HSC Physics Board Question Paper 2022 Solutions - Maharashtra Board

Physics Board Question Paper: March 2022 - Solved

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.
(2) Section B: Q. No. 3 to Q. No. 14 contain Twelve short answer type of questions carrying Two marks each.
(3) Section C: Q. No. 15 to Q. No. 26 contains Twelve short answer type of questions carrying Three marks each.
(4) Section D: Q. No. 27 to Q. No. 31 contain Five long answer type of questions carrying Four marks each.

SECTION − A

Q.1. Select and write the correct answers for the following multiple choice type of questions: [10]

(i) The first law of thermodynamics is concerned with the conservation of _______.

Answer

(b) energy

Explanation: The first law states that heat supplied is equal to the sum of the change in internal energy and work done ($Q = \Delta U + W$), which is essentially the law of conservation of energy.

(ii) The average value of alternating current over a full cycle is always _______.
[I₀ = Peak value of current]

Answer

(a) zero

Explanation: In a full cycle of AC, the positive half cycle exactly cancels out the negative half cycle.

(iii) The angle at which maximum torque is exerted by the external uniform electric field on the electric dipole is _______.

Answer

(d) 90°

Explanation: Torque $\tau = pE \sin\theta$. It is maximum when $\sin\theta = 1$, which implies $\theta = 90^\circ$.

(iv) The property of light which does not change, when it travels from one medium to another is _______.

Answer

(c) frequency

Explanation: Frequency depends on the source of light and remains constant during refraction. Velocity and wavelength change.

(v) The root mean square speed of the molecules of a gas is proportional to _______.
[T = Absolute temperature of gas]

Answer

(a) \(\sqrt{T}\)

Explanation: $v_{rms} = \sqrt{\frac{3RT}{M}}$. Therefore, $v_{rms} \propto \sqrt{T}$.

(vi) The unit Wbm^–2 is equal to _______.

Answer

(d) tesla

Explanation: Weber per square meter ($Wb/m^2$) is the unit of Magnetic Flux Density (Magnetic Induction), which is the Tesla.

(vii) When the bob performs a vertical circular motion and the string rotates in a vertical plane, the difference in the tension in the string at horizontal position and uppermost position is _______.

Answer

(c) 3 mg

Explanation: Tension at Top ($T_T$) = $\frac{mv_T^2}{r} - mg$. Tension at Horizontal ($T_H$) = $\frac{mv_H^2}{r}$. Using energy conservation: $v_H^2 = v_T^2 + 2gr$. $T_H = \frac{m(v_T^2 + 2gr)}{r} = T_T + mg + 2mg = T_T + 3mg$. Difference = $3mg$.

(viii) A liquid rises in glass capillary tube upto a height of 2.5 cm at room temperature. If another glass capillary tube having radius half that of the earlier tube is immersed in the same liquid, the rise of liquid in it will be _______.

Answer

(c) 5 cm

Explanation: Capillary rise $h = \frac{2T\cos\theta}{r\rho g}$, so $h \propto \frac{1}{r}$.
If radius is halved ($r/2$), the height doubles. $2 \times 2.5 \text{ cm} = 5 \text{ cm}$.

(ix) In young’s double slit experiment the two coherent sources have different amplitudes. If the ratio of maximum intensity to minimum intensity is 16:1, then the ratio of amplitudes of the two source will be _______.

Answer

(b) 5 : 3

Explanation: $\frac{I_{max}}{I_{min}} = \left(\frac{a_1+a_2}{a_1-a_2}\right)^2 = \frac{16}{1}$.
$\frac{a_1+a_2}{a_1-a_2} = 4$.
$a_1 + a_2 = 4a_1 - 4a_2 \Rightarrow 5a_2 = 3a_1 \Rightarrow \frac{a_1}{a_2} = \frac{5}{3}$.

(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 _______.

