Maharashtra Board Class 12 Physics Sample Paper Set 4 with Solutions

Maharashtra State Board Class 12th Physics Sample Paper Set 4 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Physics Model Paper Set 4 with Solutions

General instructions:

The question paper is divided into four sections.

1. Section A: Q. No. 1 contains Ten multiple choice type of questions carrying One mark each.
   Q. No. 2 contains Eight very short answer type of questions carrying One mark each.
2. Section B: Q. No. 3 to Q. No. 14 are Twelve short answer type of questions carrying Two marks each. (Attempt any Eight).
3. Section C: Q. No. 15 to Q. No. 26 are Twelve short answer type of questions carrying Three marks each. (Attempt any Eight).
4. Section D: Q. No. 27 to Q. No. 31 are Five long answer type of questions carrying Four marks each. (Attempt any Three).
5. Use of log table is allowed. Use of calculator is not allowed.
6. Figures to the right indicate full marks.
7. For each MCQ, correct answer must be written along with its alphabet.
   e.g., (a) / (b)…/ (c)…/ (d) Only first attempt will be consideredfor evaluation.
8. Physical constants:
   1. Stefan’s constant σ = 5.67 × 10^-8 Js^-1 m^-2 K^-4
   2. Mass of electron m = 9.1 × 10^-1 kg
   3. Charge of an electron, e = 1.6 × 10^-19 C

Section – A

Question 1.
Select and write the correct answer for the following multiple choice type of questions: [10 Marks]

i. Reactance of a coil is 157 Ω. On connecting the coil across a source of frequency 100 Hz, the current lags behind e.m.f. by 45°. The inductance of the coil is ____.
(A) 0.25
(B) 0.5 H
(C) 4H
(D) 314 H
Answer:
(A) 0.25 H

Explanation:
X_L = 157 Ω, f = 100 Hz,
X_L = 2πfL
∴ L = \(\frac{X_L}{2 \pi f}\)
= \(\frac{157}{2 \times 3.142 \times 100}\)
= 0.249 ≈ 0.25 H

ii. Choose the correct statement.
(A) Magnetic force changes speed as well as direction of motion of charged particle.
(B) Magnetic force can change speed but cannot change direction of motion of charged particle.
(C) Magnetic force can change direction of motion but cannot change speed of charged particle.
(D) Magnetic force can neither change speed not direction of motion of charged particle.
Answer:
(C) Magnetic force can change direction of motion but cannot change speed of charged particle.

iii. In potentiometer experiment, the cell balances at a length of 240 cm. When the cell is shunted by a resistance of 2 Ω, the balancing length becomes 120 cm. The internal resistance of the cell is ____.
(A) 4 Ω
(B) 2 Ω
(C) 1 Ω
(D) 0.5 Ω
Answer:
(B) 2 Ω
r = R(\(\frac{I_1}{I_2}\) – 1] = 2(\(\frac{240}{120}\) – 1) = 2Ω

iv. A graph of pressure versus volume for an ideal gas for different processes is as sho’ñ. In the graph curve OA represents ______.
(A) isochoric process
(B) isothermal process
(C) isobaric process
(D) adiabatic process
Answer:
(C) isobaric process

v. Gauss’ law helps in
(A) determination of electric field due to symmetric charge distribution.
(B) determination of electric potential due to symmetric charge distribution.
(C) determination of electric flux.
(D) situations where Coulomb’s law fails.
Answer:
(A) determination of electric field due to symmetric charge distribution.

vi. Periodic time of angular oscillations of a bar magnet is proportional to the ___.
(A) ratio of moment of inertia and magnetic field
(B) ratio of magnetic field and moment of inertia
(C) square root of the ratio of moment of inertia and magnetic field
(D) square root of the ratio of magnetic field and moment of inertia
Answer:
(C) square root of the ratio of moment of inertia and magnetic field

vii. To demonstrate the phenomenon of interference, we require
(A) two sources which emit radiation of same frequency.
(B) two sources which emit radiation of nearly same frequency.
(C) two sources which emit radiation of the same frequency and have a definite phase relationship.
(D) two sources which emit radiation of different wavelength.
Answer:
(C) two sources which emit radiation of the same frequency and have a definite phase relationship.

viii. When the temperature increases, the angle of contact of a liquid
(A) increases
(B) decreases
(C) remains unchanged
(D) first increases and then decreases
Answer:
(A) increases

ix. Bohr magneton is
(A) magnetic induction of electron when it is revolving ¡n the first Bohr orbit.
(B) magnetic moment of electron when it is revolving in the fïršt Bohr orbit.
(C) angular momentum of electron when revolving in the first Bohr orbit.
(D) angular mortientum of electron when revolving in last Bohr orbit.
Answer:
(B) maqnetic moment of electron when it is revolvinq in the first Bohr orbit.

