HSC 12th Important Board Question Paper 2026 Physics (Maharashtra Board) | Target 90+ Marks

Subject: PHYSICS (54)
Class: 12th HSC (Maharashtra Board)
Max. Marks: 70
Time: 3 Hours
Note: This is a Model Question Paper for the 2026 Board Examination, based on the latest pattern.
Physical Constants:
  • \(h = 6.63 \times 10^{-34} \text{ Js}\)
  • \(c = 3 \times 10^8 \text{ m/s}\)
  • \(\pi = 3.142\)
  • \(g = 9.8 \text{ m/s}^2\)
  • \(\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2\)
  • \(\mu_0 = 4\pi \times 10^{-7} \text{ Wb/A-m}\)
  • \(\sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\)
  • \(1 \text{ atm} = 1.013 \times 10^5 \text{ N/m}^2\)
  • \(R = 8.319 \text{ J/mol-K}\)

SECTION – A

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

(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: (a) zeroth law of thermodynamics
(ii) In Bernoulli’s theorem, which of the following is constant?
  • (a) Linear momentum
  • (b) Angular momentum
  • (c) Mass
  • (d) Energy
Answer: (d) Energy
(iii) Which of the following materials belongs to diathermanous substance?
  • (a) wood
  • (b) iron
  • (c) glass
  • (d) copper
Answer: (c) glass
(iv) Electric potential ‘V’ at a distance ‘r’ from a point charge is directly proportional to _____.
  • (a) \(r\)
  • (b) \(r^2\)
  • (c) \(1/r\)
  • (d) \(1/r^2\)
Answer: (c) \(1/r\)
(v) Which of the following equations gives correct expression for the internal resistance of a cell by using potentiometer?
  • (a) \(r = R(\frac{V}{E} - 1)\)
  • (b) \(r = R(1 - \frac{E}{V})\)
  • (c) \(r = R(1 - \frac{V}{E})\)
  • (d) \(r = R(\frac{E}{V} - 1)\)
Answer: (d) \(r = R(\frac{E}{V} - 1)\)
(vi) An electron, a proton, an \(\alpha\)-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) \(\alpha\)-particle
Answer: (b) electron
(Explanation: \(\lambda = h/\sqrt{2mE}\). Since E is same, \(\lambda \propto 1/\sqrt{m}\). Electron has smallest mass, so longest wavelength.)
(vii) The gate which produces high output, when its both inputs are high is –
  • (a) X-OR gate
  • (b) AND gate
  • (c) NAND gate
  • (d) NOR gate
Answer: (b) AND gate
(viii) The power rating of a ceiling fan rotating with a constant torque of 2 Nm with an angular speed of \(2\pi\) rad/s will be _____.
  • (a) \(\pi\) W
  • (b) \(2\pi\) W
  • (c) \(3\pi\) W
  • (d) \(4\pi\) W
Answer: (d) \(4\pi\) W
(Explanation: \(P = \tau \omega = 2 \times 2\pi = 4\pi\) W)
(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: 2 loops \(\implies L = \lambda = 2\)m. Distance between node and antinode = \(\lambda/4 = 2/4 = 0.5\)m)
(x) A transformer increases an alternating e.m.f. from 220V to 880V. If 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: \(N_s/N_p = V_s/V_p \implies N_s/1000 = 880/220 = 4 \implies N_s = 4000\))

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

(i) At what temperature the surface tension of a liquid becomes zero?
The surface tension of a liquid becomes zero at its critical temperature.
(ii) Define self inductance.
Self inductance is defined as the phenomenon of production of induced e.m.f. in a coil when the current flowing through the same coil changes.
(iii) What is the work done by an external uniform magnetic field perpendicular to the velocity of a moving charge?
The work done is zero because the magnetic force is always perpendicular to the direction of displacement (velocity).
(iv) What do you mean by a thermodynamic system?
A thermodynamic system is a collection of matter or radiation within a defined boundary (real or imaginary) that is separated from the surroundings, upon which thermodynamic observations are made.
(v) What is value of B called, when H = 0 is in the hysteresis loop?
The value of magnetic induction B left in the specimen when the magnetizing field H is reduced to zero is called Retentivity or Remanence.
(vi) State the formula for angle of banking.
\(\theta = \tan^{-1}\left(\frac{v^2}{rg}\right)\), where \(v\) is the speed limit, \(r\) is radius of curvature, and \(g\) is acceleration due to gravity.
(vii) Calculate the electric field intensity at a point just near the surface of a charged plane sheet, measured from its mid-point. [\(\sigma = 8.85 \mu \text{C/m}^2\)]
Formula: \(E = \frac{\sigma}{2\epsilon_0}\)
Given: \(\sigma = 8.85 \times 10^{-6} \text{ C/m}^2\), \(\epsilon_0 = 8.85 \times 10^{-12}\)
Calculation: \(E = \frac{8.85 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = \frac{10^6}{2} = 5 \times 10^5 \text{ N/C}\).
(viii) Find kinetic energy of 1 litre of an ideal gas at S.T.P.
Formula: \(E = \frac{3}{2}PV\)
Given: \(P = 1.013 \times 10^5 \text{ N/m}^2\), \(V = 1 \text{ litre} = 10^{-3} \text{ m}^3\)
Calculation: \(E = \frac{3}{2} \times 1.013 \times 10^5 \times 10^{-3} = 1.5 \times 101.3 \approx 152 \text{ J}\).