Answer

(d) 200 cm/s

Explanation: Comparing with $y = A \sin(kx - \omega t)$, we get $k = 0.02$ and $\omega = 4$.
Wave speed $v = \frac{\omega}{k} = \frac{4}{0.02} = \frac{400}{2} = 200 \text{ cm/s}$.

HSC Physics Board Papers with Solution

• Physics - March 2025 - English Medium View Answer Key
• Physics - March 2025 - Marathi Medium View Answer Key
• Physics - March 2025 - Hindi Medium View Answer Key
• Physics - March 2024 - English Medium View Answer Key Answer Key
• Physics - March 2024 - Marathi Medium View Answer Key
• Physics - March 2024 - Hindi Medium View Answer Key
• Physics - March 2023 - English Medium View Answer Key
• Physics - July 2023 - English Medium View Answer Key
• Physics - March 2022 - English Medium View Answer Key
• Physics - July 2022 - English Medium View Answer Key
• Physics - March 2013 View
• Physics - October 2013 View
• Physics - March 2014 View
• Physics - October 2014 View
• Physics - March 2015 View
• Physics - July 2015 View
• Physics - March 2016 View
• Physics - July 2016 View
• Physics - March 2017 View
• Physics - July 2017 View

Q.2. Answer the following questions: [8]

(i) Define potential gradient of the potentiometer wire.

Solution

The potential gradient is defined as the potential difference (or potential drop) per unit length of the potentiometer wire.

$$ K = \frac{V}{L} $$

Where V is potential difference and L is the length.

(ii) State the formula for critical velocity in terms of Reynold’s number for a flow of a fluid.

Solution $$ v_c = \frac{R_n \eta}{\rho D} $$

Where:
\( R_n \) = Reynold's number
\( \eta \) = Coefficient of viscosity
\( \rho \) = Density of fluid
\( D \) = Diameter of the tube

(iii) Is it always necessary to use red light to get photoelectric effect?

Solution

No. The photoelectric effect depends on the frequency of the incident light being greater than the threshold frequency of the metal surface. Red light has a low frequency; for many metals (like Zinc), UV light is required, while alkali metals might respond to visible light. It depends on the work function of the material.

(iv) Write the Boolean expression for Exclusive – OR (X – OR) gate.

Solution $$ Y = A \oplus B = \bar{A}B + A\bar{B} $$

(v) Write the differential equation for angular S.H.M.

Solution $$ I \frac{d^2\theta}{dt^2} + c\theta = 0 $$

Where \( I \) is the moment of inertia, \( \theta \) is angular displacement, and \( c \) is the torsion constant (restoring torque per unit twist).

(vi) What is the mathematical formula for third postulate of Bohr’s atomic model?

Solution $$ h\nu = E_n - E_p $$

Where \( E_n \) and \( E_p \) are the energies of the electron in the higher and lower orbits respectively, and \( \nu \) is the frequency of radiation emitted.

(vii) Two inductor coils with inductance 10 mH and 20 mH are connected in series. What is the resultant inductance of the combination of the two coils?

Solution

For series combination (assuming no mutual inductance coupling):

$$ L_s = L_1 + L_2 $$ $$ L_s = 10 \text{ mH} + 20 \text{ mH} = 30 \text{ mH} $$

(viii) Calculate the moment of inertia of a uniform disc of mass 10 kg and radius 60 cm about an axis perpendicular to its length and passing through its centre.

Solution

Note: The phrase "perpendicular to its length" usually applies to rods. For a disc, the standard context implies an axis perpendicular to the plane of the disc passing through the centre (geometric axis).

Given: \( M = 10 \text{ kg} \), \( R = 60 \text{ cm} = 0.6 \text{ m} \).

$$ I = \frac{1}{2}MR^2 $$ $$ I = \frac{1}{2} \times 10 \times (0.6)^2 $$ $$ I = 5 \times 0.36 = 1.8 \text{ kg m}^2 $$

SECTION − B

Attempt any EIGHT questions of the following: [16]

Q.3. Define moment of inertia of a rotating rigid body. State its SI unit and dimensions.

