HSC 12th Physics June 2026 Board Question Paper with Solutions (J-229)

Maharashtra State Board HSC 12th Standard Physics (Subject Code: 54) June / July 2026 Supplementary Exam Question Paper (J-229) with Complete Solutions and Model Answer Key.

SECTION - A

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

The MI of ring of radius 'R' and mass 'M' about its central axis is

Answer: (c) \( MR^2 \)
For a monoatomic gas enclosed in a container at a constant temperature has degrees of freedom

Answer: (b) \( 3 \)
The equation of the first law of thermodynamics for isochoric process is

Answer: (a) \( Q = \Delta U \)
Cyclotron can not accelerate

Answer: (c) neutrons
In a purely inductive and capacitive circuit, power factor is,

Answer: (b) zero
The Boolean equation for NOR gate is

Answer: (d) \( Y = \overline{A + B} \)
When n number of capacitors having each of capacity 'C' are connected in series, their equivalent capacity is

Answer: (c) \( \frac{C}{n} \)
A polarising angle for an optical material is \( 35^\circ \). Then refractive index of the material is

Answer: (b) \( 0.7002 \)
For a streamline flow, the relative velocity between the two liquid layers is \( 10\text{ cm/s} \). The distance between them is \( 5\text{ cm} \). The velocity gradient is

Answer: (d) \( 2\text{ s}^{-1} \)
The wavelength of a stationary wave on a string is \( 0.75\text{ m} \) with frequency of \( 64\text{ Hz} \). Then, the velocity of wave is

Answer: (a) \( 48\text{ m/s} \)

Q. 2. Answer the following questions:

Absorption of ink by a blotting paper is an example of which phenomenon of fluids?

Answer: Capillarity (or Capillary action).
State Kirchhoff's first law for electrical network.

Answer: Kirchhoff's first law (Current Law) states that the algebraic sum of the currents meeting at any junction (node) in an electrical network is zero, i.e., \( \sum I = 0 \).
Define 'photoelectric work function.'

Answer: The minimum amount of energy required to be provided to an electron to just pull it out of the metal surface (with zero kinetic energy) is called the photoelectric work function of that metal.
Which series of spectral lines belongs to the ultraviolet region of the hydrogen spectrum?

Answer: Lyman series.
Which semiconductor device converts solar energy into electrical energy?

Answer: Solar cell (or Photovoltaic cell).
Define the term wavefront.

Answer: The locus of all points in a medium to which light waves reach simultaneously, so that all the points are in the same phase, is called a wavefront.
A motorcyclist performs horizontal circular motion inside a well of inner radius \( 3\text{ m} \). Calculate the minimum speed to perform the stunt if coefficient of static friction is \( 0.3 \). \( [g = 10\text{ m/s}^2] \).

Answer:
Given: \( r = 3\text{ m} \), \( \mu_s = 0.3 \), \( g = 10\text{ m/s}^2 \)
Formula: \( v_{\min} = \sqrt{\frac{rg}{\mu_s}} \)
\[ v_{\min} = \sqrt{\frac{3 \times 10}{0.3}} = \sqrt{\frac{30}{0.3}} = \sqrt{100} = 10\text{ m/s} \] \( v_{\min} = 10\text{ m/s} \)
Calculate the maximum velocity of a particle having amplitude of oscillation \( 4\text{ cm} \) and period \( 4\text{ seconds} \).

Answer:
Given: \( A = 4\text{ cm} = 4 \times 10^{-2}\text{ m} \), \( T = 4\text{ s} \), \( \pi = 3.142 \)
Formula: \( v_{\max} = A\omega = A\left(\frac{2\pi}{T}\right) \)
\[ v_{\max} = 4 \times 10^{-2} \times \left(\frac{2 \times 3.142}{4}\right) = 2 \times 3.142 \times 10^{-2} = 6.284 \times 10^{-2}\text{ m/s} \] \( v_{\max} = 6.284 \times 10^{-2}\text{ m/s} \) (or \( 6.284\text{ cm/s} \))

SECTION - B

Attempt any EIGHT questions from the following:

Q. 3. Obtain an expression for the maximum speed of a vehicle along horizontal circular curved road.

Answer:

(Note to student: Draw a neat diagram of a vehicle on a horizontal curved road of radius \( r \) showing weight \( mg \) acting vertically downwards, normal reaction \( N \) acting vertically upwards, and force of static friction \( f_s \) directed horizontally towards the centre \( C \) of the circular track.)

Consider a vehicle of mass \( m \) moving with speed \( v \) along a horizontal circular track of radius \( r \).

In the reference frame of an inertial observer, the forces acting on the vehicle are:

Weight (\( mg \)): acting vertically downwards.
Normal reaction (\( N \)): acting vertically upwards, perpendicular to the road surface.
Force of static friction (\( f_s \)): acting horizontally between the road and the tyres, directed towards the centre of the circular track.

Since there is no vertical motion of the vehicle, the normal reaction balances the weight of the vehicle:

\[ N = mg \quad \text{... (Equation 1)} \]

The force of static friction provides the necessary centripetal force for circular motion:

\[ f_s = \frac{mv^2}{r} \quad \text{... (Equation 2)} \]

Dividing Equation 2 by Equation 1, we get:

\[ \frac{f_s}{N} = \frac{v^2}{rg} \quad \text{... (Equation 3)} \]

For a given track, radius \( r \) is constant. As speed \( v \) increases, the force of static friction \( f_s \) also increases up to its limiting value:

\[ (f_s)_{\max} = \mu_s N \quad \text{(where } \mu_s \text{ is the coefficient of static friction between the tyres and the road)} \]

Substituting \( (f_s)_{\max} = \mu_s N \) for maximum possible speed (\( v_{\max} \)) in Equation 3:

\[ \frac{\mu_s N}{N} = \frac{v_{\max}^2}{rg} \implies \mu_s = \frac{v_{\max}^2}{rg} \]

\[ v_{\max}^2 = \mu_s rg \]

\[ \mathbf{v_{\max} = \sqrt{\mu_s rg}} \]

This is the required expression for the maximum safe speed of a vehicle along a horizontal circular curved road.

