What is wattless current in an AC circuit?

Wattless current is the current in an AC circuit for which the average power consumed is zero.

Maharashtra Board Class 12 Physics 2025

Q: Which phenomenon proves that light is a transverse wave?

Polarization proves that light is a transverse wave.

Q: Which physical quantity of light remains unchanged when it travels from one medium to another?

The frequency of light remains unchanged when it travels from one medium to another.

Q: What is the SI unit of electric flux?

The SI unit of electric flux is newton metre squared per coulomb (N m² C⁻¹).

Q: State Bohr’s quantization condition for angular momentum.

According to Bohr, the angular momentum of an electron is quantized as mvr = nh / 2π.

Maharashtra Board Class 12 Physics 2025 – Solved Paper

Published: Jan 02, 2026 •

Section A

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

Question 1 (i) “If two systems are each in thermal equilibrium with a third system, they are also in thermal equilibrium with each other.” This statement refers to:

(a) Zeroth law of thermodynamics
(b) First law of thermodynamics
(c) Second law of thermodynamics
(d) Carnot’s law

Answer:

The correct answer is (a) Zeroth law of thermodynamics.

Explanation (Step-by-step):

Thermal equilibrium means that there is no net heat flow between systems.
The given statement establishes a relationship of temperature between systems.
If system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then A and B must have the same temperature.
This fundamental idea forms the basis of temperature measurement.

Hence, the statement defines the Zeroth Law of Thermodynamics.

Question 1 (ii) In Bernoulli’s theorem, which of the following is constant?

(a) Mass
(b) Linear momentum
(c) Angular momentum
(d) Energy

Answer:

The correct answer is (d) Energy.

Explanation (Step-by-step):

Bernoulli’s theorem is based on the principle of conservation of energy.
It applies to the steady flow of an incompressible and non-viscous fluid.
According to Bernoulli’s equation:
$$ P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant} $$
Here, pressure energy, kinetic energy, and potential energy per unit volume together remain constant.

Thus, Bernoulli’s theorem states that the total mechanical energy of the fluid remains constant.

Question 1 (iii) Which of the following materials belongs to a diathermanous substance?

(a) Wood (b) Iron (c) Glass (d) Copper

Answer:

The correct answer is (c) Glass.

Explanation (Step-by-step):

A diathermanous substance allows heat (thermal radiation) to pass through it.
Glass permits the transmission of radiant heat.
Wood, iron, and copper either absorb or conduct heat rather than allowing radiation to pass freely.

Hence, glass is a diathermanous substance.

Question 1 (iv) Electric potential V at a distance r from a point charge is directly proportional to _____.

(a) $ r $ (b) $ r^2 $ (c) $ \dfrac{1}{r} $ (d) $ \dfrac{1}{r^2} $

Answer:

The correct answer is (c) $ \dfrac{1}{r} $.

Explanation (Step-by-step):

The electric potential due to a point charge is given by:
$$ V = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r} $$
From the formula, electric potential varies inversely with distance.

Therefore, electric potential is directly proportional to $ \dfrac{1}{r} $.

Question 1 (v) Which of the following equations gives the correct expression for the internal resistance of a cell using a potentiometer?

(a) $ r = R\left(\dfrac{V}{E}-1\right) $
(b) $ r = R\left(1-\dfrac{E}{V}\right) $
(c) $ r = R\left(1-\dfrac{V}{E}\right) $
(d) $ r = R\left(\dfrac{E}{V}-1\right) $

Answer:

The correct answer is (d) $ r = R\left(\dfrac{E}{V}-1\right) $.

Explanation (Step-by-step):

In a potentiometer experiment:
$ E $ = emf of the cell (open circuit)
$ V $ = terminal potential difference (closed circuit)
$ R $ = external resistance
The internal resistance is derived using the relation:
$$ r = R\left(\frac{E}{V}-1\right) $$

Thus, the correct expression for internal resistance is option (d).

Question (vi) An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de-Broglie wavelength will be longest for:

(a) Proton
(b) Electron
(c) Hydrogen atom
(d) α-particle

Answer: (b) Electron

Explanation:

De-Broglie wavelength:

\lambda = \frac{h}{\sqrt{2mK}}

For same kinetic energy $K$ :

Smaller mass → larger wavelength

Electron has the least mass, hence maximum wavelength.

Question (Vii) The gate which produces a high output when both inputs are high is:

(a) X-OR gate
(b) AND gate
(c) NAND gate
(d) NOR gate

Answer: (b) AND gate

Explanation:

Truth table of AND gate:

A	B	Output
0	0	0
0	1	0
1	0	0
1	1	1

Output is high only when both inputs are high.

Question 1 (viii) The power rating of a ceiling fan rotating with a constant torque of 2 Nm with an angular speed of 2 rad/s will be ____.

(a) 1 W
(b) 2 W
(c) 3 W
(d) 4 W

Answer: (d) 4 W

Explanation:

Power delivered by a rotating body is given by:

\[ P = \tau \omega \]

Where,
\[ \tau = 2 \, \text{Nm} \]
\[ \omega = 2 \, \text{rad s}^{-1} \]

Substituting the values:

\[ P = 2 \times 2 = 4 \, \text{W} \]

Hence, the correct answer is 4 W.

Question 1 (ix) A string of length 2 m is vibrating with 2 loops. The distance between its node and adjacent antinode is ____.

(a) 0.5 m
(b) 1.0 m
(c) 1.5 m
(d) 2.0 m

Answer: (a) 0.5 m

Explanation:

For a string fixed at both ends:

Each loop corresponds to half a wavelength:

\[ \text{Length of string} = n \times \frac{\lambda}{2} \]

Given,
\[ L = 2 \, \text{m}, \quad n = 2 \]

\[ 2 = 2 \times \frac{\lambda}{2} \Rightarrow \lambda = 2 \, \text{m} \]

Distance between a node and the adjacent antinode is:

\[ \frac{\lambda}{4} = \frac{2}{4} = 0.5 \, \text{m} \]

Hence, the correct answer is 0.5 m.

Question 1 (x) A transformer increases an alternating e.m.f. from 220 V to 880 V. If the primary coil has 1000 turns, the number of turns in the secondary coil are ____.

(a) 1000
(b) 2000
(c) 3000
(d) 4000

Answer: (d) 4000

Explanation:

For an ideal transformer:

\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]

Given,
\[ V_p = 220 \, \text{V}, \quad V_s = 880 \, \text{V} \]
\[ N_p = 1000 \]

\[ \frac{880}{220} = \frac{N_s}{1000} \]

\[ 4 = \frac{N_s}{1000} \Rightarrow N_s = 4000 \]

Hence, the correct answer is 4000 turns.

Question 2. Answer the following questions : [8]

Question 2 (i) At what temperature does the surface tension of a liquid become zero?

Answer:

The surface tension of a liquid becomes zero at its critical temperature.

Explanation:

At the critical temperature, the distinction between liquid and vapour phases disappears. Hence, intermolecular forces become negligible, and surface tension reduces to zero.

Question 2 (ii) Define self-inductance.

Answer:

Self-inductance is the property of a coil by virtue of which an emf is induced in it when the current flowing through it changes.

Explanation:

If a changing current produces a changing magnetic flux linked with the same coil, an opposing emf is induced. Mathematically,

\[ e = -L \frac{dI}{dt} \]

where $ L $ is the coefficient of self-inductance.

