Kerala Plus Two Physics Question Paper SAY 2019 with Answers

Reviewing Kerala Syllabus Plus Two Physics Previous Year Question Papers and Answers Pdf SAY 2019 helps in understanding answer patterns.

Kerala Plus Two Physics Previous Year Question Paper SAY 2019

Time : 2 Hours
Total Score: 60

Use the following constants wherever neces-sary:

Permitivity of free space, s0= 8.854 × 10^-12 C²N^-1m^-2
Permeability of free space, JI0= 4π0 × 10^-7 Tm/A
Speed of light in vacuum, c = 3 × 10⁸ m/s

Questions 1 to 4 carry 1 score each. Answer any 3 questions. (3 × 1 = 3)

Question 1.
The value pf Electric field inside a charged conductor is ………….
Answer:
Zero

Question 2.
An e.m.f. of 16V is induced in a coil of self-inductance 4H. The rate of change of current must be
(i) 64 A/s
(ii) 32 A/s
(iii) 16 A/s
(iv) 4A/s
Answer:
(iv) 4A/s

ε = \(\frac{\mathrm{L dI}}{\mathrm{dt}}\)
\(\frac{\mathrm{dI}}{\mathrm{dt}}\) = \(\frac{\varepsilon}{L}\) = \(\frac{16}{4}\) = 4 A/s

Question 3.
The fundamental logic gate represented by the following circuit is:

Answer:
NOT

Question 4.
The type of modulation technique using frequency range 88 MHz. to 108 MHz.
Answer:
FM (Frequency modulation)

Questions 5 to 11 carry 2 score each. Answer any 6 questions. (6 × 2 = 12)

Question 5.
(a) The figure shows the Q(charge) versus V (potential) graph for a two combination of capacitors. Identify the graph representing the parallel combination.

Answer:
Graph A

(b) W rite the expression for the effective capacitance of two capacitors connected in parallel.
Answer:
C = C₁ + C₂

Question 6.
An α particle and a proton are moving in a region normal to a uniform magnetic field B. If two particles have equal linear momenta, what will be the ratio of the radii of their trajectories in the field? (2)
Answer:
Linear moment of α and proton are same
P_α = P_P
Radius R = \(\frac{m v}{q B}\)
r ∝ \(\frac{1}{q}\)
qα = 2q
q_P = q
r_α = \(\frac{1}{2 q}\)
r_P = \(\frac{1}{q}\)

Question 7.
The phasor diagram of an a.c. circuit is shown in figure.

(a) Identify the circuit. (1)
Answer:
Answer:
AC circuit containg inductor only

(b) Prove that the average power dissipated in the above circuit is zero. (1)
Answer:
P_av = V_rmsI_rms cosϕ
for pure indutor = ϕ = 90°
P_av = V_rmsI_rms cos90°
P_av = 0

Question 8.
In an electromagnetic wave, the oscillating electric field having a frequency of 3 x 10¹⁰ Hz. and an am-plitude of 30 V/m propagates in free space in the positive x-direction. Write down the expression to represent the electric field. (2)
Answer:
Wave lenght λ = \(\frac{C}{v}\) = \(\frac{3 \times 10^8}{3 \times 10^{10}}\) = \(10^{-2}\)
K = \(\frac{2 \pi}{\lambda}\) = \(\frac{2 \times 3.15}{10^{-2}}\) = 628 rad/m

Angular frequency ω = 2 π v
= 2 × 3.14 × 3 × 10⁸
18.84 × 10 rad/s

Question 9.
An unpolarized beam of light is incidewnt on water surface as in figure.

(a) Name the law relating refractive index of water with angle of incidence. (1)
Answer:
Brewsters law

(b) Find the value of refractive index of water, if the angle of incidence & 53°.
Answer:
n = tan θ
n = tan 53°
n = 1.33

Question 10.
(a) ypte Einstein’s photo electric equation. (1)
Answer:
Einstein’s photo electric equation

(b) Draw a graph showing the variation of Stopping potential with frequency of incident radiation. Mark threshold frequency on the graph.
Answer:

Question 11.
The block diagram of AM transmitter is shown in figure.