x. The equation of a simple harmonic progressive wave travelling on a string is y = 8 sin (0.02 x – 4t) cm. The speed of the wave is ____.
(A) 10 cm/s
(B) 20 cm/s
(C) 100 cm/s
(D) 200 cm/s
Answer:
(D) 200 cm/s

Explanation:
From the given equation,
k = 0.02 cm^-1, ω = 4 rad/s
∴ v = \(\frac{\omega}{\mathrm{k}}\) = \(\frac{4}{0.02}\) = 200 cm/s

Question 2.
Answer the following questions: [18 Marks]

i. What are quasi static processes?
Answer:
Processes in which changes in the state variables of a system occur infinitesimally slowly are called quasi static processes.

ii. Define linear simple harmonic motion.
Answer:
Linear S.H.M. is defined as the linear periodic motion of a body, in which force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position.

iii. The magnetic susceptibility of annealed iron at saturation is 4224. Find the permeability of annealed iron at saturation.
(µo = 4π × 10^-7 SI unit).
Answer:
μ = μo (1 + χ)
∴ μ = 4π × 10^-7 (1 + 4224)
= 5.31 × 10^-3 Hm^-1

iv. State Ampere’s law.
Answer:
The line integral of magnetic field of induction B around any closed path in free space is equal to absolute permeability of free space p0 times the total current flowing through area bounded by the path.

v. State any two sources of errors in metrebridge experiment.
Answer:
Sources of errors in a metrebridge:
a. The cross section of the wire may not be uniform.
b. The ends of the wire are soldered to the metallic strip where contact resistance is developed, which is not taken into account.
c. The measurements of I_X and I_R may not be accurate.

vi. Calculate the minimum pressure required to force the blood from the heart to the top of the head (vertical distance = 50 cm). Assume the density of blood to be 1.04 g cm^-3. Friction is to be neglected.
Answer:
P = hρg = 50 × 1.04 × 980 = 5.096 × 10⁴ dyne/cm²

vii. What are the conditions for current and impedance in parallel resonance circuit.
Answer:
In parallel resonant circuit, Impedance is maximum and current is minimum.

viii. Explain what is meant by polarization.
Answer:
The phenomenon of restriction of the vibration of light waves in a particular plane perpendicular to direction of wave motion is called polarization of light.

Section – B

Attempt any EIGHT of the following questions: [16 Marks]

Question 3.
Explain the use of potentiometer as a voltage divider.
Answer:
Potentiometer as a voltage Divider:
i. The potentiometer can be used as a voltage divider to continuously change the output voltage of a voltage supply.

ii. As shown in the above figure, potential V is set up between points A and B of a potentiometer wire.

iii. One end of a device is connected to positive point A and the other end is connected to a slider that can move along wire AB.

iv. The voltage V gets divided in proportion of lengths l₁ and l₂, such that,
V₁ = \(\frac{d V(I)}{d L}\) end V₂ = \(\frac{d V\left(L-l_1\right)}{d L}\)

Question 4.
What will be the energy of each photon in monochromatic light of frequency 3.5 × 10¹⁴ Hz?
Answer:
Given: v = 3.5 × 10¹⁴ Hz
To find: Energy of photon (E)
Formula: E = hv
Calculation: From formula
E = 6.63 × 10^-34 × 3.5 × 10¹⁴
= 2.32 × 10^-19 J
= \(\frac{2.32 \times 10^{-19}}{1.6 \times 10^{-19}}\)eV
= 1.4503 ≈ 1.45 eV
Ans: Energy of the photon is 1.45 eV.

Question 5.
Distinguish between streamline flow and turbulent flow.
Answer:

Streamline flow	Turbulent flow
i. The smooth flow of a fluid, with velocity smaller than certain critical velocity (limiting value of velocity) is called streamline flow or laminar flow of a fluid	The irregular and unsteady flow of a fluid when its velocity increases beyond critical velocity is called turbulent flow.
ii. In a streamline flow, velocity of a fluid at a given point is always constant.	In a turbulent flow, the velocity of a fluid at any point does not remain constant.
iii. Two streamlines can never intersect, i.e., they are always parallel and hence can never form eddies.	In a turbulent flow, at some points, the fluid may have rotational motion which gives rise to eddies.
iv. Streamline flow over a plane surface can be assumed to be divided into a number of plane layers. In a flow of liquid through a pipe of uniform cross sectional area, all the streamlines will be parallel to the axis of the tube.	A flow tube loses its order and particles move in random direction.