Question 2 (iii) What is the work done by an external uniform magnetic field perpendicular to the velocity of a moving charge?

Answer:

The work done is zero.

Explanation:

The magnetic force on a moving charge always acts perpendicular to its velocity. Since work done is given by

\[ W = \vec{F} \cdot \vec{s} \]

and the angle between force and displacement is $90^\circ$, we get

\[ W = Fs \cos 90^\circ = 0 \]

Question 2 (iv) What do you mean by a thermodynamic system?

Answer:

A thermodynamic system is a specified quantity of matter or a region in space chosen for thermodynamic study.

Explanation:

The system is separated from its surroundings by real or imaginary boundaries. Examples include a gas in a cylinder, water in a container, etc.

Question 2 (v) What is the value of B called when H = 0 in the hysteresis loop?

Answer:

The value of $ B $ when $ H = 0 $ is called retentivity or residual magnetism.

Explanation:

Retentivity indicates the ability of a magnetic material to retain magnetism after the magnetizing field is removed.

Question 2 (vi) State the formula for angle of banking.

Answer:

The formula for angle of banking is:

\[ \tan \theta = \frac{v^2}{rg} \]

Explanation:

where $ v $ = speed of the vehicle, $ r $ = radius of the curved road, $ g $ = acceleration due to gravity.

Question 2 (vii) Calculate the electric field intensity at a point just near the surface of a charged plane sheet, measured from its mid-point.
[ Surface charge density, $ \sigma = 8.85~\mu C/m^2 $ ]

Given:

Surface charge density, \[ \sigma = 8.85~\mu C/m^2 = 8.85 \times 10^{-6}~C/m^2 \] Permittivity of free space, \[ \varepsilon_0 = 8.85 \times 10^{-12}~C^2/Nm^2 \]

Formula:

Electric field intensity near a charged plane sheet is given by: \[ E = \frac{\sigma}{2\varepsilon_0} \]

Substitution:

\[ E = \frac{8.85 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \]

Calculation:

\[ E = \frac{10^{6}}{2} = 5 \times 10^{5}~N/C \]

Answer:

\[ \boxed{E = 5 \times 10^{5}~N/C} \]

Question 2 (viii) Find the kinetic energy of 1 litre of an ideal gas at S.T.P.

Given:

Volume of gas, \[ V = 1~\text{litre} = 10^{-3}~m^3 \] Pressure at S.T.P., \[ P = 1.013 \times 10^{5}~N/m^2 \]

Concept:

Total kinetic energy of an ideal gas is given by: \[ \text{K.E.} = \frac{3}{2}PV \]

Substitution:

\[ \text{K.E.} = \frac{3}{2} \times (1.013 \times 10^{5}) \times (10^{-3}) \]

Calculation:

\[ \text{K.E.} = \frac{3}{2} \times 101.3 = 151.95~J \]

Answer:

\[ \boxed{\text{Kinetic Energy} = 151.95~J \approx 152~J} \]

Section B

Attempt any EIGHT questions of the following : [16]

Q.3. Explain harmonics and overtones.

Answer:

When a stretched string, air column, or vibrating system produces sound, it does not vibrate with a single frequency. Instead, it vibrates with several frequencies simultaneously. These frequencies are classified as harmonics and overtones.

1) Harmonics:

Harmonics are frequencies that are integral multiples of the fundamental frequency. If the fundamental frequency is $f$ , then the harmonics are:

$$ f,\; 2f,\; 3f,\; 4f,\; \dots $$

The first harmonic is the fundamental frequency $f$ .
The second harmonic has frequency $2f$ .
The third harmonic has frequency $3f$ , and so on.

Harmonics are always whole-number multiples of the fundamental frequency.

2) Overtones:

Overtones are frequencies other than the fundamental frequency that are produced by a vibrating system.

The first overtone is the frequency next to the fundamental.
The second overtone is the frequency next to the first overtone.

Relation between harmonics and overtones:

$First harmonic =$ undamental frequency
$Second harmonic =$ First overtone
$Third harmonic =$ Second overtone

Thus, overtones may or may not be exact multiples of the fundamental frequency depending on the system.

Q.4. Using Newton’s law of viscosity for streamline flow, derive an expression for coefficient of viscosity.

Answer:

According to Newton’s law of viscosity, the tangential force required to maintain a steady relative motion between two parallel layers of a liquid is directly proportional to:

The area of the layers
The velocity gradient between the layers

Consider a liquid flowing in layers such that:

$A$ = area of each layer
$\Delta v$ = difference in velocity between two layers
$\Delta y$ = perpendicular distance between layers
$F$ = tangential force applied

According to Newton’s law of viscosity:

\[ F \propto A \frac{\Delta v}{\Delta y} \]

Introducing the constant of proportionality $\eta$, called the coefficient of viscosity:

\[ F = \eta A \frac{\Delta v}{\Delta y} \]

Rearranging the equation, we get:

\[ \eta = \frac{F}{A} \cdot \frac{\Delta y}{\Delta v} \]

Hence,

\[ \boxed{\eta = \frac{\text{Shearing stress}}{\text{Velocity gradient}}} \]

where $\frac{F}{A}$ is the shearing stress and $\frac{\Delta v}{\Delta y}$ is the velocity gradient.

SI Unit:

\text{Pa-s (Pascal-second)}

Conclusion:

The coefficient of viscosity measures the resistance offered by a liquid to the relative motion between its layers.

Q.4. Using Newton’s law of viscosity for streamline flow, derive an expression for coefficient of viscosity.

Answer:

According to Newton’s law of viscosity, the tangential force required to maintain a steady relative motion between two parallel layers of a liquid is directly proportional to:

The area of the layers
The velocity gradient between the layers

Consider a liquid flowing in layers such that:

$A$ = area of each layer
$\Delta v$ = difference in velocity between two layers
$\Delta y$ = perpendicular distance between layers
$F$ = tangential force applied

According to Newton’s law of viscosity:

\[ F \propto A \frac{\Delta v}{\Delta y} \]

Introducing the constant of proportionality $\eta$, called the coefficient of viscosity:

\[ F = \eta A \frac{\Delta v}{\Delta y} \]

Rearranging the equation, we get:

\[ \eta = \frac{F}{A} \cdot \frac{\Delta y}{\Delta v} \]

Hence,

\[ \boxed{\eta = \frac{\text{Shearing stress}}{\text{Velocity gradient}}} \]

where $\frac{F}{A}$ is the shearing stress and $\frac{\Delta v}{\Delta y}$ is the velocity gradient.

SI Unit:

\text{Pa-s (Pascal-second)}

Conclusion:

The coefficient of viscosity measures the resistance offered by a liquid to the relative motion between its layers.

Q.5 State the formula for magnetic induction produced by a current in a circular arc of a wire. Hence find the magnetic induction at the centre of a current carrying circular loop.