SECTION – B

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

Q. 3. Explain harmonics and overtones.
Harmonics: The term used to indicate the fundamental frequency and all integral multiples of the fundamental frequency (e.g., \(n, 2n, 3n...\)) are called harmonics.

Overtones: The frequencies of vibrations which are higher than the fundamental frequency and are actually present in the emitted sound are called overtones. The first frequency higher than fundamental is the 1st overtone, the next is 2nd overtone, etc.
Q. 4. Using Newton’s law of viscosity for streamline flow, derive an expression for coefficient of viscosity.
According to Newton's law of viscosity, the viscous force (\(f\)) acting between two layers is directly proportional to the area (\(A\)) and velocity gradient (\(dv/dx\)).
\(f \propto A \frac{dv}{dx}\)
\(f = \eta A \frac{dv}{dx}\)
Therefore, \(\eta = \frac{f}{A(dv/dx)}\).
This is the 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.
Magnetic induction due to an arc of radius \(r\) subtending angle \(\theta\) (radians) at center:
\(B = \frac{\mu_0 I}{4\pi r} \theta\)
For a full circular loop, \(\theta = 2\pi\).
\(B = \frac{\mu_0 I}{4\pi r} (2\pi) = \frac{\mu_0 I}{2r}\).
Q. 6. State and prove the law of conservation of angular momentum.
Statement: Angular momentum of a rotating body remains constant if the resultant external torque acting on it is zero.
Proof: We know \(\tau = \frac{dL}{dt}\).
If external torque \(\tau = 0\), then \(\frac{dL}{dt} = 0\).
This implies \(L = \text{constant}\). Hence proved.
Q. 7. State any four advantages of light emitting diode (LED).
1. Energy efficient and low power consumption.
2. Long life and ruggedness.
3. Fast switching capability (fast on-off response).
4. Environmentally friendly (no mercury).
Q. 8. Calculate the energy radiated in one minute by a perfectly black body of surface area 200 cm² when it is maintained at 127ºC.
Given: \(t = 60\)s, \(A = 200 \times 10^{-4} \text{ m}^2\), \(T = 127+273 = 400 \text{ K}\), \(\sigma = 5.67 \times 10^{-8}\).
Formula: \(Q = \sigma A T^4 t\)
Calculation:
\(Q = 5.67 \times 10^{-8} \times (200 \times 10^{-4}) \times (400)^4 \times 60\)
\(Q = 5.67 \times 10^{-8} \times 2 \times 10^{-2} \times 256 \times 10^8 \times 60\)
\(Q = 5.67 \times 2 \times 256 \times 60 \times 10^{-2}\)
\(Q \approx 1741.8 \text{ Joules}\).
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.
Given: \(L_1 = 60\) mH, \(L_2 = 60\) mH, \(k = 0.75\).
Formula: \(M = k\sqrt{L_1 L_2}\)
Calculation: \(M = 0.75 \sqrt{60 \times 60} = 0.75 \times 60 = 45\) mH.
Q. 10. In a series LCR circuit, if resistance, inductive reactance and capacitive reactance are 3\(\Omega\), 8\(\Omega\) and 4\(\Omega\) respectively, calculate phase difference between voltage and current.
Given: \(R=3\Omega, X_L=8\Omega, X_C=4\Omega\).
Formula: \(\tan \phi = \frac{X_L - X_C}{R}\)
Calculation: \(\tan \phi = \frac{8-4}{3} = \frac{4}{3}\).
\(\phi = \tan^{-1}(1.333) \approx 53.13^{\circ}\).
Q. 11. State the advantages of a potentiometer over a voltmeter.
1. Potentiometer measures e.m.f. of a cell very accurately (null deflection method draws no current).
2. It can measure very small potential differences.
3. It can be used to measure internal resistance of a cell (Voltmeter cannot directly).
Q. 12. Draw a neat and labelled circuit diagram for a full wave rectifier.
(Student should draw: Center-tapped transformer, two diodes D1 and D2, and a load resistor RL. Diodes conduct in alternate half cycles, current through RL is unidirectional.)
Q. 13. A body of mass 0.8 kg performs linear S.H.M. It experiences a restoring force of 0.4N, when its displacement from mean position is 4 cm. Determine Force constant and Period of S.H.M.
Given: \(m=0.8\) kg, \(F=0.4\) N, \(x=0.04\) m.
Force constant \(k = F/x = 0.4 / 0.04 = 10\) N/m.
Period \(T = 2\pi \sqrt{m/k} = 2\pi \sqrt{0.8/10} = 2\pi \sqrt{0.08}\) seconds.
\(T \approx 2 \times 3.142 \times 0.2828 \approx 1.77\) s.
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?
Given: \(n=0.5, T=300, V_1=2, V_2=6\).
Formula: \(W = nRT \ln(V_2/V_1) = 2.303 nRT \log_{10}(V_2/V_1)\)
Calculation: \(W = 2.303 \times 0.5 \times 8.319 \times 300 \times \log(3)\)
\(W = 2.303 \times 0.5 \times 8.319 \times 300 \times 0.4771 \approx 1371\) Joules.