Q. 4. For a thermodynamic system: Define: (a) Mechanical equilibrium (b) Chemical equilibrium

Answer:

Mechanical equilibrium: When there are no unbalanced forces within the system and between the system and its surroundings, the system is said to be in mechanical equilibrium. In this state, the pressure throughout the system and between the system and its surroundings is the same and does not change with time.
Chemical equilibrium: A system is said to be in chemical equilibrium when there are no chemical reactions going on within the system, or there is no transfer of matter from one part of the system to the other due to diffusion. In this state, the chemical composition of the system is the same throughout and does not change with time.

Q. 5. What are polar molecules and non polar molecules?

Answer:

Polar molecules: A molecule in which the centre of mass of positive charges (protons) does not coincide with the centre of mass of negative charges (electrons), because of the asymmetric shape of the molecules, is called a polar molecule. They possess a permanent electric dipole moment. Examples: \( \text{HCl}, \text{H}_2\text{O}, \text{NH}_3, \text{N}_2\text{O} \).
Non-polar molecules: A molecule in which the centre of mass of positive charges coincides with the centre of mass of negative charges is called a non-polar molecule. These molecules have symmetrical shapes and have zero dipole moment in the normal state. Examples: \( \text{H}_2, \text{N}_2, \text{O}_2, \text{CO}_2, \text{CH}_4 \).

Q. 6. Explain how will you convert moving coil galvanometer into voltmeter?

Answer:

(Note to student: Draw a neat circuit diagram showing a Moving Coil Galvanometer of resistance \( G \) connected in series with a high resistance \( X \) across potential difference \( V \), carrying full-scale deflection current \( I_g \).)

To convert a moving coil galvanometer (MCG) into a voltmeter, its effective resistance must be increased and its voltage measuring capacity must be increased to the desired value. This is achieved by connecting a very high resistance in series with the galvanometer.

Let \( G \) be the resistance of the galvanometer coil.
Let \( I_g \) be the current required for full-scale deflection of the galvanometer.
Let \( V \) be the maximum potential difference (voltage) to be measured.
Let \( X \) (or \( R \)) be the high resistance connected in series with the galvanometer.

When resistance \( X \) is connected in series with the galvanometer, the total resistance of the circuit becomes \( (G + X) \). By Ohm's law, if \( V \) is the potential difference across the series combination when current \( I_g \) flows through it:

\[ V = I_g (G + X) \]

\[ \frac{V}{I_g} = G + X \]

\[ \mathbf{X = \frac{V}{I_g} - G} \]

This equation gives the value of the high series resistance required to convert a moving coil galvanometer into a voltmeter of range \( 0 \) to \( V \).

Q. 7. Distinguish between free and forced vibrations.

Answer:

Free Vibrations	Forced Vibrations
1. Free vibrations are produced when a body is disturbed from its equilibrium position and released to oscillate on its own.	1. Forced vibrations are produced by the continuous action of an external periodic force on a body.
2. The body vibrates with its own natural frequency, which depends on the intrinsic elastic and inertial properties of the body.	2. The body vibrates with a frequency equal to the frequency of the applied external periodic force.
3. No external periodic force is required to maintain the vibrations once initiated.	3. A continuous external periodic force is required to maintain the vibrations.
4. The amplitude of vibrations continuously decreases with time due to damping forces.	4. The amplitude of vibrations remains constant as long as the external periodic force acts on the body.

Q. 8. State any two postulates of Bohr's theory of hydrogen atom.

Answer:

Any two of the following postulates:

First Postulate (Circular Orbit): In a hydrogen atom, the electron revolves around the nucleus in a fixed circular orbit with constant speed. The necessary centripetal force is provided by the electrostatic force of attraction between the positively charged nucleus and the negatively charged electron: \[ \frac{m_e v^2}{r} = \frac{e^2}{4\pi\varepsilon_0 r^2} \]
Second Postulate (Quantization of Angular Momentum): The radius of the orbit of an electron can only take certain fixed values such that the angular momentum of the electron in these stationary orbits is an integral multiple of \( \frac{h}{2\pi} \), where \( h \) is Planck's constant: \[ L = m_e v r = \frac{nh}{2\pi} \quad (\text{where } n = 1, 2, 3, \dots) \]
Third Postulate (Energy Radiation): An electron does not radiate energy while revolving in a stationary orbit. Energy is radiated in the form of a photon only when the electron jumps from a higher energy orbit (\( E_n \)) to a lower energy orbit (\( E_p \)): \[ h\nu = E_n - E_p \]

Q. 9. Explain the cyclic process with the help of a neat p-V diagram.

Answer:

(Note to student: Draw a \( p\text{-}V \) diagram with Volume \( V \) on the X-axis and Pressure \( p \) on the Y-axis showing a closed loop formed by two paths between state \( A(p_1, V_1) \) and state \( B(p_2, V_2) \) with directional arrows indicating the cycle.)

Cyclic Process: A thermodynamic process that returns a system to its initial state is called a cyclic process.

In a cyclic process, the initial and the final states are the same. Since internal energy is a state variable, the total change in internal energy of the system for a cyclic process is zero (\( \Delta U = 0 \)).
According to the first law of thermodynamics: \( Q = \Delta U + W \).
Since \( \Delta U = 0 \), we get \( \mathbf{Q = W} \). Thus, the total heat supplied to the system in a cyclic process is completely converted into work done by the system (or vice versa).
\( p\text{-}V \) Diagram Representation: On a pressure-volume (\( p\text{-}V \)) diagram, a cyclic process is represented by a closed curve (loop), and the area enclosed by the closed curve gives the net work done during the cycle.

