size of the antenna can be reduced.
ii) Effectiven power radiated from the source can be increased.

Answer any 2 questions from 24 to 26, Each carries 4 scores. (2 × 4 = 8)

Question 24.
Three capacitors of capacitances 2 pF, 3pF are connected in parallel.
a) Write the SI unit of capacitance.
Answer:
Farad

b) Calculate the effective capacitance of the combination.
Answer:
C = C₁ + C₂ + C₃
C = (2+3+4) PF
C = 9 PF

c) Determine the charge on each capacitor if the combination is connected to a 100V supply
Q = CV
Q₁ = C₁V
= 2 PF × 100 = 2 × 10^-10 C
Q₂ = C₂V = 3 PF × 100 = 3 × 10^-10 C
Q₃ = C₃V = 4 PF × 100 = 4 × 10^-10 C

Question 25.
A rectangular loop o ara A and carrying a steady current I is placed in a uniform magnetic field.
a) Derive the expression of torque, acting on the loop.
Answer:
Torque on a rectangular current loop in uniform magnetic filed

Consider a rectangular coil PQRS of N turns which is suspended in a magnetic field, so that it can rotate (about yy1). Let l be the length (PQ) and ‘b’ be the breadth (QR).
When a current ‘l’ flows in the coil, each side produces a force. The forces on the QR and PS will not produce torque. But the forces on PQ and RS will produce a Torque.
Which can be written as
τ = Force × ⊥ distance …….(1)
But, force = BIl ……..(2) [since θ = 90°]
And from AQTR, we get
⊥ distance (QT) = b sin θ ………..(3)
Substituting the vales of eq (2) and eq (3) in eq(1)
we get
τ = BIl b sin θ
= BIA sin θ [since lb = A (area)]
τ = IAB sin θ
τ = m B sin θ [since m = IA]
If there are N turns in the coil, then
τ = NIAB sin θ

b) Increasing the current sensitivity may not necessarily increase the voltage sensitivity, of a galvanometer. Justify.
Answer:
Current sensitivtity can be increased by incraesing number of turns. When number of turns doubble, the resistance of the wire will also be doubbled. Hence the voltage sensitivity does not change.

Question 26.
The work function of caesium metal is 2.14eV. When light of frequency 6 × 10¹⁴ Hz is incident On the metal surface, photoemission of electrons occurs. (h = 6.6 × 10¹⁴)
a) Define work function
Answer:
Work function is the amount of energy required fora electron tojust escape from a metal surface.

b) Calculate the maximum kinetic energy of the emitted electrons.
Answer:
hυ = ϕ₀ + KE
6.6 × 10³⁴ × 6 × 10^-20 = 2.14 × 1.6 × 10^-19 + KE
39.6 × 10^-20 = 3.424 × 10^-19 + KE
KE = 5.36 × 10^-19 J
KE = 3.35 ev
KE = ev₀
3.35 ev = ev₀
v₀ = 3.35 v

Answer any 3 questions from 27 to 30, Each carries 5 scores. (3×5 = 15)

Question 27.
A Wheatstone bridge is shown in figure.

a) Derive a relation connecting the four resistors for the galvanometer to give zero or null deflection.
Answer:
Four resistances P,Q,R and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I, 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 loop ABDA and BCDB, weget
V_AB = V_AD ………(3)
and V_BC = V_DC —-(4)
Substituting the values from eq(1) into (3) and (4), weget
I₁P = I₂R …….(5)
I₃Q = I₄S …….(6)
Dividing Eq(5)byEq(6)
\(\frac{\mathrm{I}_1 P}{\mathrm{I}_3 Q}\) = \(\frac{I_2 R}{I_4 S}\)
\(\frac{P}{Q}\) = \(\frac{R}{S}\)
This is called Wheatstone condition,