Answer:

According to the Biot–Savart law, the magnetic induction produced at the centre of a circular arc of radius $R$, carrying current $I$, and subtending an angle $\theta$ (in radians) at the centre is given by:

$$ B = \frac{\mu_0 I \theta}{4\pi R} $$

where:

$\mu_0 $ = permeability of free space
$I$ = current in the wire
$R$ = radius of the circular arc
$\theta$ = angle subtended by the arc at the centre (in radians)

Magnetic induction at the centre of a complete circular loop:

For a complete circular loop, $\theta = 2\pi$. Substituting this value in the above formula:

$$ B = \frac{\mu_0 I (2\pi)}{4\pi R} $$

Simplifying,

$$ B = \frac{\mu_0 I}{2R} $$

Thus, the magnetic induction at the centre of a current carrying circular loop is:

$$ \boxed{B = \frac{\mu_0 I}{2R}} $$

Q.6 State and prove the law of conservation of angular momentum.

Answer:

Statement:

The law of conservation of angular momentum states that:

If the net external torque acting on a system is zero, then the total angular momentum of the system remains constant.

Mathematically,

$$ \text{If } \tau_{\text{ext}} = 0 \Rightarrow L = \text{constant} $$

Proof:

Angular momentum $L$ of a particle is defined as:

$$ \vec{L} = \vec{r} \times \vec{p} $$

where $\vec{r}$ is position vector and $\vec{p}$ is linear momentum.

The rate of change of angular momentum is given by:

$$ \frac{d\vec{L}}{dt} = \vec{\tau} $$

where $\vec{\tau}$ is the external torque acting on the system.

If no external torque acts on the system, then:

$$ \vec{\tau}_{\text{ext}} = 0 $$

Hence,

$$ \frac{d\vec{L}}{dt} = 0 $$

Integrating with respect to time,

$$ \boxed{\vec{L} = \text{constant}} $$

Thus, the angular momentum of the system remains conserved when no external torque acts on it.

Hence proved.

Q.7 State any four advantages of Light Emitting Diode (LED).

Answer:

The advantages of Light Emitting Diodes (LEDs) are:

Low Power Consumption: LEDs consume very less electrical power compared to conventional bulbs.
High Efficiency: A large fraction of electrical energy is converted into light energy.
Long Life: LEDs have a very long operational life (typically more than 50,000 hours).
Instant Response: LEDs glow immediately when switched on and do not require warm-up time.

Q.8 Calculate the energy radiated in one minute by a perfectly black body of surface area 200 cm² when it is maintained at 127 °C.

Answer:

Given:

Surface area, $$ \A = 200\ \text{cm}^2 = 200 \times 10^{-4} = 0.02\ \text{m}^2 $$

Temperature, $$ T = 127^\circ C = (127 + 273)\ \text{K} = 400\ \text{K} $$

Time, $$ t = 1\ \text{minute} = 60\ \text{s} $$

Stefan–Boltzmann constant, $$ \sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4} $$

Formula used:

Energy radiated by a perfectly black body is given by $$ E = \sigma A T^4 t $$

Substituting the values:

$$ E = (5.67 \times 10^{-8}) \times (0.02) \times (400)^4 \times (60) $$

$$ E = (5.67 \times 10^{-8}) \times 0.02 \times 2.56 \times 10^{10} \times 60 $$

$$ E = 1740\ \text{J} $$

Final Answer:

$$ \boxed{E = 1.74 \times 10^3\ \text{J}} $$

Q.9 Two coils having self inductances 60 mH each are coupled with each other. If the coefficient of coupling is 0.75, calculate the mutual inductance between them.

Answer:

Given:

Self inductance of first coil, $$ L_1 = 60\ \text{mH} = 60 \times 10^{-3}\ \text{H} $$

Self inductance of second coil, $$ L_2 = 60\ \text{mH} = 60 \times 10^{-3}\ \text{H} $$

Coefficient of coupling, $$ k = 0.75 $$

Formula used:

Mutual inductance is given by $$ M = k \sqrt{L_1 L_2} $$

Substituting the values:

$$ M = 0.75 \sqrt{(60 \times 10^{-3})(60 \times 10^{-3})} $$

$$ M = 0.75 \times 60 \times 10^{-3} $$

$$ M = 45 \times 10^{-3}\ \text{H} $$

Final Answer:

$$ \boxed{M = 45\ \text{mH}} $$

Q.10 In a series LCR circuit, if resistance, inductive reactance and capacitive reactance are 3 Ω, 8 Ω and 4 Ω respectively, calculate the phase difference between voltage and current.

Answer:

Given:

Resistance, $ R = 3\,\Omega $
Inductive reactance, $ X_L = 8\,\Omega $
Capacitive reactance, $ X_C = 4\,\Omega $

Step 1: Calculate net reactance

\[ X = X_L - X_C = 8 - 4 = 4\,\Omega \]

Step 2: Use the formula for phase difference

\[ \tan \phi = \frac{X}{R} \]

\[ \tan \phi = \frac{4}{3} \]

Step 3: Calculate phase angle

\[ \phi = \tan^{-1}\left(\frac{4}{3}\right) \]

\[ \phi \approx 53.1^\circ \]

Conclusion:

The phase difference between voltage and current is approximately 53.1°.

Since $ X_L > X_C $, the circuit is inductive and current lags behind voltage.

Q.11 State the advantages of a potentiometer over a voltmeter.

Answer:

Explanation:

A potentiometer is a null-deflection instrument used to measure potential difference accurately, whereas a voltmeter draws current from the circuit.

Advantages:

No current is drawn from the circuit:
A potentiometer works on the null method, so it does not disturb the circuit under measurement.
More accurate than a voltmeter:
Since no current flows through the potentiometer during measurement, errors due to internal resistance are eliminated.
Can measure very small potential differences:
Potentiometers are highly sensitive and suitable for precise measurements.
Used to compare EMFs of cells:
A voltmeter cannot compare EMFs accurately, whereas a potentiometer can.
Can determine internal resistance of a cell:
This is not possible using a simple voltmeter.

Conclusion:

A potentiometer is preferred over a voltmeter for precision measurements due to its high accuracy and non-loading effect on the circuit.

Q.13 A body of mass 0.8 kg performs linear simple harmonic motion. It experiences a restoring force of 0.4 N when its displacement from the mean position is 4 cm. Determine:

Force constant of the SHM
Time period of SHM

Answer:

Given:

Mass of the body, $ m = 0.8 \, \text{kg} $
Restoring force, $ F = 0.4 \, \text{N} $
Displacement from mean position, $ x = 4 \, \text{cm} = 0.04 \, \text{m} $

Step 1: Find the Force Constant

In simple harmonic motion, the restoring force is given by:

\[ F = kx \]

Where $ k $ is the force constant.

Rearranging the formula:

\[ k = \frac{F}{x} \]

Substituting the given values:

\[ k = \frac{0.4}{0.04} \] \[ k = 10 \, \text{N m}^{-1} \]

Force constant of the SHM is: \[ \boxed{k = 10 \, \text{N m}^{-1}} \]

Step 2: Find the Time Period of SHM

The time period of a body performing linear SHM is given by:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

Substituting the known values:

\[ T = 2\pi \sqrt{\frac{0.8}{10}} \] \[ T = 2\pi \sqrt{0.08} \] \[ T = 2\pi \times 0.283 \] \[ T \approx 1.78 \, \text{s} \]

Time period of SHM is: \[ \boxed{T \approx 1.78 \, \text{s}} \]

Final Answers:

Force constant, $ k = 10 \, \text{N m}^{-1} $
Time period, $ T \approx 1.78 \, \text{s} $

Exam Tip:
Always convert displacement into meters before substituting values in SHM formulas. Use $ F = kx $ for force constant and $ T = 2\pi\sqrt{\frac{m}{k}} $ for time period.