Solution

Definition: Moment of inertia of a rigid body about an axis of rotation is defined as the sum of the product of the mass of each particle of the body and the square of its perpendicular distance from the axis of rotation.

$$ I = \sum_{i=1}^{n} m_i r_i^2 $$

• SI Unit: \( \text{kg m}^2 \)
• Dimensions: \( [L^2 M^1 T^0] \)

Q.4. What are polar dielectrics and non polar dielectrics?

Solution

• Polar Dielectrics: Substances made of molecules that possess a permanent electric dipole moment even in the absence of an external electric field. The center of positive charge and negative charge do not coincide (e.g., HCl, H₂O).
• Non-polar Dielectrics: Substances made of molecules where the center of positive charge coincides with the center of negative charge. They have zero dipole moment in the absence of an external field (e.g., H₂, O₂, CO₂).

Q.5. What is a thermodynamic process? Give any two types of it.

Solution

Definition: A thermodynamic process is a procedure by which the state of a system changes from its initial equilibrium state to a final equilibrium state due to change in thermodynamic coordinates like Pressure (P), Volume (V), and Temperature (T).

Types (Any two):

1. Isothermal Process: Temperature remains constant ($\Delta T = 0$).
2. Adiabatic Process: No exchange of heat with surroundings ($Q = 0$).
3. (Other options: Isobaric, Isochoric).

Q.6. Derive an expression for the radius of the n^th Bohr orbit of the electron in hydrogen atom.

Solution

Consider an electron of mass \(m\) and charge \(-e\) revolving around a nucleus of charge \(+e\) with velocity \(v\) in an orbit of radius \(r\).

1. Centripetal force is provided by Electrostatic force:

$$ \frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} \implies mv^2 = \frac{e^2}{4\pi\epsilon_0 r} \quad \text{---(i)} $$

2. Bohr's Quantization condition:

$$ mvr = \frac{nh}{2\pi} \implies v = \frac{nh}{2\pi mr} \quad \text{---(ii)} $$

Substitute (ii) in (i):

$$ m \left( \frac{nh}{2\pi mr} \right)^2 = \frac{e^2}{4\pi\epsilon_0 r} $$ $$ m \frac{n^2 h^2}{4\pi^2 m^2 r^2} = \frac{e^2}{4\pi\epsilon_0 r} $$

Rearranging for \(r\):

$$ r = \frac{\epsilon_0 n^2 h^2}{\pi m e^2} $$

For the \(n^{th}\) orbit: \( r_n \propto n^2 \).

Q.7. What are harmonics and overtones (Two points)?

Solution

Harmonics:

• The fundamental frequency and all its integral multiples are called harmonics.
• The first harmonic is the fundamental frequency ($n$), the second harmonic is $2n$, etc.

Overtones:

• The frequencies higher than the fundamental frequency that are actually produced by the instrument are called overtones.
• The first frequency higher than fundamental is the 1st overtone. Note: Overtones may not always be integral multiples of the fundamental in all systems.

Q.8. Distinguish between potentiometer and voltmeter.

Solution

Potentiometer	Voltmeter
It does not draw any current from the circuit at the null point.	It always draws some current from the circuit to deflect the needle.
It measures the potential difference (emf) very accurately.	It measures the potential difference approximately.
It is very sensitive.	It is less sensitive compared to a potentiometer.

Q.9. What are mechanical equilibrium and thermal equilibrium?

Solution

• Mechanical Equilibrium: A system is said to be in mechanical equilibrium if there are no unbalanced forces within the system or between the system and its surroundings. The pressure is uniform throughout the system.
• Thermal Equilibrium: A system is in thermal equilibrium if the temperature is uniform throughout the system and does not change with time. There is no net flow of heat.

Q.10. An electron in an atom is revolving round the nucleus in a circular orbit of radius 5.3 × 10^–11 m with a speed of 3 × 10⁶ m/s. Find the angular momentum of electron.