(a) Identify X and Y.
Answer:
x – amplitude modulator
y – power amplifier

(b) The amplitudes of modulating signal and carrier wave are 9V and 12V respectively. Find the value of modulation index.
Answer:
Modulation index M = \(\frac{A_M}{A_C}\) = \(\frac{9}{12}\) = 0.75

Questions 12 to 18 carry 3 scores each. Answer any 6 questions. (6 × 3 = 18)

Question 12.
(a) Name the theorem used to find the electric flux through a surface enclosing charge, (1)
Answer:
Guess theorem

(b) Using this theorem derive an expression for the electric field due to an infinite plane sheet of charge with surface charge density o. (2)
Answer:

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

Question 13.
(a)With suitable circuit diagram. show how emfs of two cells can be compared using a potentiometer. (2)
Answer:

Principle : Potential difference between two points of a current carrying conductor (having uniform thickness) is directly proportional to the length’of the wire between two points.

Circuit details: A battery (B₁), Rheostat and key are connected in between A and B. This circuit is called primary circuit. Positive end of E₁ and E₂ are connected to A and other ends are connected to a-two way key. Jockey is connected to a two key through galvanometer. This circuit is called secondary circuit.

(b) Why do we prefer potentiometer to measure the e.m.f. a cell than a voltmeter? (1)
Answer:
Potentio meter does not take current at null point and hence measure actual emf. Hence potentio meter is more accurate than volt meter.

Question 14.
Based on the magnetic property, materials are classified into diamagnetic, paramagnetic and ferromagnetic.

(a) Compare paramagnetic and ferromagnetic materials in view of magnetic susceptibility. (1)
Answer:
For paramagnetic susceptibility is small and positive but for ferromagnetic its value is positive and high.

(b) State the law expalining the variation of magnetic susceptibility with temperature in the case of ferromagnetic material. What happens to its magnetic property when the temperature is increased? (2)
Answer:
(i) Curie law
(ii) When temparature increases, at a particular temparature ferromagnetic becomes

Question 15.
Motion of a conductor in a magnetic field induces’ an emf across its ends.

(a) Write an expression for the emf induced across the ends of a conductor moving right angles to a uniform magnetic field.
Answer:
ε = Bl v

(b) An aeroplane is flying horizontally from west to east with a velocity of 720 km/h. Calculate the e.m.f. induced between the ends of its wings having a span of 25m. The vertical component of earth’s magnetic field at the place is 2.19 × 10 5T.
Answer:
Velocity = 720 × \(\frac{5}{18}\) = 200 m/s
emf ε = Bl v
= 2.19 × 10^-5 × 200
0.1090 v

Question 16.
The figure below shows the transmission of light signals through an optical fibre.
(Refractive index of the core – n₁, Refractive index of cladding-n₂)

(a) Name the optical phenomenon happening at the point A. (1)
Answer:
Total internal reflection

(b) Write the necessary conditions for this to happen. (1)
Answer:
i) Light should travel from denser to rarer medium,
ii) incident angle must

(c) Write the relationship between critical angle and refractive indices of the material of the core and cladding of the optical fibre. (1)
Answer:
\(\frac{n_2}{n_1}\) = \(\frac{1}{sin c}\)

Question 17.
(a) Draw a graph showing the variation of de-Broglie’s wavelength (A) with accelerating potential (V).(1)
Answer:
Out of syllabus
λ = \(\frac{1}{\sqrt{2 \mathrm{mev}}}\)
λ ∝ \(\frac{1}{\sqrt{r}}\)
This equation is in the form of y ∝ \(\frac{1}{\sqrt{x}}\) hence, we get a graph as shown below

(b) Express de-Broglie’s wavelength in terms of ac-celerating voltage. (2)
Answer:
de-Broglie’s wavelength λ = \(\frac{h}{P}\)
biet P = \(\sqrt{2 \mathrm{mev}}\)
λ = \(\frac{\mathrm{h}}{\sqrt{2 \mathrm{mev}}}\)

Question 18.
The energy of an electron in the nth electronic orbit of hydrogen atom is
E_n = \(\frac{-m e^4}{8 \epsilon_0^2 n^2 h^2}\)
(a) What does the negative sign signify? (1)
Answer:
Negative sigyis shows that electrons are bound to the nucleus.