Q.14 A gas of 0.5 mole at 300 K expands isothermally from an initial volume of 2.0 litre to a final volume of 6.0 litre. What is the work done by gas?

Answer:

For an isothermal expansion of an ideal gas, the work done is given by:

\[ W = nRT \ln\!\left(\frac{V_2}{V_1}\right) \]

Step 1: Write the given data

\[ n = 0.5 \text{ mole} \] \[ T = 300 \, \text{K} \] \[ V_1 = 2.0 \, \text{L} \] \[ V_2 = 6.0 \, \text{L} \] \[ R = 8.314 \, \text{J mol}^{-1}\text{K}^{-1} \]

Step 2: Substitute the values in the formula

\[ W = (0.5)(8.314)(300)\ln\!\left(\frac{6}{2}\right) \]

\[ W = (0.5)(8.314)(300)\ln(3) \]

Step 3: Calculate numerical values

\[ \ln(3) \approx 1.098 \]

\[ W = (0.5)(8.314)(300)(1.098) \]

\[ W = (4.157)(300)(1.098) \]

\[ W \approx 1369 \, \text{J} \]

\[ W \approx 1.37 \times 10^3 \, \text{J} \]

Final Answer:

\[ \boxed{W \approx 1370 \, \text{J}} \]

Hence, the correct option is (C) 1370 J.

Explanation:

In an isothermal process, the temperature remains constant. The work done depends on the ratio of final and initial volumes, not on pressure directly. Since the gas expands ($V_2 > V_1$), the work done by the gas is positive.

Section C

Attempt any EIGHT questions of the following : [24]

Q.15 Obtain an expression for the period of a bar magnet vibrating in a uniform magnetic field, performing S.H.M.

Answer:

Consider a bar magnet of magnetic moment M and moment of inertia I, placed in a uniform magnetic field of magnetic induction B. Let the magnet be slightly displaced through a small angle θ from its equilibrium position.

Step 1: Torque acting on the magnet

When the magnet is deflected, a restoring torque acts on it which tends to bring it back to its original position. The magnitude of the torque is given by:

$$ \tau = MB \sin \theta $$

For small angular displacement,

$$ \sin \theta \approx \theta $$

Hence, the restoring torque becomes:

$$ \tau = MB \theta $$

Step 2: Relation between torque and angular acceleration

Torque is also related to angular acceleration by:

$$ \tau = I \alpha $$

where angular acceleration,

$$ \alpha = \frac{d^2 \theta}{dt^2} $$

Since the torque is restoring in nature, it acts opposite to the displacement:

$$ I \frac{d^2 \theta}{dt^2} = - MB \theta $$

Step 3: Differential equation of motion

Rewriting the equation:

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

This equation is of the standard form of angular simple harmonic motion:

$$ \frac{d^2 \theta}{dt^2} + \omega^2 \theta = 0 $$

Comparing, we get:

$$ \omega^2 = \frac{MB}{I} $$

Step 4: Expression for time period

The time period of SHM is given by:

$$ T = \frac{2\pi}{\omega} $$

Substituting the value of ω:

$$ T = 2\pi \sqrt{\frac{I}{MB}} $$

Thus, the period of oscillation of a bar magnet vibrating in a uniform magnetic field is:

$$ \boxed{T = 2\pi \sqrt{\frac{I}{MB}}} $$

Q.16 In a thermodynamic system, define:

(a) Mechanical equilibrium

(b) Chemical equilibrium

(c) Thermal equilibrium

Answer:

(a) Mechanical Equilibrium:

A thermodynamic system is said to be in mechanical equilibrium if there is no unbalanced force acting within the system or between the system and its surroundings.

This condition is achieved when the pressure is uniform throughout the system and remains constant with time.

(b) Chemical Equilibrium:

A system is in chemical equilibrium when there is no net chemical reaction taking place within the system.

The chemical composition of the system remains constant, and there is no transfer of matter between the system and its surroundings.

(c) Thermal Equilibrium:

A system is said to be in thermal equilibrium if there is no exchange of heat between the system and its surroundings.

This occurs when the temperature of the system is uniform and remains constant with time.

Q.17 What is Brewster’s law? Derive the formula for Brewster’s angle.

Answer:

Brewster’s Law:

Brewster’s law states that when unpolarized light is incident on a transparent surface at a particular angle (called Brewster’s angle), the reflected light is completely plane polarized.

According to Brewster’s law:

$$ \mu = \tan i_B $$

where,
$$ \mu = \text{refractive index of the medium} $$ $$ i_B = \text{Brewster’s angle} $$

Derivation:

Let a ray of unpolarized light be incident at angle $$i_B$$ on a transparent medium of refractive index $$\mu$$.

At Brewster’s angle, the reflected ray and refracted ray are perpendicular to each other.

Therefore,

$$ i_B + r = 90^\circ $$

Using Snell’s law,

$$ \mu = \frac{\sin i_B}{\sin r} $$

Substitute $$r = 90^\circ - i_B$$,

$$ \mu = \frac{\sin i_B}{\sin (90^\circ - i_B)} $$

Since,

$$ \sin (90^\circ - i_B) = \cos i_B $$

We get,

$$ \mu = \frac{\sin i_B}{\cos i_B} $$ $$ \boxed{\mu = \tan i_B} $$

This is the required expression for Brewster’s angle.

Q.18 Derive an expression for the law of radioactive decay. Define one becquerel (Bq).

Answer:

Law of Radioactive Decay:

The law of radioactive decay states that the rate of disintegration of radioactive nuclei at any instant is directly proportional to the number of undecayed nuclei present at that instant.

Let,

$$ N = \text{number of undecayed nuclei at time } t $$ $$ \frac{dN}{dt} = \text{rate of decay} $$

According to the law,

$$ \frac{dN}{dt} \propto N $$

Introducing proportionality constant $$\lambda$$ (decay constant),

$$ \frac{dN}{dt} = -\lambda N $$

Negative sign indicates decrease in number of nuclei with time.

Rearranging,

$$ \frac{dN}{N} = -\lambda dt $$

Integrating both sides,

$$ \int \frac{dN}{N} = -\lambda \int dt $$ $$ \ln N = -\lambda t + C $$

Let at $$t = 0$$, number of nuclei $$= N_0$$.

$$ \ln N_0 = C $$

Substituting value of $$C$$,

$$ \ln N = \ln N_0 - \lambda t $$

Taking antilog,

$$ \boxed{N = N_0 e^{-\lambda t}} $$

This is the mathematical expression for the law of radioactive decay.

Definition of Becquerel (Bq):

One becquerel is defined as the activity of a radioactive substance in which one nucleus decays per second.

$$ \boxed{1 \text{ Bq } = 1 \text{ disintegration per second}} $$

Q.19 State and prove Kirchhoff’s law of heat radiation.

Answer:

Statement of Kirchhoff’s Law of Heat Radiation:

At a given temperature and for a given wavelength, the ratio of the emissive power (E) of a body to its absorptive power (a) is the same for all bodies and is equal to the emissive power of a perfectly black body (E_b) at that temperature.

Mathematically,

$$ \frac{E}{a} = E_b $$

or,

$$ E = a \, E_b $$

This law shows that a good absorber is also a good emitter.