Solution

Given:
Mass of electron \( m = 9.1 \times 10^{-31} \text{ kg} \)
Radius \( r = 5.3 \times 10^{-11} \text{ m} \)
Speed \( v = 3 \times 10^6 \text{ m/s} \)

Angular Momentum \( L = mvr \)

$$ L = (9.1 \times 10^{-31}) \times (3 \times 10^6) \times (5.3 \times 10^{-11}) $$ $$ L = 9.1 \times 3 \times 5.3 \times 10^{-36} $$ $$ L = 27.3 \times 5.3 \times 10^{-36} $$ $$ L = 144.69 \times 10^{-36} $$ $$ L \approx 1.45 \times 10^{-34} \text{ kg m}^2/\text{s} $$

Q.11. Plane wavefront of light of wavelength 6000 Å is incident on two slits on a screen perpendicular to the direction of light rays. If the total separation of 10 bright fringes on a screen 2 m away is 2 cm, find the distance between the slits.

Solution

Given:
\( \lambda = 6000 \text{ \AA} = 6 \times 10^{-7} \text{ m} \)
\( D = 2 \text{ m} \)
Total width of 10 fringes \( X_{10} = 2 \text{ cm} = 0.02 \text{ m} \)
Fringe width \( W = \frac{X_{10}}{10} = \frac{0.02}{10} = 0.002 \text{ m} \)

Formula: \( W = \frac{\lambda D}{d} \)

$$ d = \frac{\lambda D}{W} $$ $$ d = \frac{6 \times 10^{-7} \times 2}{2 \times 10^{-3}} $$ $$ d = 6 \times 10^{-4} \text{ m} $$

Distance between slits = 0.6 mm

Q.12. Eight droplets of water each of radius 0.2 mm coalesce into a single drop. Find the decrease in the surface area.

Solution

Given:
\( n = 8 \), \( r = 0.2 \text{ mm} = 0.02 \text{ cm} \).

1. Find Radius of big drop (R):
Volume is conserved. \( V_{big} = n \times V_{small} \)
\( \frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3 \)
\( R^3 = 8 r^3 \implies R = 2r = 0.4 \text{ mm} \)

2. Change in Area:
Initial Area \( A_1 = n \times 4\pi r^2 = 8 \times 4\pi (0.2)^2 \)
Final Area \( A_2 = 4\pi R^2 = 4\pi (2r)^2 = 16\pi r^2 \)
Decrease \( \Delta A = A_1 - A_2 = 32\pi r^2 - 16\pi r^2 = 16\pi r^2 \)

$$ \Delta A = 16 \times 3.142 \times (0.2 \times 10^{-3})^2 $$ $$ \Delta A = 50.272 \times 0.04 \times 10^{-6} $$ $$ \Delta A \approx 2.01 \times 10^{-6} \text{ m}^2 $$

Q.13. A 0.1 H inductor, a 25 × 10^–6 F capacitor and a 15 Ω resistor are connected in series to a 120 V, 50 Hz AC source. Calculate the resonant frequency.

Solution

Resonant frequency depends only on L and C, not on R or the source frequency.

Given:
\( L = 0.1 \text{ H} \)
\( C = 25 \times 10^{-6} \text{ F} \)

$$ f_r = \frac{1}{2\pi\sqrt{LC}} $$ $$ f_r = \frac{1}{2\pi\sqrt{0.1 \times 25 \times 10^{-6}}} $$ $$ f_r = \frac{1}{2\pi\sqrt{2.5 \times 10^{-6}}} $$ $$ f_r = \frac{1}{2\pi \times 1.58 \times 10^{-3}} $$ $$ f_r = \frac{1000}{9.92} \approx 100.8 \text{ Hz} $$

Q.14. The difference between the two molar specific heats of a gas is 9000 J/kg K. If the ratio of the two specific heats is 1.5, calculate the two molar specific heats.