Q. 10. A star is emitting light at the wavelength of \( 4000\text{ \AA} \). Determine the limit of angular resolution of a telescope having an objective of diameter of \( 400\text{ cm} \).

Answer:

Given:
Wavelength of light (\( \lambda \)) \( = 4000\text{ \AA} = 4000 \times 10^{-10}\text{ m} = 4 \times 10^{-7}\text{ m} \)
Diameter of objective (\( D \)) \( = 400\text{ cm} = 4\text{ m} \)

Formula:
Limit of angular resolution: \[ d\theta = \frac{1.22 \lambda}{D} \]

Calculation: \[ d\theta = \frac{1.22 \times 4 \times 10^{-7}}{4} = 1.22 \times 10^{-7}\text{ rad} \]

The limit of angular resolution of the telescope is \( 1.22 \times 10^{-7}\text{ rad} \).

Q. 11. Calculate the value of \( \lambda_{\max} \) for solar radiation assuming the temperature on the surface of the sun is \( 5800\text{ K} \). [Wien's constant \( = 2.897 \times 10^{-3}\text{ m}\cdot\text{K} \)]

Answer:

Given:
Temperature (\( T \)) \( = 5800\text{ K} \)
Wien's constant (\( b \)) \( = 2.897 \times 10^{-3}\text{ m}\cdot\text{K} \)

Formula:
According to Wien's displacement law: \[ \lambda_{\max} = \frac{b}{T} \]

Calculation: \[ \lambda_{\max} = \frac{2.897 \times 10^{-3}}{5800} = 0.00049948 \times 10^{-3}\text{ m} = 4.995 \times 10^{-7}\text{ m} \text{ (or } 4995\text{ \AA)} \]

The value of \( \lambda_{\max} \) for solar radiation is \( 4.995 \times 10^{-7}\text{ m} \) (or \( 4995\text{ \AA} \)).

Q. 12. Calculate the radius of water drop, so that the excess pressure inside it is \( 24\text{ N/m}^2 \). [Surface tension of water \( = 7.2 \times 10^{-2}\text{ N/m} \)]

Answer:

Given:
Excess pressure (\( \Delta P \)) \( = 24\text{ N/m}^2 \)
Surface tension (\( T \)) \( = 7.2 \times 10^{-2}\text{ N/m} \)

Formula:
Excess pressure inside a liquid drop: \[ \Delta P = \frac{2T}{r} \]

Calculation: \[ r = \frac{2T}{\Delta P} = \frac{2 \times 7.2 \times 10^{-2}}{24} = \frac{14.4 \times 10^{-2}}{24} = 0.6 \times 10^{-2}\text{ m} = 6 \times 10^{-3}\text{ m} \]

The radius of the water drop is \( 6 \times 10^{-3}\text{ m} \) (or \( 6\text{ mm} \)).

Q. 13. The susceptibility of magnetic material is \( \chi \) at \( 27^\circ\text{C} \). At what temperature its susceptibility will be \( \chi/3 \)?

Answer:

Given:
Initial susceptibility (\( \chi_1 \)) \( = \chi \)
Initial temperature (\( T_1 \)) \( = 27^\circ\text{C} = 27 + 273 = 300\text{ K} \)
Final susceptibility (\( \chi_2 \)) \( = \frac{\chi}{3} \)

Formula:
According to Curie's law: \[ \chi \propto \frac{1}{T} \implies \chi_1 T_1 = \chi_2 T_2 \]

Calculation: \[ T_2 = \frac{\chi_1 \times T_1}{\chi_2} = \frac{\chi \times 300}{\chi / 3} = 300 \times 3 = 900\text{ K} \] In Celsius: \( T_2 = 900 - 273 = 627^\circ\text{C} \)

Its susceptibility will be \( \chi/3 \) at \( 900\text{ K} \) (or \( 627^\circ\text{C} \)).

Q. 14. An alternating voltage source given by \( e = 140 \sin(314.2 t) \) is connected across a pure resistor of \( 50\text{ }\Omega \). Determine: (i) Frequency of the source. (ii) RMS value of emf.

Answer:

Given:
Equation of alternating voltage: \( e = 140 \sin(314.2 t) \)
Comparing with the standard equation \( e = e_0 \sin(\omega t) \):
Peak emf (\( e_0 \)) \( = 140\text{ V} \)
Angular frequency (\( \omega \)) \( = 314.2\text{ rad/s} \)

(i) Frequency of the source (\( f \)):
Formula: \( \omega = 2\pi f \) \[ f = \frac{\omega}{2\pi} = \frac{314.2}{2 \times 3.142} = \frac{314.2}{6.284} = 50\text{ Hz} \] Frequency of the source is \( 50\text{ Hz} \).

(ii) RMS value of emf (\( e_{\text{rms}} \)):
Formula: \[ e_{\text{rms}} = \frac{e_0}{\sqrt{2}} = 0.707 \times e_0 = 0.707 \times 140 = 98.98\text{ V} \] RMS value of emf is \( 98.98\text{ V} \).

SECTION - C

Attempt any EIGHT questions of the following:

Q. 15. State and prove perpendicular axes theorem.

Answer:

Statement: The moment of inertia (\( I_z \)) of a laminar object about an axis (\( Z \)) perpendicular to its plane is the sum of its moments of inertia about two mutually perpendicular axes (\( X \) and \( Y \)) in its plane, all the three axes being concurrent.

Mathematically: \( \mathbf{I_z = I_x + I_y} \)

(Note to student: Draw a neat diagram of a plane lamina in the X-Y plane showing mutually perpendicular axes \( OX, OY \) in the plane and \( OZ \) perpendicular to the plane, with a mass element \( dm \) at point \( P(x, y) \) at perpendicular distances \( y \) from \( OX \), \( x \) from \( OY \), and \( r \) from \( OZ \).)