Proof:

Consider two bodies A and B placed inside an enclosure maintained at the same uniform temperature T.

Let:

Emissive power of A = $E_1$
Absorptive power of A = $a_1$
Emissive power of B = $E_2$
Absorptive power of B = $a_2$

Since both bodies are at thermal equilibrium, the heat absorbed by one body from the other must be equal to the heat emitted by it.

Heat absorbed by A from B:

$$ = a_1 E_2 $$

Heat absorbed by B from A:

$$ = a_2 E_1 $$

At thermal equilibrium,

$$ a_1 E_2 = a_2 E_1 $$

Rearranging,

$$ \frac{E_1}{a_1} = \frac{E_2}{a_2} $$

Since this ratio is the same for any body, it must be equal to that of a perfectly black body:

$$ \frac{E}{a} = E_b $$

Hence, Kirchhoff’s law of heat radiation is proved.

Q.20 Obtain an expression for practical determination of end correction (i) for a pipe open at both ends and (ii) for a pipe closed at one end.

Answer:

End Correction:

In air columns, the antinode is formed slightly outside the open end of the pipe. This additional length added to the actual length of the pipe is called end correction.

Let end correction = $$e$$.

(i) Pipe open at both ends

Let:

Length of pipe = $$L$$
Velocity of sound in air = $$v$$
Frequency of fundamental note = $$f$$

For an open pipe, antinodes are formed at both ends. Effective length of the air column is:

$$ L_{\text{eff}} = L + 2e $$

For fundamental mode:

$$ L_{\text{eff}} = \frac{\lambda}{2} $$

So,

$$ \lambda = 2(L + 2e) $$

Using relation $$v = f\lambda$$,

$$ v = f \cdot 2(L + 2e) $$

Rearranging,

$$ L + 2e = \frac{v}{2f} $$

Hence, end correction for an open pipe:

$$ e = \frac{1}{2}\left(\frac{v}{2f} - L\right) $$

(ii) Pipe closed at one end

Let:

Length of pipe = $$L$$
Velocity of sound in air = $$v$$
Frequency of fundamental note = $$f$$

In a closed pipe, one end is a node and the other is an antinode. Effective length of the air column is:

$$ L_{\text{eff}} = L + e $$

For fundamental mode:

$$ L_{\text{eff}} = \frac{\lambda}{4} $$

So,

$$ \lambda = 4(L + e) $$

Using $$v = f\lambda$$,

$$ v = f \cdot 4(L + e) $$

Rearranging,

$$ L + e = \frac{v}{4f} $$

Hence, end correction for a closed pipe:

$$ e = \frac{v}{4f} - L $$

These expressions are used for the practical determination of end correction in resonance tube experiments.

Q.21 A conducting bar is rotating with constant angular speed around a pivot at one end in a uniform magnetic field perpendicular to its plane of rotation. Obtain an expression for the rotational e.m.f. induced between the ends of the bar.

Answer:

Given:

Length of the conducting bar = l
Angular speed of rotation = ω
Uniform magnetic field = B (perpendicular to plane of rotation)

Step 1: Consider a small element of the rod

Let us consider a small element of length dx at a distance x from the pivot.

Step 2: Velocity of the element

The linear velocity of this element is:

$$ v = \omega x $$

Step 3: Induced e.m.f. in the small element

The induced e.m.f. across the small length dx is:

$$ d\varepsilon = B \, v \, dx $$

Substituting the value of v:

$$ d\varepsilon = B \, \omega x \, dx $$

Step 4: Total induced e.m.f.

Integrate from x = 0 to x = l:

$$ \varepsilon = \int_0^l B \omega x \, dx $$ $$ \varepsilon = B \omega \int_0^l x \, dx $$ $$ \varepsilon = B \omega \left[ \frac{x^2}{2} \right]_0^l $$ $$ \varepsilon = \frac{1}{2} B \omega l^2 $$

Final Expression:

$$ \boxed{\varepsilon = \frac{1}{2} B \omega l^2} $$

Q.22 An electron in hydrogen atom stays in its second orbit for $10^{-8}\,\text{s}$. How many revolutions will it make around the nucleus in that time?
[ Given: $e = 1.6 \times 10^{-19}\,\text{C},\; m = 9.1 \times 10^{-31}\,\text{kg}$ ]

Answer:

Step 1: Formula for time period of electron in nth orbit

From Bohr’s atomic model, the time period of an electron in the n^th orbit is:

$$ T_n = \frac{2\pi r_n}{v_n} $$

Using Bohr’s expressions, the time period can be written as:

$$ T_n = \frac{2\pi \varepsilon_0 h^3 n^3}{\pi m e^4} $$

Hence,

$$ T_n \propto n^3 $$

Step 2: Time period of electron in second orbit

For hydrogen atom, the time period of the first orbit is:

$$ T_1 = 1.52 \times 10^{-16} \, \text{s} $$

For second orbit:

$$ T_2 = T_1 \times 2^3 $$ $$ T_2 = 1.52 \times 10^{-16} \times 8 $$ $$ T_2 = 1.216 \times 10^{-15} \, \text{s} $$

Step 3: Number of revolutions

Time spent by electron in second orbit:

$$ t = 10^{-8} \, \text{s} $$

Number of revolutions:

$$ N = \frac{t}{T_2} $$ $$ N = \frac{10^{-8}}{1.216 \times 10^{-15}} $$ $$ N \approx 8.2 \times 10^{6} $$

Final Answer:

$$ \boxed{N \approx 8.2 \times 10^{6} \text{ revolutions}} $$

Question 23 A flywheel of a motor has mass 100 kg and radius 1.5 m. The motor develops a constant torque of 2000 Nm. The flywheel starts rotating from rest. Calculate the work done during the first 4 revolutions.

Answer:

Given:

$$ \text{Mass of flywheel, } m = 100 \text{ kg} $$ $$ \text{Radius of flywheel, } r = 1.5 \text{ m} $$ $$ \text{Torque applied, } \tau = 2000 \text{ Nm} $$ $$ \text{Number of revolutions, } n = 4 $$

Step 1: Convert revolutions into angular displacement

One complete revolution corresponds to an angular displacement of:

$$ 2\pi \text{ radians} $$

For 4 revolutions:

$$ \theta = 4 \times 2\pi = 8\pi \text{ radians} $$

Step 2: Use formula for work done by torque

Work done by a constant torque is given by:

$$ W = \tau \theta $$

Step 3: Substitute values

$$ W = 2000 \times 8\pi $$ $$ W = 16000\pi \text{ J} $$

Step 4: Numerical value

$$ W = 16000 \times 3.14 $$ $$ W = 50240 \text{ J} $$

Final Answer:

$$ \boxed{W = 5.02 \times 10^4 \text{ J}} $$

Question 24 A galvanometer has a resistance of 50 Ω and a current of 2 mA is needed for its full scale deflection. Calculate the resistance required to convert it:
(i) into an ammeter of 0.5 A range
(ii) into a voltmeter of 10 V range

Answer:

Given:

$$ \text{Galvanometer resistance, } G = 50 \, \Omega $$ $$ \text{Full scale deflection current, } I_g = 2 \text{ mA } = 0.002 \text{ A} $$

(i) Conversion into an Ammeter of range 0.5 A

Step 1: Calculate shunt current

$$ I = 0.5 \text{ A} $$ $$ I_s = I - I_g $$ $$ I_s = 0.5 - 0.002 = 0.498 \text{ A} $$