Question 6.
Energy of an electron in the second Bohr orbit is -3.4 eV. Calculate the energy of an electron in the third Bohr orbit.
Answer:
Given: E = -3.4 eV
To find: Energy of electron in third orbit (E₃)
Formula: E ∝ \(\frac{1}{n^2}\)
Calculation: From formula,
\(\frac{E_3}{E_2}\) = \(\frac{n_2^2}{n_3^2}\)
∴ E₃ = \(\left(\frac{2}{3}\right)^2\) × (-3.4) = -1.51 eV
Ans: Energy of an electron in the third orbit ¡s -1 .51 eV.

Question 7.
The charge per unit area of a large flat sheet of charge is 3 μC/m². Calculate the electric field intensity at a point just near the surface of the sheet, measured from its midpoint.
Answer:
given: σ = 3 × 10^-6 Cm^-2
To find: Electric field intensity (E)
Formula: E = \(\frac{\sigma}{2 \varepsilon_0}\)
Calculation: From formula,
E = \(\frac{3 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}\) = \(\frac{3 \times 10^{-6}}{17.7 \times 10^{-12}}\) = 1.695 × 10⁵ N/C
Ans: Electric field intensity is 1.695 × 10⁵ N/C.

Question 8.
Describe V-I characteristics of a solar cell.
Answer:

1. I-V characteristic of solar cell is drawn in fourth quadrant because a solar cell supplies current to the load.
2. The power delivered to the load is zero when the load is short-circuited.
3. The intersection of the curve with the I-axis is the short-circuit current (I_sc) corresponding to a given light intensity. The intersection of the curve with the V- axis is the open circuit voltage (V_OC) corresponding to given light intensity.
4. Power delivered to the load is also zero when the load is open. However, there is a point on the curve where power delivered P_L = (V_0L × I_sc) is maximum.

Question 9.
Derive an expression for the energy stored in a magnetic field in terms of induced current.
Answer:
i. Changing magnetic flux in a coil causes an induced emf.
ii. The induced emf so produced opposes the change and hence the energy has to be spent to overcome it to build up the magnetic field.
iii. This energy may be recovered as heat in a resistance of the circuit.
iv. The induced emf is given as, e = -L \(\frac{d i}{d t}\)
v. The work done in moving a charge dq against this emf is,
dW = -e.dq = L\(\frac{d i}{d t}\).dq = L.\(\frac{\mathrm{di} \cdot \mathrm{dq}}{\mathrm{~d} t}\)
∴ dW = L⋅i⋅di …….. [∴ \(\frac{\mathrm{dq}}{\mathrm{~d} t}\) = i]
Therefore total work,
W = ∫dW = \(\int d W\) = \(\int_0^i \text { Lidi }\) = \(\frac{1}{2} L i^2\) = U_B ….(1)
vi. Equation (1) gives the energy stored (U_B) in magnetic field and is analogous to the energy stored (U_E) in the electric field in a capacitor

Question 10.
A coin placed on a revolving disc, with its centre at a distance of 6 cm from the axis of rotation just slips off when the speed of the revolving disc exceeds 45 r.p.m. What should be the maximum angular speed (in r.p.m.) of the disc, so that when the coin is at a distance of 12 cm from the axis of rotation, it does pot slip?
Answer:
Given: r₁ = 6 cm, r₂ = 12 cm, n₁ = 45 r.p.m
To find: Maximum angular speed (n₂)
Formula: Max.C.F = mrω²

Ans: The maximum angular speed of the disc should be 31.8 r.p.m.

Question 11.
Explain the term free vibration.
Answer:

1. In forced vibrations an external periodic force is applied on a body whose natural period is different from the period of the force
2. The body is made to vibrate with a frequency equal to that of the externally impressed force
3. The amplitude of forced vibrations depends upon the difference between the frequency of external periodic force and the natural frequency of the body.
4. If this difference is small, the amplitude of forced vibrations is large and vice versa.
5. If the frequencies exactly match, it is termed as resonance and the amplitude of vibration is maximum.
6. An object vibrating with its natural frequency can cause another nearby object to vibrate. The second object absorbs the energy transmitted by the first object and starts vibrating if the natural frequencies of the two objects match. The second object is said to undergo forced vibrations.
7. Strings or air columns can also undergo forced oscillations if the frequency of the external source of sound is close to the natural frequency of the system. Resonance is said to occur and we hear a louder sound.
8. Examples: vibrations in a drill machine and washing machine.