Solution

Given:
\( C_p - C_v = 9000 \) ---(1)
\( \frac{C_p}{C_v} = 1.5 \) ---(2)

From (2), \( C_p = 1.5 C_v \). Substitute in (1):
\( 1.5 C_v - C_v = 9000 \)
\( 0.5 C_v = 9000 \)
\( C_v = \frac{9000}{0.5} = 18000 \text{ J/kg K} \)

Now find \( C_p \):
\( C_p = 1.5 \times 18000 = 27000 \text{ J/kg K} \)

Answer: \( C_v = 18000 \text{ J/kg K} \), \( C_p = 27000 \text{ J/kg K} \).

SECTION − C

Attempt any EIGHT questions of the following: [24]

Q.15. With the help of a neat diagram, explain the reflection of light on a plane reflecting surface.

Solution

Explanation based on Wave Theory (Huygens' Principle):

1. Consider a plane wavefront AB incident obliquely on a plane reflecting surface MN.
2. At t=0, point A touches the surface. Point B is at distance 'ct' from the surface (point C).
3. While B travels to C, a secondary wavelet originates from A and grows into a hemisphere of radius 'ct'.
4. Draw a tangent CD from point C to this hemisphere. CD represents the reflected wavefront.
5. By geometry of congruent triangles (Triangle ADC and Triangle ABC), the angle of incidence (i) equals the angle of reflection (r).

Q.16. What is magnetization, magnetic intensity and magnetic susceptibility?

Solution

• Magnetization (M): The net magnetic dipole moment per unit volume of a material. \( M = \frac{m_{net}}{V} \). Unit: A/m.
• Magnetic Intensity (H): The capability of a magnetic field to magnetize a material. It relates to the external current. \( B = \mu(H+M) \). Unit: A/m.
• Magnetic Susceptibility (\( \chi \)): The measure of how easily a substance can be magnetized. It is the ratio of the magnitude of magnetization to the magnetic intensity. \( \chi = \frac{M}{H} \). It is dimensionless.

Q.17. Prove that the frequency of beats is equal to the difference between the frequencies of the two sound notes giving rise to beats.

Solution

Let two sound waves be \( y_1 = A \sin(2\pi n_1 t) \) and \( y_2 = A \sin(2\pi n_2 t) \), where \( n_1 \) and \( n_2 \) are frequencies ($n_1 > n_2$).

By superposition: \( y = y_1 + y_2 \)

$$ y = A [\sin(2\pi n_1 t) + \sin(2\pi n_2 t)] $$

Using \( \sin C + \sin D = 2 \sin\frac{C+D}{2} \cos\frac{C-D}{2} \):

$$ y = 2A \cos\left[ 2\pi \left(\frac{n_1-n_2}{2}\right) t \right] \sin\left[ 2\pi \left(\frac{n_1+n_2}{2}\right) t \right] $$

The resultant amplitude is \( R = 2A \cos [2\pi (\frac{n_1-n_2}{2}) t] \).

Intensity is max when \( \cos \theta = \pm 1 \). This occurs when argument is \( 0, \pi, 2\pi \)...
Time interval between two maxima is \( T = \frac{1}{n_1 - n_2} \).
Frequency of beats = \( \frac{1}{T} = n_1 - n_2 \).

Q.18. Define: (a) Inductive reactance (b) Capacitive reactance (c) Impedance

Solution

• (a) Inductive Reactance (\(X_L\)): The opposition offered by an inductor to the flow of alternating current. \( X_L = \omega L = 2\pi f L \).
• (b) Capacitive Reactance (\(X_C\)): The opposition offered by a capacitor to the flow of alternating current. \( X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \).
• (c) Impedance (Z): The total opposition offered by an AC circuit containing combinations of Resistors, Inductors, and Capacitors to the flow of current. \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).

Q.19. Derive an expression for the kinetic energy of a body rotating with a uniform angular speed.