Proof:

Consider a rigid laminar object lying in the X-Y plane. Let \( OX \) and \( OY \) be two mutually perpendicular axes in the plane of the lamina, and \( OZ \) be the axis perpendicular to its plane intersecting at origin \( O \).
Let \( I_x, I_y, \) and \( I_z \) be the moments of inertia of the lamina about the \( X, Y, \) and \( Z \) axes respectively.
Consider an infinitesimal mass element \( dm \) situated at point \( P(x, y) \).
The perpendicular distance of \( P \) from the X-axis is \( y \), from the Y-axis is \( x \), and from the Z-axis is \( OP = r \).
By definitions of moment of inertia: \[ I_x = \int y^2 \, dm, \quad I_y = \int x^2 \, dm, \quad I_z = \int r^2 \, dm \]
From Pythagoras theorem in the right-angled triangle in the X-Y plane: \[ r^2 = x^2 + y^2 \]
Substituting \( r^2 = x^2 + y^2 \) in the expression for \( I_z \): \[ I_z = \int (x^2 + y^2) \, dm = \int y^2 \, dm + \int x^2 \, dm \]
Therefore, \( \mathbf{I_z = I_x + I_y} \).

Hence, the theorem of perpendicular axes is proved.

Q. 16. On the basis of molecular theory, explain the phenomenon of surface tension.

Answer:

(Note to student: Draw a neat diagram of a liquid in a container showing the free surface \( XY \), surface film \( XY\text{-}X'Y' \) of thickness equal to the molecular range, and three molecules \( A, B, \) and \( C \) along with their spheres of influence.)

The phenomenon of surface tension arises due to intermolecular cohesive forces and can be explained by considering three molecules \( A, B, \) and \( C \) in a liquid:

Molecule A (deep inside the liquid): The sphere of influence of molecule \( A \) lies completely inside the liquid. As a result, it is acted upon by equal cohesive forces in all directions. Therefore, the net cohesive force acting on molecule \( A \) is zero.
Molecule B (just below the free surface, within the surface film): A larger part of its sphere of influence lies inside the liquid and a smaller part lies in air. Since the number of air molecules is negligible compared to liquid molecules, the downward cohesive force due to liquid molecules is much stronger than the upward adhesive force. Hence, molecule \( B \) experiences a net downward force.
Molecule C (on the free surface of the liquid): Exactly half of its sphere of influence lies in the liquid and the upper half lies in air. Therefore, molecule \( C \) experiences the maximum net downward force directed into the liquid.

Conclusion: Thus, all molecules lying within the surface film experience a net inward (downward) pull. Whenever a molecule is brought from the interior of the liquid to the surface film, work has to be done against this inward force. This work is stored in the surface molecules in the form of potential energy. To achieve stable equilibrium with minimum potential energy, the free surface of the liquid tends to contract to have the minimum possible surface area and behaves like a stretched elastic membrane under tension. This phenomenon is called surface tension.

Q. 17. Show that all harmonics are present in the case of an air column vibrating in a pipe open at both the ends.

Answer:

(Note to student: Draw neat diagrams showing the first three modes of vibration of an air column in a pipe open at both ends: (a) Fundamental mode with 1 node and 2 antinodes, (b) First overtone with 2 nodes and 3 antinodes, and (c) Second overtone with 3 nodes and 4 antinodes.)

When a longitudinal stationary wave is set up in an air column inside a pipe open at both ends, antinodes are formed at both open ends and nodes are formed in between. Let \( L \) be the corrected length of the air column and \( v \) be the speed of sound in air.

Fundamental mode (First harmonic):
In the simplest mode of vibration, two antinodes are formed at the open ends and one node is formed at the centre.
Length of air column: \( L = \frac{\lambda_1}{2} \implies \lambda_1 = 2L \)
Fundamental frequency (\( n \)) is: \[ n = \frac{v}{\lambda_1} = \mathbf{\frac{v}{2L}} \quad \text{... (Equation 1)} \]
First overtone (Second harmonic):
In the next mode of vibration, three antinodes and two nodes are formed.
Length of air column: \( L = \lambda_2 \implies \lambda_2 = L \)
Frequency of first overtone (\( n_1 \)) is: \[ n_1 = \frac{v}{\lambda_2} = \frac{v}{L} = 2\left(\frac{v}{2L}\right) = \mathbf{2n} \quad \text{... (Equation 2)} \]
Second overtone (Third harmonic):
In this mode of vibration, four antinodes and three nodes are formed.
Length of air column: \( L = \frac{3\lambda_3}{2} \implies \lambda_3 = \frac{2L}{3} \)
Frequency of second overtone (\( n_2 \)) is: \[ n_2 = \frac{v}{\lambda_3} = \frac{3v}{2L} = 3\left(\frac{v}{2L}\right) = \mathbf{3n} \quad \text{... (Equation 3)} \]

Conclusion: From Equations 1, 2, and 3, the frequencies of the fundamental tone and overtones are in the ratio \( n : n_1 : n_2 : \dots = 1 : 2 : 3 : \dots \) Thus, both even and odd harmonics (i.e., all harmonics) are present as overtones in the vibrations of an air column in a pipe open at both ends.

Q. 18. What is interference of light? State any four conditions for obtaining well defined and steady interference pattern.

Answer:

Interference of light: The modification in the intensity of light (redistribution of light energy) produced by the superposition of two or more light waves from coherent sources is called interference of light.

Conditions for obtaining a well-defined and steady interference pattern (Any four):

The two sources of light must be coherent (i.e., they must maintain a constant phase difference).
The two sources of light must be monochromatic (having the same wavelength and frequency).
The two interfering waves must have the same amplitude (or equal intensity) for good contrast between bright and dark fringes.
The two light sources must be narrow.
The separation between the two light sources should be small, and the distance between the sources and the screen should be large.
The two interfering light waves must be in the same state of polarization.