(b) Obtain the Rydberg formula for the spectral series of hydrogen atom. (2).
Answer:
Energy of hydrogen atom, when elctron is in the ground state
E_i = \(\frac{-\mathrm{me}^4}{8 \varepsilon_0^2 \mathrm{n}_2^2 \mathrm{~h}^2}\)
Energy of hydrogon atom, when electron is in excited state.
E_f = \(\frac{-m e^4}{8 \varepsilon_0^2 n_f^2 h^2}\)

Energy emitted, when an electron comes from 1 excited state into ground state.
∆ E = E_f – E_i

Questions 19 to 22 carry 4 scores each. Answer any 3questions. (3x 4 = 12)

Question 19.
An electric dipole is placed in a uniform electric field of intensity ‘E’. Let ‘0’ be the angle between dipole moment and electric field.

(a) Derive an expression for the torque acting on the electric dipole in vector form. (2)
Answer:
Torque \(\overrightarrow{\mathrm{\tau}}=\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{E}}\)

(b) What orientation of the dipole corresponds to stable and unstable equillibrium. (1)
Answer:
for stable equilibrium, θ = 0°
for unstable equilibrium, θ = 180°

(c) Write the equation of potential energy corre-sponds to stable equilibrium.
Answer:
Ptential energy U = – PE cos Q or U = \(\overrightarrow{\mathrm{P}} . \overrightarrow{\mathrm{E}}\)

Question 20.
An alternating voltage is applied across an LCR cir¬cuit as in figure.
(L = 100mH. C = 10OuF, R = 50Q)

(a) Draw the phasor diagram of the circuit. (1)
Answer:

(b) In the.circuit, if V = 20sin (100t) volts
(i) Impedance
(ii) Peak value of current
(iii) Resonant frequency of the circuits
Answer:

Question 21.
Coherent sources are necessary for producing sus-tained interference.

(a) Write the condition of coherence.
Answer:
Two light sources are said to be coherant, if they emit light waves of same frequency, same amplitude and same phase.

(b) With the help of a suitable diagram, find the dis-tance between two consecutive bright brands pro-duced by Young’s double slit.
Answer:
Young’s double slit experiment

The experiment consists of a slit ‘S’. A monochromatic light illuminates this slit. S₁ and S₂ are two slits in front of the slit ‘S’. A screen is placed at a suitable distance from S₁ and S₂.
Light from S₁ and S₂ falls bn the screen. On the screen interference bands can be seen.

Explanation : If crests (or troughs) from S₁ and S₂. meet at certain points on the screen, the interference of these points will be constructive and we get bright bands on the screen.

At certain points on the screen, crest and trough meet together. Destructive interference takes place at those points. So we get dark bands

Expression for band width

S₁ and S₂ are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S₁ and S₂.
Hence the path difference, S₁ O – S₂ O = 0
So at ‘O’ maximum brightness is obtained.
Let ‘P’ be the position of n,h bright band at a distance xn from O. Draw S₁ A and S₂ B as shown in figure.
From the right angle AS₁AP
we get, S₁P² = S₁A² + AP²
S₁P² = D² + (X_n – d/2)²
= D² + X_n² + X_nd + \(\frac{d^2}{2}\)

Question 22.
(a) What do you mean by half life period of a radio-active substance. (1)
Answer:
Half life is the time taken to reduce half of its initial value.

(b) Write the relation between mean life and half life. (1)
Answer:
T_mean = \(\frac{T_{1 / 2}}{0.693}\)