Solution

Consider a rigid body rotating with constant angular speed \( \omega \).
The body consists of particles of masses \( m_1, m_2, ... m_n \) at distances \( r_1, r_2, ... r_n \) from axis.

Linear velocity of particle 1 is \( v_1 = r_1 \omega \).
KE of particle 1: \( E_1 = \frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 r_1^2 \omega^2 \).

Total Rotational KE is sum of individual KEs:

$$ E_{rot} = \frac{1}{2}m_1 r_1^2 \omega^2 + \frac{1}{2}m_2 r_2^2 \omega^2 + ... $$ $$ E_{rot} = \frac{1}{2} \omega^2 (\sum m_i r_i^2) $$

Since \( \sum m_i r_i^2 = I \) (Moment of Inertia):

$$ E_{rot} = \frac{1}{2} I \omega^2 $$

Q.20. Derive an expression for emf (e) generated in a conductor of length (l) moving in uniform magnetic field (B) with uniform velocity (v) along x-axis.

Solution

Consider a conductor of length \( l \) moving with velocity \( v \) perpendicular to a magnetic field \( B \).

Lorentz Force Method:
A charge \( q \) inside the conductor experiences a magnetic force \( F_m = q(v \times B) \). Magnitude \( F = qvB \).
This force pushes electrons to one end, creating an electric field \( E \).
Equilibrium is reached when Electric Force \( F_e = qE \) balances Magnetic Force.

$$ qE = qvB \implies E = vB $$

Potential difference (EMF) \( e = E \times l \).
Therefore, \( e = Blv \).

Q.21. Derive an expression for terminal velocity of a spherical object falling under gravity through a viscous medium.

Solution

Forces acting on the sphere of radius \(r\) and density \(\rho\) falling in fluid of density \(\sigma\) and viscosity \(\eta\):

1. Weight acting down: \( W = \frac{4}{3}\pi r^3 \rho g \)
2. Buoyant Force acting up: \( F_b = \frac{4}{3}\pi r^3 \sigma g \)
3. Viscous Drag (Stokes' Law) acting up: \( F_v = 6\pi \eta r v \)

At terminal velocity \(v\), Upward forces = Downward forces:

$$ 6\pi \eta r v + \frac{4}{3}\pi r^3 \sigma g = \frac{4}{3}\pi r^3 \rho g $$ $$ 6\pi \eta r v = \frac{4}{3}\pi r^3 g (\rho - \sigma) $$ $$ v = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta} $$

Q.22. Determine the shortest wavelengths of Balmer and Paschen series. Given the limit for Lyman series is 912 Å.

Solution

Formula: \( \frac{1}{\lambda} = R (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \).

1. Lyman Limit (Shortest wavelength): \( n_1=1, n_2=\infty \).
\( \frac{1}{\lambda_L} = R(1 - 0) = R \). Given \( \lambda_L = 912 \text{ \AA} = 1/R \).

2. Balmer Limit (Shortest wavelength): \( n_1=2, n_2=\infty \).
\( \frac{1}{\lambda_B} = R (\frac{1}{2^2}) = \frac{R}{4} \).
\( \lambda_B = \frac{4}{R} = 4 \times 912 = 3648 \text{ \AA} \).

3. Paschen Limit (Shortest wavelength): \( n_1=3, n_2=\infty \).
\( \frac{1}{\lambda_P} = R (\frac{1}{3^2}) = \frac{R}{9} \).
\( \lambda_P = \frac{9}{R} = 9 \times 912 = 8208 \text{ \AA} \).

Q.23. Calculate the value of magnetic field at a distance of 3 cm from a very long, straight wire carrying a current of 6A.