Question 12.
The current of 1 A is flowing through an external resistance of 10 Q when it is connected to the terminals of a cell. This current reduces to 0.5 A when the external resistance is 25 Q. Find the internal resistance of the cell.
Answer:
Given: I₁ = 1 A, R₁ = 10 Ω, I₂ = 0.5 A, R₂ = 25 Ω
To find: Internal resistance (r)
Formula: Σ I(R + r) = E
Calculation:
From first condition,
E = I₁ (R₁ + r) ……..(i)
From second condition,
E = I₂ (R₂ + r) ….(ii)
Equating equation (i) and (ii),
I₁ (R₁ + r) = I₂ (R₂ + r)
∴ 1 (10 + r) = 0.5 (25 + r)
∴ 10 + r = 12.5 + 0.5 r
∴ r – 0.5 r = 12.5 – 10
∴ 0.5 r = 2.5
∴ r = \(\frac{2.5}{0.5}\) = 5 Ω
Ans: The internal resistance of the cell is 5 Ω

Question 13.
An ideal monatomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. What is the ratio of the final pressure to its initial pressure?
Answer:
Given: As the gas is monatomic, γ = \(\frac{5}{3}\)
T₂ = 2T₁ ⇒ \(\frac{T_2}{T_1}\) = 2
To find: Ratio of final pressure to its initial pressure \(\left(\frac{\mathrm{p}_2}{\mathrm{p}_1}\right)\)
Formula: \(p_1^{1-\gamma} T_1^\gamma\) = \(\mathrm{p}_2^{1-\gamma} \mathrm{T}_2^{\mathrm{v}}\)
Calculation: From formula,

Question 14.
State the factors on which resolving power of microscope depends. How can it be increased?
Answer:
i. Resolving power of microscope depends on numerical aperture of objective of microscope and wavelength of light used.
ii. Resolving power of a microscope can be increased by,
a. Reducing distance of separation between two objects,
b. Filling eyepiece with material of high refractive index (n) such as oil,
c. Using a source with small value of wavelength (λ).

Section – C

Attempt any EIGHT of the following questions: [24 Marks]

Question 15.
In a single slit diffraction pattern, the distance between first minima on the right and first minima on the left of central maximum is 4 mm. The screen, on which the pattern is displaced, is 2 m from the slit and wavelength of light used is 6000 A. Calculate width of the slit and width of the central, maximum.
Answer:
given: λ = 6000 Å = 6 × 10^-7 m, D = 2 m, 2y_1d = 4mm = 4 × 10^-3m
To find: Width of slit (a)
Width of central maximum (W)

Formulae:
i. 2Y_1d = \(\frac{2 \lambda D}{a}\)
ii. W = \(2 y_{1 d}\)
Calculation: From formula (i),
4 × 10^-3 = \(\frac{2 \times 6 \times 10^{-7} \times 2}{a}\)
∴ a = 6 × 10^-4 m
From formula (ii),
Width of central maximum = 4 × 10^-3 m

Ans:
i. The width of the slit is 6 × 10^-4 m.
ii. The width of the central maximum is 4 × 10^-3 m.

Question 16.
Draw a neat diagram of a full wave rectifier and explain its working.
Answer:
Working of full wave rectifier:
i. The full wave rectifier circuit consists of two diodes conducting alternately as shown in the given circuit diagram.

ii. The circuit consists of a centre tapped transformer and diodes D₁ and D₂ connected such that D₁ conducts in the positive half cycle and D₂ conducts in the negative half cycle of the input voltage.

iii. During the positive half cycle of the input voltage, the point A is at a higher potential than that of the point B and the diode Di conducts. The current through the load resistance RL follows the path APQRC as shown in figure.

iv. During the negative half cycle of the input voltage, point B is at higher potential than point A and the diode D₂ conducts. The current through the load resistance R_L follows the path BPQRC.

v. Thus, the current flowing through the load resistance is in the same direction during both the cycles and DC output voltage obtained across R_L in the form of continuous pulses.