Step 2: Use shunt resistance formula

$$ S = \frac{I_g G}{I_s} $$

Step 3: Substitute values

$$ S = \frac{0.002 \times 50}{0.498} $$ $$ S = \frac{0.1}{0.498} $$ $$ S \approx 0.20 \, \Omega $$

Final Answer (Ammeter):

$$ \boxed{S \approx 0.20 \, \Omega} $$

(ii) Conversion into a Voltmeter of range 10 V

Step 1: Use series resistance formula

$$ R = \frac{V}{I_g} - G $$

Step 2: Substitute values

$$ R = \frac{10}{0.002} - 50 $$ $$ R = 5000 - 50 $$ $$ R = 4950 \, \Omega $$

Final Answer (Voltmeter):

$$ \boxed{R = 4950 \, \Omega} $$

Q.25 Diameter of a water drop is 0.6 mm. Calculate the pressure inside a liquid drop. (T = 72 dyne/cm, atmospheric pressure = 1.013 × 10⁵ N/m²)

Answer:

Given:

Diameter of water drop, $$ d = 0.6 \text{ mm} = 0.6 \times 10^{-3} \text{ m} $$
Radius of drop, $$ r = \frac{d}{2} = \frac{0.6 \times 10^{-3}}{2} = 0.3 \times 10^{-3} \text{ m} $$
Surface tension, $$ T = 72 \text{ dyne/cm} $$

Unit conversion:

$$ 1 \text{ dyne/cm} = 10^{-3} \text{ N/m} $$ $$ T = 72 \times 10^{-3} = 0.072 \text{ N/m} $$

Formula:

Excess pressure inside a liquid drop is given by:

$$ \Delta P = \frac{2T}{r} $$

Substituting values:

$$ \Delta P = \frac{2 \times 0.072}{0.3 \times 10^{-3}} $$ $$ \Delta P = \frac{0.144}{0.0003} $$ $$ \Delta P = 480 \text{ N/m}^2 $$

Total pressure inside the drop:

$$ P = P_{\text{atmospheric}} + \Delta P $$ $$ P = 1.013 \times 10^{5} + 480 $$ $$ P = 1.0178 \times 10^{5} \text{ N/m}^2 $$

Final Answer:

$$ \boxed{P = 1.018 \times 10^{5} \text{ N/m}^2} $$

Q.26 A solenoid of length 1 m and 5 cm in diameter has a winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its centre along the axis.

Answer:

Given:

Length of solenoid, $$ l = 1 \text{ m} $$
Number of turns, $$ N = 1000 $$
Current through solenoid, $$ I = 5 \text{ A} $$
Permeability of free space, $$ \mu_0 = 4\pi \times 10^{-7} \text{ T m/A} $$

Formula:

Magnetic field at the centre of a long solenoid is:

$$ B = \mu_0 n I $$

where number of turns per unit length,

$$ n = \frac{N}{l} $$

Calculation:

$$ n = \frac{1000}{1} = 1000 \text{ turns/m} $$ $$ B = (4\pi \times 10^{-7}) \times 1000 \times 5 $$ $$ B = 20\pi \times 10^{-4} $$ $$ B = 2\pi \times 10^{-3} \text{ T} $$

Numerical value:

$$ B = 6.28 \times 10^{-3} \text{ T} $$

Final Answer:

$$ \boxed{B = 6.28 \times 10^{-3} \text{ Tesla}} $$

Section D

Attempt any THREE questions of the following : [12]

Q.27 What is Ferromagnetism? Explain it on the basis of domain theory.

Answer:

Ferromagnetism is the property of certain materials like iron, cobalt and nickel by which they become strongly magnetized even in the presence of a weak external magnetic field and retain magnetism after the field is removed.

Ferromagnetic substances show very large magnetic susceptibility and relative permeability.

Explanation based on Domain Theory:

According to Weiss Domain Theory, a ferromagnetic material is divided into a large number of small regions called magnetic domains.

Each domain is magnetized to saturation due to the alignment of atomic magnetic moments.
Within a domain, all magnetic moments are aligned in the same direction.
In the absence of an external magnetic field, the domains are randomly oriented.

Hence, the net magnetization of the specimen is zero.

When an external magnetic field is applied:

Domains oriented parallel to the field grow in size.
Domains opposite to the field shrink.
Magnetic moments rotate towards the direction of the applied field.

As a result, strong magnetization is produced.

When the magnetic field is removed, some domains remain aligned, resulting in residual magnetism.

This explains permanent magnetism in ferromagnetic materials.

Q.28 Obtain an expression for average power dissipated in a series LCR circuit.

Answer:

Consider a series LCR circuit connected to an alternating voltage source.

Given:

Applied voltage, $$ v = V_0 \sin \omega t $$

Instantaneous current, $$ i = I_0 \sin (\omega t - \phi) $$

where $\phi$ is the phase difference between voltage and current.

Instantaneous Power:

Instantaneous power is given by: $$ p = vi $$

Substituting values, $$ p = V_0 I_0 \sin \omega t \sin (\omega t - \phi) $$

Using trigonometric identity, $$ \sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)] $$

$$ p = \frac{V_0 I_0}{2} [\cos \phi - \cos (2\omega t - \phi)] $$

Average Power:

Over one complete cycle, the average value of $\cos (2\omega t - \phi)$ is zero.

Hence, average power, $$ P_{\text{avg}} = \frac{V_0 I_0}{2} \cos \phi $$

Using RMS values, $$ V = \frac{V_0}{\sqrt{2}}, \quad I = \frac{I_0}{\sqrt{2}} $$

$$ P_{\text{avg}} = VI \cos \phi $$

In a series LCR circuit, $$ \cos \phi = \frac{R}{Z} $$

where $Z$ is the impedance of the circuit.

$$ Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} $$

Therefore, $$ P_{\text{avg}} = I^2 R $$

This shows that average power is dissipated only in the resistor.

Q.29 (a) Distinguish between interference and diffraction of light.

Answer:

Interference	Diffraction
Interference is due to superposition of light waves coming from two or more coherent sources.	Diffraction is due to bending and spreading of light waves from different parts of the same wavefront.
It requires two coherent sources (like double slit).	It requires a single slit or obstacle.
Fringes are equally spaced.	Fringes are not equally spaced.
All bright fringes have nearly equal intensity.	Central maximum is brightest and widest.
Fringe width depends on slit separation.	Fringe width depends on slit width.

Q.29 (b) A double slit arrangement produces interference fringes for sodium light of wavelength 589 nm, that are 0.20° apart. What is the angular fringe separation if the entire arrangement is immersed in water? (Refractive index of water = 1.33)

Answer:

Given:

Wavelength in air, $$ \lambda = 589 \text{ nm} $$

Angular fringe separation in air, $$ \theta_{\text{air}} = 0.20^\circ $$

Refractive index of water, $$ \mu = 1.33 $$

Concept Used:

Angular fringe separation is directly proportional to the wavelength of light: $$ \theta \propto \lambda $$

When light enters a medium of refractive index $\mu$, its wavelength becomes: $$ \lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{\mu} $$

Hence, angular fringe separation in water is: $$ \theta_{\text{water}} = \frac{\theta_{\text{air}}}{\mu} $$

Calculation:

$$ \theta_{\text{water}} = \frac{0.20^\circ}{1.33} $$

$$ \theta_{\text{water}} = 0.150^\circ \; (\text{approximately}) $$

Final Answer:

The angular fringe separation when the arrangement is immersed in water is: $$ \boxed{0.15^\circ} $$

Question 30

Q.30 State Einstein’s photoelectric equation and mention the physical significance of each term involved in it. The wavelength of incident light is 4000 Å. Calculate the energy of the incident photon.