Solution

Given:
\( I = 6 \text{ A} \)
\( r = 3 \text{ cm} = 0.03 \text{ m} \)
\( \mu_0 = 4\pi \times 10^{-7} \)

$$ B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 6}{2\pi \times 0.03} $$ $$ B = \frac{2 \times 10^{-7} \times 6}{0.03} = \frac{12 \times 10^{-7}}{3 \times 10^{-2}} $$ $$ B = 4 \times 10^{-5} \text{ T} $$

Q.24. A parallel plate capacitor filled with air has an area of 6 cm² and plate separation of a 3 mm. Calculate its capacitance.

Solution

Given:
\( A = 6 \text{ cm}^2 = 6 \times 10^{-4} \text{ m}^2 \)
\( d = 3 \text{ mm} = 3 \times 10^{-3} \text{ m} \)
\( \epsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \)

$$ C = \frac{\epsilon_0 A}{d} $$ $$ C = \frac{8.85 \times 10^{-12} \times 6 \times 10^{-4}}{3 \times 10^{-3}} $$ $$ C = 8.85 \times 10^{-12} \times 2 \times 10^{-1} $$ $$ C = 17.7 \times 10^{-13} \text{ F} = 1.77 \text{ pF} $$

Q.25. An emf of 91 mV is induced in the windings of a coil, when the current in a nearby coil is increasing at the rate of 1.3 A/s, what is the mutual inductance (M) of the two coils in mH?

Solution

Given:
\( e = 91 \text{ mV} = 91 \times 10^{-3} \text{ V} \)
\( \frac{dI}{dt} = 1.3 \text{ A/s} \)

Formula: \( |e| = M \frac{dI}{dt} \)

$$ M = \frac{e}{dI/dt} = \frac{91 \times 10^{-3}}{1.3} $$ $$ M = \frac{91}{1.3} \times 10^{-3} = 70 \times 10^{-3} \text{ H} $$

Answer: \( M = 70 \text{ mH} \).

Q.26. Two cells of emf 4V and 2V having respective internal resistance of 1 Ω and 2 Ω are connected in parallel, so as to send current in the same direction through an external resistance of 5 Ω. Find the current through the external resistance.

Solution

Given: \( E_1=4V, r_1=1\Omega \); \( E_2=2V, r_2=2\Omega \); \( R=5\Omega \).

Equivalent EMF (\(E_{eq}\)) and Internal Resistance (\(r_{eq}\)) for parallel combination:

$$ r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \Omega $$ $$ E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} = \frac{4(2) + 2(1)}{3} = \frac{10}{3} \text{ V} $$

Current \( I \):

$$ I = \frac{E_{eq}}{R + r_{eq}} = \frac{10/3}{5 + 2/3} $$ $$ I = \frac{10/3}{17/3} = \frac{10}{17} \text{ A} \approx 0.588 \text{ A} $$

SECTION − D

Attempt any THREE questions of the following: [12]

Q.27. Derive an expression for a pressure exerted by a gas on the basis of kinetic theory of gases.

Solution

Consider a cube of side \( L \) containing \( N \) molecules each of mass \( m \).

1. A molecule moving with velocity \( v_x \) hits the wall. Change in momentum = \( -mv_x - (mv_x) = -2mv_x \). Momentum imparted to wall = \( 2mv_x \).
2. Time between successive collisions with same wall \( \Delta t = \frac{2L}{v_x} \).
3. Force by one molecule \( f = \frac{\Delta p}{\Delta t} = \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L} \).
4. Total Force \( F_x = \frac{m}{L} \sum v_{x}^2 \).
5. Pressure \( P = \frac{F}{A} = \frac{F}{L^2} = \frac{m}{L^3} \sum v_x^2 \).
6. Using mean square velocity and isotropy (\( \overline{v_x^2} = \frac{1}{3}\overline{v^2} \)):

$$ P = \frac{1}{3} \frac{Nm}{V} \overline{v^2} = \frac{1}{3} \rho \overline{v^2} $$

Q.28. What is a rectifier? With the help of a neat circuit diagram, explain the working of a half wave rectifier.

Solution

Rectifier: A device that converts Alternating Current (AC) into Direct Current (DC).