Question 17.
A mercury drop of radius 0.5 cm falls from a height on a glass plate and breaks up into a million droplets, all of the same size. Find the height from which the drop must have fallen. (Density of mercury = 13600 kg/m³, Surface tension of water = 0.465 N/m)
Answer:
Given: R = 0.5 cm = 0.5 × 10^-2 m, n = 10⁶, ρ = 13600 kg/m³, T = 0.465 N/m
To find: Height (h)

Formulae:
i. W = T(dA)
ii. P.E. = mgh

Calculation:
Volume of single big drop, V = \(\frac{4}{3} \pi R^3\)
Volume of sinqle small droplet = \(\frac{4}{3} \pi r^3\)
∴ Volume of n droplets = n × \(\frac{4}{3} \pi r^3\)
∴ \(\frac{4}{3} \pi R^3\) = \(n \frac{4}{3} \pi r^3\)
∴ R³ = nr³
∴ r = \(\frac{R}{\sqrt[3]{n}}\) = \(\frac{0.5 \times 10^{-2}}{\sqrt[3]{10^6}}\) = \(\frac{0.5 \times 10^{-2}}{10^2}\)
∴ r = 0.5 × 10^-4 m
The single drop falls from height h, hence its P.E. = mgh
This P.E. is equal to the work done due to change in area of drop.
Change in surface area, ∆ A = (4πr²n – 4πR²).
From formula (i),
W = T∆A = T (n4πr² – 4πR²)
W = 4πT (nr² – R²)
From formula (ii),
P.E = Work done

Ans: The height from which the drop must have fallen is 0.2072 m.

Question 18.
On the basis of de Broglie hypothesis, obtain the relation for wavelength of an electron accelerated by a potential difference of ‘ V’ volt.
Answer:
i. Consider an electron of mass m and charge q is accelerated through potential difference of V volts.

ii. According to de Broglie hypothesis, wavelength of any particle is,
λ = \(\frac{h}{p}\) = \(\frac{h}{m v}\)
∵ p = mv
∴ m²v² = p²
∴ \(\left(\frac{1}{2} m v^2\right) m\) = \(\frac{1}{2} p^2\)
∴ E = \(\frac{p^2}{2} m\)
∴ p = \(\sqrt{2 m E}\) ⇒ λ = \(\frac{h}{\sqrt{2 m E}}\)

iii. Now, for an electron accelerated through potential difference V, the work done on the electron is stored in it in the form of energy.
∴ E = qV
∴ λ = \(\frac{h}{\sqrt{2 m E}}\) = \(\frac{h}{\sqrt{2 m q V}}\)

iv. But, q = 1.6 × 10^-19 C, m = 9.1 × 10^-31 kg, h = 6.63 × 10^-34 J-s

Question 19.
Explain superposition of two wave pulses moving towards each other with equal amplitude and opposite phase.
Answer:

1. The propagation of approaching wave pulses, their successive positions after every second, their superposition and propagation after superposition are shown in figure (a) to figure (e).
2. These wave pulses superimpose at t = 2 s and the resultant displacement (full line) is zero, due to individual displacements (dashed lines) differing in phase exactly by 180°. This is destructive interference.
3. Displacement due to one wave pulse is cancelled by the displacement due to the other wave pulse when they cross each other as shown in figure (c).
4. After crossing each other, both the wave pulses continue and maintain their individual shapes.

Question 20.
An electron in an atom revolves around the nucleus in an orbit of radius 0.53 A. If the frequency of revolution of an electron is 9 × 10⁹ MHz, calculate the orbital angular momentum.
[Given: Charge on an electron = 1.6 × 10^-19 C; Gyromagnetic ratio = 8.8 × 10¹⁰ C/kg; π = 3.142]
Answer:
Given: r = 0.53 A = 0.53 × 10^-10 m, f = 9 × 10⁹ MHz = 9 × 10¹⁵ Hz
To find: Angular momentum (L)
Formula: L = \(=\frac{m_{\text {orb }}}{\text { gyromagnetic ratio }}\)
Calculation:
m_orb = IA
Since, I = \(\frac{1}{T}\) e = fe
From formula,
m_orb = feA = feπr²
= 9 × 10¹⁵ × 1.6 × 10^-19 × π × (0.53 × 10^-10)²
= 1.6 × πx × 0.25 × 10^-23
∴ m_0rb = 1.270 × 10^-23 Am²
Using formula,
L = \(\frac{1.270 \times 10^{-23}}{8.8 \times 10^{10}}\)
= {antilog(log 1.27 – log 8.8)} × 10^-33
= {antilog(0.1038 – 0.9445)} × 10^-33
= (antilog(\(\overline{1} .1593\))} × 10^-33
= 1.443 × 10^-1 × 10^-33
∴ L = 0.1443 × 10^-33 kgm²/s
Ans: The orbital angular momentum of the electron is 0.1443 × 10^-33 kgm²/s.