Answer:

Einstein’s Photoelectric Equation:

According to Einstein, when light of suitable frequency falls on a metal surface, electrons are emitted. The energy of an incident photon is used in two ways:

To overcome the work function of the metal
To provide kinetic energy to the emitted photoelectrons

The mathematical form of Einstein’s photoelectric equation is:

$$ h\nu = \phi + K_{\text{max}} $$

Physical Significance of Each Term:

$h\nu = $ Energy of the incident photon, where $h$ is Planck’s constant and $\nu$ is the frequency of incident radiation.
$\phi = $ Work function of the metal. It is the minimum energy required to remove an electron from the metal surface.
$K_{\text{max}} = $ Maximum kinetic energy of the emitted photoelectrons.

Calculation of Energy of Incident Photon:

Given:

$$ \lambda = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} $$

We know the relation between energy and wavelength:

$$ E = \frac{hc}{\lambda} $$

Where:

$ h = 6.63 \times 10^{-34} \, \text{J s} $
$ c = 3 \times 10^{8} \, \text{m/s}$

Substituting the values:

$$ E = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-10}} $$ $$ E = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}} $$ $$ E = 4.97 \times 10^{-19} \, \text{J} $$

Final Answer:

Energy of the incident photon is:

$$ \boxed{E = 4.97 \times 10^{-19} \, \text{J}} $$

Question 31
(a) State any four uses of Van de Graaff generator.
(b) In a parallel plate air capacitor, intensity of electric field is changing at the rate of $2 \times 10^{11}\ \text{V/m-s}.$ If the area of each plate is $20\ \text{cm}^2,$ calculate the displacement current.

Answer:

(a) Uses of Van de Graaff Generator:

It is used to accelerate charged particles such as protons and ions to very high energies.
It is used in nuclear physics experiments for studying nuclear reactions.
It is used to produce high-energy X-rays and gamma rays.
It is used for testing the insulating properties of materials under high voltage.

(b) Calculation of Displacement Current:

Displacement current is given by the formula:

$$ I_d = \varepsilon_0 \frac{d\Phi_E}{dt} $$

For a parallel plate capacitor, electric flux $$\Phi_E = EA$$

Hence,

$$ I_d = \varepsilon_0 A \frac{dE}{dt} $$

Given:

$$ \varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\text{/N-m}^2 $$ $$ \frac{dE}{dt} = 2 \times 10^{11}\ \text{V/m-s} $$ $$ A = 20\ \text{cm}^2 = 20 \times 10^{-4}\ \text{m}^2 = 2 \times 10^{-3}\ \text{m}^2 $$

Substituting values:

$$ I_d = (8.85 \times 10^{-12}) \times (2 \times 10^{-3}) \times (2 \times 10^{11}) $$ $$ I_d = 35.4 \times 10^{-4} $$ $$ I_d = 3.54 \times 10^{-3}\ \text{A} $$

Final Answer:

$$ \boxed{I_d = 3.54\ \text{mA}} $$

✦✦✦ End of Solution ✦✦✦

Key Takeaways:

Conceptual clarity is more important than memorization: The Maharashtra Board emphasizes understanding of physical concepts, especially in derivations, numericals, and reasoning-based MCQs.
Formulas must be supported with proper derivation steps: Writing only the final formula fetches fewer marks. Always show intermediate steps and substitutions clearly.
SI units and symbols are compulsory: Incorrect or missing units can lead to loss of marks even if the numerical answer is correct.
MCQs test concept application, not definitions: Most MCQs involve elimination of incorrect options using physics principles rather than direct recall.
Ray optics and wave optics require neat diagrams: Well-labelled ray diagrams and wavefront constructions significantly improve presentation and scoring.
Electricity and magnetism numericals follow standard formula patterns: Questions often repeat similar models based on Gauss’s law, Ohm’s law, Faraday’s law, and LCR circuits.
AC circuits focus on phase difference and power concepts: Students must clearly understand rms values, power factor, resonance, and wattless current.
Modern physics is highly scoring: Photoelectric effect, Bohr’s model, semiconductors, and logic gates mostly involve direct formulas and reasoning.
Derivations should be written step-by-step: Jumping directly to the final result is discouraged. Logical progression is necessary for full marks.
Graphs are frequently asked and carry easy marks: Energy vs frequency, current vs voltage, and magnetic field graphs should be practiced carefully.
Use proper physical terms and standard definitions: Board examiners expect textbook language, especially in theory-based answers.
Numericals should follow a fixed format: Given → Formula → Substitution → Calculation → Final Answer (with unit).
Time management is critical: Avoid spending too much time on lengthy derivations; attempt short answers and MCQs first.
Previous years’ questions repeat conceptually: Practicing past papers helps identify frequently tested laws, principles, and numericals.
Presentation directly impacts marks: Use proper spacing, underlined formulas, and clear headings for answers.

Frequently Asked Questions (FAQ) - Maharashtra Board Class 12 Physics 2025 – Solved Paper

Which phenomenon proves that light is a transverse wave?

(A) Reflection
(B) Interference
(C) Diffraction
(D) Polarization

Answer:

Correct option: (D) Polarization

Which physical quantity of light remains unchanged when it travels from one medium to another?

(A) Velocity
(B) Wavelength
(C) Frequency
(D) Amplitude

Answer:

Correct option: (C) Frequency

What is the SI unit of electric flux?

(A) N C⁻¹
(B) C m²
(C) N m² C⁻¹
(D) V m⁻¹

Correct option: (B) C m²

Define wattless current in an AC circuit.

Wattless current is the current flowing in an AC circuit in which the average power consumed is zero.

$$ \phi = 90^\circ $$

Average power is given by:

$$ P_{\text{avg}} = VI \cos\phi $$

Since:

$$ \cos 90^\circ = 0 $$

Therefore:

$$ P_{\text{avg}} = 0 $$

Hence, the current does not consume power and is called wattless current.

State Bohr’s quantization condition for angular momentum.

According to Bohr, the angular momentum of an electron revolving around the nucleus is quantized.

The quantization condition is:

$$ mvr = \frac{nh}{2\pi} $$

Where:

m = mass of electron
v = velocity of electron
r = radius of orbit
h = Planck’s constant
n = principal quantum number (1, 2, 3, ...)

This condition explains the stability of atomic orbits and discrete energy levels in hydrogen atom.

What is the condition for resonance in a series LCR circuit?

Resonance occurs when the inductive reactance equals the capacitive reactance.

Condition for resonance:

$$ X_L = X_C $$

Where:

$$ X_L = \omega L \quad \text{and} \quad X_C = \frac{1}{\omega C} $$

At resonance:

$$ \omega_0 = \frac{1}{\sqrt{LC}} $$

At this frequency, impedance is minimum and current is maximum.