Q. 19. With the help of a neat circuit diagram, explain the working of a half wave rectifier. Draw input and output waveforms.

Answer:

(Note to student: Draw (1) the circuit diagram of a half-wave rectifier showing AC mains connected to primary coil \( P_1P_2 \) of a transformer, secondary coil \( S_1S_2 \) connected in series with junction diode \( D \) and load resistor \( R_L \), and (2) the vertically aligned Input AC voltage waveform and Output pulsating DC voltage waveform plotted against time.)

Working of a Half Wave Rectifier:

The AC input voltage to be rectified is applied across the primary coil (\( P_1P_2 \)) of a transformer. The secondary coil (\( S_1S_2 \)) is connected in series with a p-n junction diode (\( D \)) and a load resistance (\( R_L \)).
During the positive half cycle of input AC: Terminal \( S_1 \) of the secondary coil is positive with respect to terminal \( S_2 \). Hence, the diode \( D \) is forward-biased and conducts electric current. Current flows through the load resistance \( R_L \) from \( A \) to \( B \), and an output voltage is obtained across \( R_L \).
During the negative half cycle of input AC: Terminal \( S_1 \) becomes negative with respect to terminal \( S_2 \). Hence, the diode \( D \) is reverse-biased and does not conduct any current. No current flows through \( R_L \), and the output voltage across \( R_L \) is zero.
Since the diode conducts only during the positive half cycles of the AC input, the current through the load resistance flows only in one direction. Thus, the circuit acts as a half-wave rectifier producing a pulsating DC output.

Q. 20. Obtain an expression for orbital magnetic moment of an electron revolving around the nucleus of an atom.

Answer:

(Note to student: Draw a circular orbit of radius \( r \) showing an electron revolving anticlockwise with linear speed \( v \), conventional current \( I \) clockwise, orbital angular momentum vector \( \vec{L} \) upwards, and orbital magnetic moment vector \( \vec{M}_o \) downwards.)

Consider an electron of mass \( m_e \) and charge \( e \) revolving with constant tangential speed \( v \) in a circular orbit of radius \( r \) around the positively charged nucleus.

The revolving electron constitutes a tiny circular current loop. The magnitude of current \( I \) associated with the revolving electron is:

\[ I = \frac{e}{T} \quad \text{... (Equation 1)} \]

Where \( T \) is the time period of revolution of the electron. Since \( T = \frac{2\pi r}{v} \), substituting this in Equation 1 gives:

\[ I = \frac{e}{2\pi r / v} = \frac{ev}{2\pi r} \quad \text{... (Equation 2)} \]

The area enclosed by the circular orbit is:

\[ A = \pi r^2 \quad \text{... (Equation 3)} \]

The orbital magnetic dipole moment (\( M_o \)) associated with the current loop has a magnitude given by:

\[ M_o = I \times A \]

Substituting Equations 2 and 3:

\[ M_o = \left(\frac{ev}{2\pi r}\right) \times (\pi r^2) \]

\[ \mathbf{M_o = \frac{evr}{2}} \quad \text{... (Equation 4)} \]

Multiplying and dividing the right-hand side of Equation 4 by the mass of the electron (\( m_e \)):

\[ M_o = \frac{e}{2m_e} (m_e v r) \]

Since the orbital angular momentum of the electron is \( L = m_e v r \):

\[ \mathbf{M_o = \frac{eL}{2m_e}} \]

In vector form, \( \mathbf{\vec{M}_o = -\frac{e}{2m_e} \vec{L}} \), showing that the orbital magnetic moment is directed opposite to the orbital angular momentum.

Q. 21. Define the terms: (a) Inductive reactance (b) Capacitive reactance (c) Impedance

Answer:

Inductive reactance (\( X_L \)): The effective opposition offered by an inductor to the flow of alternating current in an AC circuit is called inductive reactance. It is given by \( X_L = \omega L = 2\pi f L \), and its SI unit is ohm (\( \Omega \)).
Capacitive reactance (\( X_C \)): The effective opposition offered by a capacitor to the flow of alternating current in an AC circuit is called capacitive reactance. It is given by \( X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \), and its SI unit is ohm (\( \Omega \)).
Impedance (\( Z \)): The total effective opposition offered by an AC circuit containing a resistor, an inductor, and a capacitor (or their combination) to the flow of alternating current is called impedance. It is the ratio of the RMS voltage to the RMS current, i.e., \( Z = \frac{e_{\text{rms}}}{i_{\text{rms}}} = \sqrt{R^2 + (X_L - X_C)^2} \), and its SI unit is ohm (\( \Omega \)).

Q. 22. A parallel plate capacitor filled with air has an area \( 6\text{ cm}^2 \) and the plate separation is \( 2\text{ mm} \). Calculate its capacity.

Answer:

Given:
Area of each plate (\( A \)) \( = 6\text{ cm}^2 = 6 \times 10^{-4}\text{ m}^2 \)
Plate separation (\( d \)) \( = 2\text{ mm} = 2 \times 10^{-3}\text{ m} \)
Permittivity of free space (\( \varepsilon_0 \)) \( = 8.85 \times 10^{-12}\text{ C}^2/\text{Nm}^2 \)

Formula:
Capacity of an air-filled parallel plate capacitor: \[ C = \frac{\varepsilon_0 A}{d} \]

Calculation: \[ C = \frac{8.85 \times 10^{-12} \times 6 \times 10^{-4}}{2 \times 10^{-3}} = 8.85 \times 10^{-12} \times 3 \times 10^{-1} = 26.55 \times 10^{-13}\text{ F} = 2.655 \times 10^{-12}\text{ F} \]

The capacity of the parallel plate capacitor is \( 2.655 \times 10^{-12}\text{ F} \) (or \( 2.655\text{ pF} \)).