Question 21.
Why should a Carnot cycle have two isothermal and two adiabatic processes?
Answer:
In a Carnot engine, there are basically two processes:

i. Exchange of heat:
a. For an ideal condition, the process of exchange of heat to be reversible, it should be carried out isothermally. i.e., the temperature of the working substance must be T_H while absorbing heat from hot reservoir and it should be T_C while rejecting heat to a cold reservoir.
b. The curves AB and CD represent the isothermal changes in the given p-V diagram.

ii. Work done:
a. The work done to be reversible, the processes must be adiabatic.
b. The curves BC and DA represent the adiabatic changes in the given p-V diagram.

iii. Thus, the cycle includes two isothermal and two adiabatic processes for maximum efficiency.

Question 22.
The speeds of a particle performing linear S.H.M. are 4 cm/s and 3 cm/s at respective displacements of 3 cm and 4 cm. Find its period and amplitude.
Answer:
Given: v₁ = 4 cm/s when x₁ = 3 cm, v₂ = 3 cm/s when x₂ = 4 cm

To find:
i. Amplitude (A)
ii. Time period (T)

Formulae:
i. v = \(\omega \sqrt{A^2-x^2}\)
ii. ω = \(\frac{2 \pi}{T}\)

Calculation:
From formula (i),

Ans: The amplitude and time period of oscillation are respectively 5 cm and 6.284 s.

Question 23.
A current 10 A in the primary of a transformer is reduced to zero at the uniform rate in 0.1 second. If the coefficient of mutual inductance be 3 H, what is the e.m.f induced in the secondary and change in the magnetic flux per turn in the secondary if it has 50 turns?
Answer:
Given: \(\mathrm{I}_{\mathrm{P}_1}\) = 10 A, \(\mathrm{I}_{\mathrm{P}_2}\) = 0 A, dt = 0.1 s, N_s = 50, M = 3 H

To find:
i. e.m.f. induced (es)
ii. Change in magnetic flux per +urn(dφ_s)
Formula: e_s = M\(\frac{d I_p}{d t}\)

Calculation:
Since,
\(\frac{d I_p}{d t}\) = \(\left|\frac{0-10}{0.1}\right|\) = 100 A
From formula,
e_s = 3 × 100 = 300 V
The e.m.f induced in the secondary is given by,
|e_s | = N_s\(\frac{d \phi_s}{d t}\)
∴ dφ_s = \(\frac{e_s d t}{N_s}\) = \(\frac{300 \times 0.1}{50}\) = 0.6 Wb

Ans:
i. The e.m.f induced in the secondary is 300 V.
ii. The change in the magnetic flux per turn in the secondary is 0.6 Wb.

Question 24.
A small-blackened solid copper sphere of radius 2 cm is placed in an evacuated chamber. The temperature of the chamber is maintained at 127 °C. At what rate energy must be supplied to the copper sphere to maintain its temperature at 137 °C? (Treat the sphere as blackbody.)
Answer:
Given: r = 2 cm, T₀ = 127 °C = 127 + 273 = 400 K,
T = 137 °C = 137 + 273 = 410 K, σ = 5.67 × 10^-8 J s^-1 m^-2 K^-4
To find: Rate of energy supplied \(\left(\frac{d Q}{d t}\right)\)

Formulae:
i. A = 4πr²
ii. \(\frac{d Q}{d t}\) = σA(T⁴ – \(T_0^4\))

Calculation:
From formula (i),
A = 4 × 3.142 × (2)² × 10^-4
= 4 × 3.142 × 4 × 10^-4
= 3.142 × 16 × 10^-4 = 5.027 × 10^-3 m²
From formula (ii),
\(\frac{d Q}{d t}\) = 5.67 × 10^-8 × 5.027 × 10^-3 × (410⁴ – 400⁴) dt
= 5.67 × 5.027 × 10^-11 × (2.826 × 10¹⁰ – 2.56 × 10¹⁰)
= 28.5 × 0.266 × 10¹
= 0.7581 W
Ans: Rate of energy supplied is 0.7581 W.

Question 25.
Obtain an expression for energy of a charged capacitor and express it in different forms.
Answer:
i. Consider a capacitor of capacitance C being charged by a DC source of V volts as shown in figure below.

ii. During the process of charging, let q’ be the charge on the capacitor and V be the potential difference between the plates. Hence C = \(\frac{q}{V}\)

iii. A small amount of work is done if a small charge dq is further transferred between the plates.
∴ dW = V dq = \(\frac{q^{\prime}}{c} d q\)

iv. Total work done in transferring the charge

v. This work done is stored as electrical potential energy U of the capacitor. This work done can be expressed in different forms as follows.
∴ U = \(\frac{1}{2} \frac{Q^2}{C}\) = \(\frac{1}{2} \mathrm{CV}^2\) = \(\frac{1}{2} \mathrm{QV}\) …… (∵ Q = CV)

Question 26.
State and explain the principle of conservation of angular momentum.
Answer:
Principle: Angular momentum of an isolated system is conserved in the absence of an external unbalanced torque.