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

Question 1 (i) “If two systems are each in thermal equilibrium with a third system, they are also in thermal equilibrium with each other.” This statement refers to:

Question 1 (ii) In Bernoulli’s theorem, which of the following is constant?

Question 1 (iii) Which of the following materials belongs to a diathermanous substance?

Question 1 (iv) Electric potential V at a distance r from a point charge is directly proportional to _____.

Question 1 (v) Which of the following equations gives the correct expression for the internal resistance of a cell using a potentiometer?

Question (vi) An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de-Broglie wavelength will be longest for:

Question (Vii) The gate which produces a high output when both inputs are high is:

Question 1 (viii) The power rating of a ceiling fan rotating with a constant torque of 2 Nm with an angular speed of 2 rad/s will be ____.

Question 1 (ix) A string of length 2 m is vibrating with 2 loops. The distance between its node and adjacent antinode is ____.

Question 1 (x) A transformer increases an alternating e.m.f. from 220 V to 880 V. If the primary coil has 1000 turns, the number of turns in the secondary coil are ____.

Question 2. Answer the following questions : [8]

Question 2 (i) At what temperature does the surface tension of a liquid become zero?

Question 2 (ii) Define self-inductance.

Question 2 (iii) What is the work done by an external uniform magnetic field perpendicular to the velocity of a moving charge?

Question 2 (iv) What do you mean by a thermodynamic system?

Question 2 (v) What is the value of B called when H = 0 in the hysteresis loop?

Question 2 (vi) State the formula for angle of banking.

Question 2 (vii) Calculate the electric field intensity at a point just near the surface of a charged plane sheet, measured from its mid-point. [ Surface charge density, $ \sigma = 8.85~\mu C/m^2 $ ]

Question 2 (viii) Find the kinetic energy of 1 litre of an ideal gas at S.T.P.

Attempt any EIGHT questions of the following : [16]

Q.3. Explain harmonics and overtones.

Q.4. Using Newton’s law of viscosity for streamline flow, derive an expression for coefficient of viscosity.

Q.4. Using Newton’s law of viscosity for streamline flow, derive an expression for coefficient of viscosity.

Q.5 State the formula for magnetic induction produced by a current in a circular arc of a wire. Hence find the magnetic induction at the centre of a current carrying circular loop.

Q.6 State and prove the law of conservation of angular momentum.

Q.7 State any four advantages of Light Emitting Diode (LED).

Q.8 Calculate the energy radiated in one minute by a perfectly black body of surface area 200 cm2 when it is maintained at 127 °C.

Q.9 Two coils having self inductances 60 mH each are coupled with each other. If the coefficient of coupling is 0.75, calculate the mutual inductance between them.

Q.10 In a series LCR circuit, if resistance, inductive reactance and capacitive reactance are 3 Ω, 8 Ω and 4 Ω respectively, calculate the phase difference between voltage and current.

Q.11 State the advantages of a potentiometer over a voltmeter.

Q.13 A body of mass 0.8 kg performs linear simple harmonic motion. It experiences a restoring force of 0.4 N when its displacement from the mean position is 4 cm. Determine:

Q.14 A gas of 0.5 mole at 300 K expands isothermally from an initial volume of 2.0 litre to a final volume of 6.0 litre. What is the work done by gas?

Attempt any EIGHT questions of the following : [24]

Q.15 Obtain an expression for the period of a bar magnet vibrating in a uniform magnetic field, performing S.H.M.

Q.16 In a thermodynamic system, define:

(a) Mechanical equilibrium

(b) Chemical equilibrium

(c) Thermal equilibrium

Q.17 What is Brewster’s law? Derive the formula for Brewster’s angle.

Q.18 Derive an expression for the law of radioactive decay. Define one becquerel (Bq).

Q.19 State and prove Kirchhoff’s law of heat radiation.

Q.20 Obtain an expression for practical determination of end correction (i) for a pipe open at both ends and (ii) for a pipe closed at one end.

Q.21 A conducting bar is rotating with constant angular speed around a pivot at one end in a uniform magnetic field perpendicular to its plane of rotation. Obtain an expression for the rotational e.m.f. induced between the ends of the bar.

Q.22 An electron in hydrogen atom stays in its second orbit for $10^{-8}\,\text{s}$. How many revolutions will it make around the nucleus in that time? [ Given: $e = 1.6 \times 10^{-19}\,\text{C},\; m = 9.1 \times 10^{-31}\,\text{kg}$ ]

Question 23 A flywheel of a motor has mass 100 kg and radius 1.5 m. The motor develops a constant torque of 2000 Nm. The flywheel starts rotating from rest. Calculate the work done during the first 4 revolutions.

Question 24 A galvanometer has a resistance of 50 Ω and a current of 2 mA is needed for its full scale deflection. Calculate the resistance required to convert it: (i) into an ammeter of 0.5 A range (ii) into a voltmeter of 10 V range

Q.25 Diameter of a water drop is 0.6 mm. Calculate the pressure inside a liquid drop. (T = 72 dyne/cm, atmospheric pressure = 1.013 × 105 N/m2)

Q.26 A solenoid of length 1 m and 5 cm in diameter has a winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its centre along the axis.

Attempt any THREE questions of the following : [12]

Q.27 What is Ferromagnetism? Explain it on the basis of domain theory.

Q.28 Obtain an expression for average power dissipated in a series LCR circuit.

Q.29 (a) Distinguish between interference and diffraction of light.

Q.29 (b) A double slit arrangement produces interference fringes for sodium light of wavelength 589 nm, that are 0.20° apart. What is the angular fringe separation if the entire arrangement is immersed in water? (Refractive index of water = 1.33)

Question 30

Question 31 (a) State any four uses of Van de Graaff generator. (b) In a parallel plate air capacitor, intensity of electric field is changing at the rate of $2 \times 10^{11}\ \text{V/m-s}.$ If the area of each plate is $20\ \text{cm}^2,$ calculate the displacement current.

Frequently Asked Questions (FAQ) - Maharashtra Board Class 12 Physics 2025 – Solved Paper

Question 2 (vii) Calculate the electric field intensity at a point just near the surface of a charged plane sheet, measured from its mid-point.
[ Surface charge density, $ \sigma = 8.85~\mu C/m^2 $ ]

Q.8 Calculate the energy radiated in one minute by a perfectly black body of surface area 200 cm² when it is maintained at 127 °C.

Q.22 An electron in hydrogen atom stays in its second orbit for $10^{-8}\,\text{s}$. How many revolutions will it make around the nucleus in that time?
[ Given: $e = 1.6 \times 10^{-19}\,\text{C},\; m = 9.1 \times 10^{-31}\,\text{kg}$ ]

Question 24 A galvanometer has a resistance of 50 Ω and a current of 2 mA is needed for its full scale deflection. Calculate the resistance required to convert it:
(i) into an ammeter of 0.5 A range
(ii) into a voltmeter of 10 V range

Q.25 Diameter of a water drop is 0.6 mm. Calculate the pressure inside a liquid drop. (T = 72 dyne/cm, atmospheric pressure = 1.013 × 10⁵ N/m²)

Question 31
(a) State any four uses of Van de Graaff generator.
(b) In a parallel plate air capacitor, intensity of electric field is changing at the rate of $2 \times 10^{11}\ \text{V/m-s}.$ If the area of each plate is $20\ \text{cm}^2,$ calculate the displacement current.

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