Q. 23. Two resistances \( X \) and \( Y \) in the two gaps of a meterbridge give a null point dividing the wire in the ratio \( 4:6 \). If both resistances are increased by \( 30\text{ }\Omega \), the null point divides the wire in the ratio \( 5:6 \). Calculate the value of \( X \) and \( Y \).

Answer:

Given:
First balancing ratio: \( \frac{l_x}{l_y} = \frac{4}{6} = \frac{2}{3} \)
Second balancing ratio (when \( X' = X + 30 \) and \( Y' = Y + 30 \)): \( \frac{l'_x}{l'_y} = \frac{5}{6} \)

Formula:
For a balanced meter bridge: \[ \frac{X}{Y} = \frac{l_x}{l_y} \]

Calculation:
Case 1: \[ \frac{X}{Y} = \frac{2}{3} \implies Y = 1.5 X \quad \text{... (Equation 1)} \]

Case 2: \[ \frac{X + 30}{Y + 30} = \frac{5}{6} \] \[ 6(X + 30) = 5(Y + 30) \] \[ 6X + 180 = 5Y + 150 \] \[ 5Y - 6X = 30 \quad \text{... (Equation 2)} \]

Substituting \( Y = 1.5 X \) from Equation 1 into Equation 2: \[ 5(1.5 X) - 6X = 30 \implies 7.5 X - 6X = 30 \implies 1.5 X = 30 \implies X = \frac{30}{1.5} = 20\text{ }\Omega \]

Substituting \( X = 20\text{ }\Omega \) in Equation 1: \[ Y = 1.5 \times 20 = 30\text{ }\Omega \]

The values of the two resistances are \( X = 20\text{ }\Omega \) and \( Y = 30\text{ }\Omega \).

Q. 24. Calculate the shortest wavelength of Paschen series and the longest wavelength of Balmer series for hydrogen atom.

Answer:

Given:
Rydberg's constant (\( R \)) \( = 1.097 \times 10^7\text{ m}^{-1} \)

Formula: \[ \frac{1}{\lambda} = R \left(\frac{1}{p^2} - \frac{1}{n^2}\right) \]

1. Shortest wavelength of Paschen series (\( \lambda_s \)):
For the Paschen series, \( p = 3 \), and for the shortest wavelength (series limit), \( n = \infty \). \[ \frac{1}{\lambda_s} = R \left(\frac{1}{3^2} - \frac{1}{\infty^2}\right) = \frac{R}{9} \] \[ \lambda_s = \frac{9}{R} = \frac{9}{1.097 \times 10^7} = 8.204 \times 10^{-7}\text{ m} \text{ (or } 8204\text{ \AA)} \]

2. Longest wavelength of Balmer series (\( \lambda_l \)):
For the Balmer series, \( p = 2 \), and for the longest wavelength, \( n = 3 \). \[ \frac{1}{\lambda_l} = R \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36} \] \[ \lambda_l = \frac{36}{5R} = \frac{36}{5 \times 1.097 \times 10^7} = \frac{36}{5.485 \times 10^7} = 6.563 \times 10^{-7}\text{ m} \text{ (or } 6563\text{ \AA)} \]

The shortest wavelength of the Paschen series is \( 8.204 \times 10^{-7}\text{ m} \) (\( 8204\text{ \AA} \)) and the longest wavelength of the Balmer series is \( 6.563 \times 10^{-7}\text{ m} \) (\( 6563\text{ \AA} \)).

Q. 25. Calculate the current flowing through two long parallel wires carrying equal currents and separated by a distance of \( 1.35\text{ cm} \) experiencing a force per unit length of \( 4.76 \times 10^{-2}\text{ N/m} \).

Answer:

Given:
Distance between the wires (\( d \)) \( = 1.35\text{ cm} = 1.35 \times 10^{-2}\text{ m} \)
Force per unit length (\( \frac{F}{L} \)) \( = 4.76 \times 10^{-2}\text{ N/m} \)
Permeability of free space (\( \mu_0 \)) \( = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A} \)
Currents in the wires: \( I_1 = I_2 = I \)

Formula: \[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{\mu_0 I^2}{2\pi d} \]

Calculation: \[ 4.76 \times 10^{-2} = \frac{4\pi \times 10^{-7} \times I^2}{2\pi \times 1.35 \times 10^{-2}} = \frac{2 \times 10^{-7} \times I^2}{1.35 \times 10^{-2}} \] \[ I^2 = \frac{4.76 \times 10^{-2} \times 1.35 \times 10^{-2}}{2 \times 10^{-7}} = 3.213 \times 10^3 = 3213 \] \[ I = \sqrt{3213} = 56.68\text{ A} \]

The current flowing through each wire is \( 56.68\text{ A} \).

Q. 26. An emf of \( 90\text{ mV} \) is induced in the winding of a coil, when current in a nearby coil is increasing at the rate of \( 4.5\text{ A/s} \). What is the mutual inductance of the two coils?

Answer:

Given:
Induced emf (\( e \)) \( = 90\text{ mV} = 90 \times 10^{-3}\text{ V} \)
Rate of change of current (\( \frac{dI}{dt} \)) \( = 4.5\text{ A/s} \)

Formula: \[ e = M \frac{dI}{dt} \]

Calculation: \[ M = \frac{e}{dI / dt} = \frac{90 \times 10^{-3}}{4.5} = 20 \times 10^{-3}\text{ H} = 20\text{ mH} \]

The mutual inductance of the two coils is \( 20 \times 10^{-3}\text{ H} \) (or \( 20\text{ mH} \)).

SECTION - D

Attempt any THREE questions of the following:

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

Answer:

(Note to student: Draw a neat diagram of a cubical box of side \( L \) along the \( X, Y, \) and \( Z \) axes with a shaded wall perpendicular to the X-axis and a gas molecule moving with velocity \( \vec{v} \).)