Proof:
i. Angular momentum or the moment of linear momentum of a system is given by, \(\overrightarrow{\mathrm{L}}\) = \(\vec{r} \times \vec{p}\) where \(\overrightarrow{\mathrm{r}}\) the position vector from the axis of rotation and \(\overrightarrow{\mathrm{p}}\) the linear momentum.

ii. Differentiating with respect to time, we get.
\(\frac{d \vec{L}}{d t}\) = \(\frac{d}{d t}(\vec{r} \times \vec{p})\) = \(\vec{r} \times \frac{d \vec{p}}{d t}+\frac{d \vec{r}}{d t} \times \vec{p}\)

iii.

iv. But \(\vec{r} \times \vec{F}\) is the moment of force or torque \(\vec{\tau}\).
∴ \(\vec{\tau}\) = \(\frac{d \vec{L}}{d t}\)
Thus, if \(\vec{\tau}\) = 0, \(\frac{d \vec{L}}{d t}\) = o or \(\vec{L}\) constant.
Hence, angular momentum \(\vec{L}\) is conserved in the absence of external unbalanced torque \(\vec{\tau}\).

v. This is the principle of conservation of angular momentum, analogous to the conservation of linear momentum.
Examples of conservation of angular momentum: During some shows of ballet dance,
acrobat in a circus, sports like ice skating, diving in a swimming pool, etc., the principle
of conservation of angular momentum is realized.

Section – D

Attempt any THREE of the following questions: [12 Marks]

Question 27.
i. Why are curved roads banked?
ii. If a copper disc swings between the poles of a magnet, the pendulum comes to rest very quickly. Explain the reason. What happens to the mechanical energy of the pendulum?
Answer:
i. a. While taking a turn on a horizontal road, the force of static friction between the tyres of the vehicle and the road provides the necessary centripetal force (or balances the centrifugal force).
b. However, the frictional force is having an upper limit. Also, its value is usually not constant as the road surface is not uniform.
c. Thus, in real life, we should not depend upon it, as it is not reliable.
d. For this purpose, the surfaces of curved roads are tilted with the horizontal with some angle θ.
e. This is called banking of a road or the road is said to be banked.
f. When the road is banked, the horizontal component of the normal reaction provides the necessary centripetal force required for circular motion of vehicle.

ii.
a. While swinging, when the copper disc is in the mean position, the magnetic flux linked with it is maximum.
b. As the plate starts oscillating, the magnetic flux linked with the plate goes on decreasing. Due to this, induced current is setup in the copper plate in the form of a closed loop. These current oppose the oscillatory motion of the copper plate. These currents are eddy currents.
c. This induced current is also responsible for the transfer of the mechanical energy of the pendulum into heat energy which is then dissipated in the solid plate making it warm.

Question 28.
i. Define half life period. Derive expression for it.
ii. The decay constant of radioactive substance is 4.16 × 10^-4 per year. Calculate its half life period.
Answer:
i. Definition:
Half life period of a radioactive substance is defined as the time in which the half substance is disintegrated.
Expression for half life period:
From law of radioactive decay, N(t) = N_o e^-λT

This is the required expression for half life period of radioactive substance.

ii.
Solution:
Given: λ = 4.16 × 10^-4 year^-1
To find’. Half life period (T_1/2)
Formula. T_1/2 = \(\frac{0.693}{\lambda}\)
Calculation:
From formula,
T_1/2 = \(\frac{0.693}{4.16 \times 10^{-4}}\) = \(\frac{6.93}{4.16}\) × 10³ = 1666 × 10³ = 1666 years.
Ans: Half-life period of a radioactive element is 1666 years.