Working of Half Wave Rectifier:

• Construction: Consists of a transformer, a p-n junction diode, and a load resistor \( R_L \).
• Positive Half Cycle: The diode is forward biased (Anode positive wrt Cathode). It conducts current. Output appears across \( R_L \).
• Negative Half Cycle: The diode is reverse biased. Ideally, no current flows. No output voltage.
• Result: Output is unidirectional pulsating DC.

Q.29. Draw a neat, labelled diagram of a suspended coil type moving coil galvanometer. The initial pressure and volume of a gas enclosed in a cylinder are 2 × 10⁵ N/m² and 6 × 10^–3 m³ respectively. If the work done in compressing the gas at constant pressure is 150 J, find the final volume of the gas.

Solution

Part 1: Diagram

Part 2: Calculation

Given:
\( P = 2 \times 10^5 \text{ N/m}^2 \)
\( V_i = 6 \times 10^{-3} \text{ m}^3 \)
Work done on the gas (compression) \( W = 150 \text{ J} \). (Note: Work done by gas is -150J, or simply use magnitude for volume change).

$$ W = P (V_f - V_i) $$

Since it is compression, \( V_f < V_i \). Work is done ON the gas.

$$ 150 = P(V_i - V_f) $$ $$ V_i - V_f = \frac{150}{2 \times 10^5} = 75 \times 10^{-5} = 0.75 \times 10^{-3} \text{ m}^3 $$ $$ V_f = V_i - 0.75 \times 10^{-3} $$ $$ V_f = 6 \times 10^{-3} - 0.75 \times 10^{-3} = 5.25 \times 10^{-3} \text{ m}^3 $$

Q.30. Define second’s pendulum. Derive a formula for the length of second’s pendulum. A particle performing linear S.H.M. has maximum velocity 25 cm/s and maximum acceleration 100 cm/s². Find period of oscillations.

Solution

Definition: A simple pendulum whose time period is equal to 2 seconds is called a second’s pendulum.

Derivation:
Time period \( T = 2\pi \sqrt{\frac{L}{g}} \).
For second's pendulum \( T = 2 \).
\( 2 = 2\pi \sqrt{\frac{L_s}{g}} \implies 1 = \pi \sqrt{\frac{L_s}{g}} \implies 1 = \pi^2 \frac{L_s}{g} \)
\( L_s = \frac{g}{\pi^2} \).

Problem:
\( v_{max} = A\omega = 25 \) ---(1)
\( a_{max} = A\omega^2 = 100 \) ---(2)
Divide (2) by (1):
\( \frac{A\omega^2}{A\omega} = \frac{100}{25} \implies \omega = 4 \text{ rad/s} \).
Period \( T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ s} \approx 1.57 \text{ s} \).

Q.31. Explain de Broglie wavelength. Obtain an expression for de Broglie wavelength of wave associated with material particles. The photoelectric work function for a metal is 4.2 eV. Find the threshold wavelength.

Solution

De Broglie Wavelength: Louis de Broglie suggested that moving material particles exhibit wave-like properties. The wavelength associated with a particle of momentum \( p \) is \( \lambda = h/p \).

Derivation:
From Planck's theory: \( E = h\nu = hc/\lambda \).
From Einstein's relativity: \( E = mc^2 \).
Equating: \( hc/\lambda = mc^2 \implies \lambda = h/mc = h/p \).

Problem:
Work function \( \phi_0 = 4.2 \text{ eV} = 4.2 \times 1.6 \times 10^{-19} \text{ J} \).
Threshold wavelength \( \lambda_0 = \frac{hc}{\phi_0} \).

$$ \lambda_0 = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4.2 \times 1.6 \times 10^{-19}} $$ $$ \lambda_0 = \frac{19.89}{6.72} \times 10^{-7} \text{ m} $$ $$ \lambda_0 \approx 2.96 \times 10^{-7} \text{ m} = 2960 \text{ \AA} $$