Consider an ideal gas enclosed in a cubical container of side \( L \) and volume \( V = L^3 \). Let \( N \) be the total number of gas molecules, each of mass \( m \).

Let a molecule move with velocity \( v \) having components \( v_x, v_y, \) and \( v_z \) along the \( X, Y, \) and \( Z \) axes respectively. Therefore, \( v^2 = v_x^2 + v_y^2 + v_z^2 \).

Change in Momentum: When this molecule collides elastically with the shaded wall perpendicular to the X-axis, its X-component of velocity reverses to \( -v_x \) while \( v_y \) and \( v_z \) remain unchanged.
Initial momentum along X-axis \( = m v_x \)
Final momentum along X-axis \( = -m v_x \)
Change in momentum of the molecule (\( \Delta p_x \)) \( = -m v_x - m v_x = -2m v_x \)
Therefore, momentum imparted to the wall during one collision \( = +2m v_x \).
Time between successive collisions: The molecule travels a distance of \( 2L \) (to the opposite wall and back) before colliding with the same wall again. The time interval between two successive collisions with the same wall is: \[ \Delta t = \frac{2L}{v_x} \]
Average force exerted by one molecule: By Newton's second law, the average force exerted by this single molecule on the wall is: \[ f_{x1} = \frac{\Delta p_x}{\Delta t} = \frac{2m v_{x1}}{2L / v_{x1}} = \frac{m v_{x1}^2}{L} \]
Total force along X-axis: The total force exerted by all \( N \) molecules on the wall is: \[ F_x = \frac{m}{L} (v_{x1}^2 + v_{x2}^2 + \dots + v_{xN}^2) \]
Pressure along X-axis: Pressure is force per unit area (\( \text{Area} = L^2 \)): \[ P = \frac{F_x}{L^2} = \frac{m}{L^3} (v_{x1}^2 + v_{x2}^2 + \dots + v_{xN}^2) \] Since \( V = L^3 \) and mean square velocity component \( \langle v_x^2 \rangle = \frac{v_{x1}^2 + v_{x2}^2 + \dots + v_{xN}^2}{N} \): \[ P = \frac{Nm}{V} \langle v_x^2 \rangle \]
Symmetry of motion: Since the gas molecules move randomly in all directions, by symmetry: \[ \langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle \] Since \( \langle v^2 \rangle = \langle v_x^2 \rangle + \langle v_y^2 \rangle + \langle v_z^2 \rangle = 3\langle v_x^2 \rangle \), we get: \[ \langle v_x^2 \rangle = \frac{1}{3} \langle v^2 \rangle = \frac{1}{3} v_{\text{rms}}^2 \]
Final Expression: Substituting \( \langle v_x^2 \rangle \) into the pressure equation: \[ \mathbf{P = \frac{1}{3} \frac{Nm}{V} v_{\text{rms}}^2} \] Since total mass of the gas \( M = Nm \) and density \( \rho = \frac{M}{V} \): \[ \mathbf{P = \frac{1}{3} \rho v_{\text{rms}}^2} \]

Q. 28. State differential equation of linear S.H.M. hence obtain an expression for acceleration, velocity and displacement of a particle performing S.H.M.

Answer:

Differential Equation of Linear S.H.M.: \[ \mathbf{\frac{d^2x}{dt^2} + \omega^2 x = 0} \] (where \( x \) is the displacement from the mean position at time \( t \), and \( \omega = \sqrt{\frac{k}{m}} \) is the angular frequency).

1. Expression for Acceleration (\( a \)):
From the differential equation of linear S.H.M.: \[ \frac{d^2x}{dt^2} = -\omega^2 x \] Since \( \frac{d^2x}{dt^2} \) is the acceleration (\( a \)) of the particle: \[ \mathbf{a = -\omega^2 x} \] (The negative sign shows that acceleration is directed opposite to displacement).

2. Expression for Velocity (\( v \)):
Acceleration can be written as: \[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \] Substituting \( a = -\omega^2 x \): \[ v \frac{dv}{dx} = -\omega^2 x \implies v \, dv = -\omega^2 x \, dx \] Integrating both sides: \[ \int v \, dv = -\omega^2 \int x \, dx \] \[ \frac{v^2}{2} = -\frac{\omega^2 x^2}{2} + C \quad \text{... (Equation 1)} \] Where \( C \) is the constant of integration. At the extreme position, \( x = \pm A \) and \( v = 0 \): \[ 0 = -\frac{\omega^2 A^2}{2} + C \implies C = \frac{\omega^2 A^2}{2} \] Substituting the value of \( C \) in Equation 1: \[ \frac{v^2}{2} = -\frac{\omega^2 x^2}{2} + \frac{\omega^2 A^2}{2} \implies v^2 = \omega^2(A^2 - x^2) \] \[ \mathbf{v = \pm \omega \sqrt{A^2 - x^2}} \]

3. Expression for Displacement (\( x \)):
Since \( v = \frac{dx}{dt} \), we can write: \[ \frac{dx}{dt} = \omega \sqrt{A^2 - x^2} \implies \frac{dx}{\sqrt{A^2 - x^2}} = \omega \, dt \] Integrating both sides: \[ \int \frac{dx}{\sqrt{A^2 - x^2}} = \int \omega \, dt \] \[ \sin^{-1}\left(\frac{x}{A}\right) = \omega t + \phi \] (where \( \phi \) is the constant of integration representing the initial phase or epoch) \[ \frac{x}{A} = \sin(\omega t + \phi) \] \[ \mathbf{x = A \sin(\omega t + \phi)} \]

Q. 29. With the help of neat diagram explain construction of a transformer. Calculate energy required to build up current of \( 1\text{ A} \) in an inductor of inductance \( 20\text{ mH} \).