Question 29.
i. Obtain an expression for magnetic induction along the axis of toroid on the basis of Ampere’s circuital law.
ii. What is a toroid?
Answer:
Expression for magnetic induction at a point along the axis passing through centre of a toroid and perpendicular to its plane:
i. Consider a toroid carrying steady current I and having turns N.

ii. The magnetic field around the toroid consists of concentric circular lines of force around it. Magnetic field is produced, when a steady current flows through toroid.

iii. The direction of magnetic field at a point is along the tangent to the circular path at that point.

iv. Let R be the radius of the Amperian loop. This loop is concentric with the axis of toroid. P is a point on the loop.

v. Applying Ampere’s law,
\(\oint \vec{B} \cdot \overrightarrow{d I}\) = µ_o = µ_oNi ……….(1)
where,
N = total number of turns in the toroid.
I = total current flowing through toroid.
i = Current flowing in each turn
Now, \(\oint \vec{B} \cdot \overrightarrow{d l}\) = \(\oint\)B dl cosθ …(2)

vi. But, \(\vec{B}\) and \(\overrightarrow{\mathrm{d}} l\) are in same direction
∴ θ = 0°
∴ cos θ = 1
∴ Equation (2) can be written as,
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = \(\oint \mathrm{Bd} l\) = \(\mathrm{B} \oint \mathrm{~d} l\) ………. (3)

vii. Also \(\oint \mathrm{d} l\) = 2πR ….(total length of circle is its circumference)
From equation (3),
\(\oint \vec{B} \cdot \overrightarrow{d I}\) = B (2πR) …(4)

viii. From equation (1) and (4),
B (2πR) = μ_oNi
∴ B = \(\frac{\mu_0 N i}{2 \pi R}\) …….(5)
This is the required expression. [3 Marks]

ii.
Answer:
A toroid is a solenoid of finite length bent into a hollow circular tube

Question 30.
i. Discuss magnetic field around solenoid carrying current.
ii. A galvanometer has a resistance of 300 Ω and its full scale deflection current is 200 μA. What shunt resistance should be added so that the ammeter can have a range of 0 to 20 mA?
Answer:
i. a. Consider a long, closely wound helical coil of a conducting wire such that the diameter of the coil is much smaller than its length. This forms a solenoid.

b. The density of the magnetic field lines along the axis of the solenoid within the solenoid and at a certain distance away from the wire, is uniform as shown in figure below.

Hence the magnetic field B is parallel to the axis of the solenoid.

c. The lines are widely spaced outside the solenoid and hence the magnetic field is weak there.

d. For a real solenoid of finite length, magnetic field is uniform and has a good strength at the centre and comparatively weak at the outside of the coil. [2 Marks]

ii. Solution:
Given: I_g = 200 μA = 0.2 mA, I = 20 mA, G = 300 Ω
To find: Shunt resistance (S)

Formulae:
i. n = \(\frac{I}{I_9}\)
ii. S = \(\frac{G}{n-1}\)

Calculation:
From formula (i),
n = \(\frac{I}{I_9}\) = \(\frac{20}{0.2}\) = 200
From formula (ii),
S = \(\frac{300}{200-1}\) = \(\frac{300}{199} \Omega\) = 1.51 Ω

Ans: The shunt resistance that should be added is 1.51 Ω.

Question 31.
i. State and prove Kirchhoff’s law of heat radiation,
ii. If r = 1 then, it is a white body. Is it true?
Answer:
i. Statement: At a given temperature, the ratio of emissive power to coefficient of absorption of a body is equal to the emissive power of a perfect blackbody at the same temperature for all wavelengths.

OR

For a body emitting and absorbing thermal radiation in thermal equilibrium, the emissivity is equal to its absorptivity.
Symbolically, a = e or a(λ) = e(λ)

Theoretical proof:
i. Consider an ordinary body A and a perfect blackbody B of identical geometric shapes placed in an enclosure. In thermal equilibrium, both bodies will be at same temperature as that of the enclosure.

ii. Let R = emissive power of body A,
R_B = emissive power of blackbody B
a = coefficient of absorption of body A.
Q = quantity of radiant heat incident on each body in unit time and
Q_a = quantity of radiant heat absorbed by the body A, then Q_a = a Q.

iii. As the temperatures of the body A and blackbody B remain the same, both must emit the same amount as they absorb in unit time.
∴ Quantity of radiant heat absorbed by body A = Quantity of heat emitted by body A
∴ aQ = R ……….(1)

iv. For the perfect blackbody B,
Q = R_B …….(2)

v. Dividing equation (1) by equation (2), we get,
a = \(\frac{R}{R_B}\)

vi. But by definition of emissivity, e = \(\frac{R}{R_3}\)
∴ a = e
Hence, Kirchhoff’s law is theoretically proved.

ii.
Answer:
Given condition, r = 1
⇒ a = 0 (∵ a + r + t = 1)
⇒ e = 0 (∵ a = e)
Such a situation is impossible as every body at a temperature above 0 K emits thermal radiation.
Hence, given statement is false.

Maharashtra Board Class 12 Physics Previous Year Question Papers