Answer:

Construction of a Transformer:

(Note to student: Draw a neat labeled diagram of a transformer showing the laminated soft iron core, primary coil \( P \) with input AC source, and secondary coil \( S \) with load impedance.)

A transformer consists of two sets of insulated copper coils, known as the primary coil (\( P \)) and the secondary coil (\( S \)), wound on a soft iron core.
Soft Iron Core: The core is constructed from thin rectangular laminated sheets (stampings) of soft iron insulated from each other by a layer of varnish. Soft iron offers high magnetic permeability and reduces hysteresis loss, while the laminations reduce eddy current losses.
Primary Coil (\( P \)): The coil to which the alternating input voltage source is connected is called the primary coil. Let \( N_p \) be the number of turns in the primary coil.
Secondary Coil (\( S \)): The coil across which the transformed alternating output voltage is obtained is called the secondary coil. Let \( N_s \) be the number of turns in the secondary coil.
The primary and secondary coils are electrically insulated from each other and from the iron core, but they are magnetically coupled through the core.

Numerical — Calculation of Energy in Inductor:
Given:
Inductance (\( L \)) \( = 20\text{ mH} = 20 \times 10^{-3}\text{ H} \)
Current (\( I \)) \( = 1\text{ A} \)

Formula:
Energy stored in the magnetic field of an inductor: \[ U = \frac{1}{2} L I^2 \]

Calculation: \[ U = \frac{1}{2} \times 20 \times 10^{-3} \times (1)^2 = 10 \times 10^{-3}\text{ J} = 10\text{ mJ} \]

The energy required to build up the current is \( 10 \times 10^{-3}\text{ J} \) (or \( 10\text{ mJ} \)).

Q. 30. Draw a neat schematic diagram of a photo cell. The photoelectric work function for a metal is \( 4.2\text{ eV} \). Find threshold wavelength.

Answer:

Schematic Diagram of a Photo Cell:
(Note to student: Draw a neat labeled schematic diagram of a photo cell showing and labeling: (1) Evacuated glass/quartz bulb, (2) Semi-cylindrical photosensitive cathode/emitter \( C \), (3) Wire anode/collector \( A \), (4) Incident light/UV radiation, and (5) External circuit containing a microammeter (\( \mu\text{A} \)) and DC source/battery.)

Numerical — Calculation of Threshold Wavelength (\( \lambda_0 \)):
Given:
Photoelectric work function (\( \phi_0 \)) \( = 4.2\text{ eV} = 4.2 \times 1.6 \times 10^{-19}\text{ J} = 6.72 \times 10^{-19}\text{ J} \)
Planck's constant (\( h \)) \( = 6.63 \times 10^{-34}\text{ J}\cdot\text{s} \)
Speed of light (\( c \)) \( = 3 \times 10^8\text{ m/s} \)

Formula: \[ \phi_0 = \frac{hc}{\lambda_0} \implies \lambda_0 = \frac{hc}{\phi_0} \]

Calculation: \[ \lambda_0 = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6.72 \times 10^{-19}} = \frac{19.89 \times 10^{-26}}{6.72 \times 10^{-19}} = 2.9598 \times 10^{-7}\text{ m} \approx 2.96 \times 10^{-7}\text{ m} \text{ (or } 2960\text{ \AA)} \]

The threshold wavelength for the metal is \( 2.96 \times 10^{-7}\text{ m} \) (or \( 2960\text{ \AA} \)).

Q. 31. Obtain an expression for magnetic induction of a toroid of \( N \) no. of turns about an axis passing through its centre and perpendicular to plane. A Carnot refrigerator operates between \( 150\text{ K} \) and \( 200\text{ K} \), calculate its coefficient of performance.

Answer:

Expression for Magnetic Induction of a Toroid:

(Note to student: Draw a neat diagram of a toroid carrying current \( I \) showing the concentric circular Amperian loop of radius \( r \) along its central axis and the tangential magnetic field \( \vec{B} \).)

A toroid is a hollow circular ring of finite thickness on which a large number (\( N \)) of insulated wire turns are closely wound.
Let \( I \) be the current flowing through the toroidal winding. To find the magnetic induction (\( B \)) inside the toroid at a distance \( r \) from its centre \( O \), consider a concentric circular Amperian loop of radius \( r \) passing through the centre of the turns.
By symmetry, the magnitude of the magnetic field \( B \) is the same at all points on the Amperian loop and is directed tangentially along the loop at every point (i.e., angle \( \theta \) between \( \vec{B} \) and \( d\vec{l} \) is \( 0^\circ \)).
Applying Ampere's Circuital Law to the Amperian loop: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}} \]
Here, \( \oint \vec{B} \cdot d\vec{l} = \oint B \, dl \cos 0^\circ = B \oint dl = B (2\pi r) \).
Since there are \( N \) turns and each turn carries current \( I \), the total current enclosed by the Amperian loop is: \[ I_{\text{enclosed}} = NI \]
Substituting these into Ampere's Circuital Law: \[ B (2\pi r) = \mu_0 NI \] \[ \mathbf{B = \frac{\mu_0 NI}{2\pi r}} \]
If \( n \) is the number of turns per unit length of the toroid (\( n = \frac{N}{2\pi r} \)), then: \[ \mathbf{B = \mu_0 n I} \]

Numerical — Coefficient of Performance of Carnot Refrigerator:
Given:
Temperature of cold reservoir (\( T_C \)) \( = 150\text{ K} \)
Temperature of hot reservoir (\( T_H \)) \( = 200\text{ K} \)

Formula:
Coefficient of performance: \[ K = \frac{T_C}{T_H - T_C} \]

Calculation: \[ K = \frac{150}{200 - 150} = \frac{150}{50} = 3 \]

The coefficient of performance of the Carnot refrigerator is \( 3 \).