SECTION – C

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

Q. 15. Obtain an expression for the period of a bar magnet vibrating in a uniform magnetic field, performing S.H.M.
Torque \(\tau = -mB \sin \theta\). For small \(\theta\), \(\tau \approx -mB\theta\).
Also \(\tau = I\alpha\). So, \(I\alpha = -mB\theta \implies \alpha = -(mB/I)\theta\).
Comparing with \(\alpha = -\omega^2 \theta\), we get \(\omega^2 = mB/I\).
Period \(T = 2\pi/\omega = 2\pi \sqrt{I/mB}\).
Q. 16. In a thermodynamic system, define – (a) Mechanical equilibrium (b) Chemical equilibrium and (c) Thermal equilibrium
(a) Mechanical equilibrium: When there are no unbalanced forces within the system and between the system and surroundings.
(b) Chemical equilibrium: When there are no chemical reactions going on within the system and no transfer of matter.
(c) Thermal equilibrium: When the temperature of the system is uniform and same as surroundings (no heat flow).
Q. 17. What is Brewster’s law? Derive the formula for Brewster’s angle.
Law: The tangent of the polarizing angle is equal to the refractive index of the reflecting medium.
Derivation: At polarizing angle \(\theta_p\), reflected and refracted rays are perpendicular. \(r + \theta_p = 90^\circ \implies r = 90 - \theta_p\).
Snell's law: \(\mu = \sin i / \sin r = \sin \theta_p / \sin(90-\theta_p) = \sin \theta_p / \cos \theta_p = \tan \theta_p\).
Q. 18. Derive an expression for law of radioactive decay. Define one becquerel (Bq).
Derivation starts from \(dN/dt \propto -N\), leads to \(N = N_0 e^{-\lambda t}\).
Becquerel: One Bq is defined as one decay per second.
Q. 19. State and prove Kirchhoff’s law of heat radiation.
Statement: At a given temperature, the coefficient of absorption is equal to the coefficient of emission for any body. (\(a = e\)).
Proof: Consider a body in thermal equilibrium inside an enclosure. Energy absorbed = \(aQ\). Energy emitted = \(eQ_b\) (where \(Q_b\) is blackbody emission). At equilibrium, absorbed = emitted. \(aQ = E\). For blackbody \(Q = E_b\). Leads to \(a=e\).
Q. 20. Obtain an expression for practical determination of end correction for (i) pipe open at both ends (ii) pipe closed at one end.
Let \(v = 2n_1(l_1 + 2e)\) and \(v = 2n_2(l_2 + 2e)\) for open pipe.
Solving for \(e\): \(e = \frac{n_1 l_1 - n_2 l_2}{2(n_2 - n_1)}\).
For closed pipe: \(e = \frac{n_1 l_1 - n_2 l_2}{2(n_2 - n_1)}\) (Usually derived as \(e = \frac{n_1 l_1 - n_2 l_2}{n_2 - n_1}\) depending on harmonics used).
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.
Consider element \(dr\) at distance \(r\). Velocity \(v = r\omega\).
Induced emf in element \(de = Bv dr = B(r\omega)dr\).
Total EMF \(E = \int_0^L B\omega r dr = B\omega [r^2/2]_0^L = \frac{1}{2} B\omega L^2\).
Q. 22. An electron in hydrogen atom stays in its second orbit for \(10^{-8}\) s. How many revolutions will it make around the nucleus in that time?
Using Bohr's theory:
\(r_2 = n^2 r_1 = 4 \times 0.53 \text{ \AA} = 2.12 \times 10^{-10}\) m.
\(v_2 = v_1 / n = (2.18 \times 10^6) / 2 = 1.09 \times 10^6\) m/s.
Time period \(T = \frac{2\pi r_2}{v_2} = \frac{2 \times 3.142 \times 2.12 \times 10^{-10}}{1.09 \times 10^6} \approx 1.22 \times 10^{-15}\) s.
Number of revolutions \(N = \frac{\text{Time}}{T} = \frac{10^{-8}}{1.22 \times 10^{-15}} \approx 8.2 \times 10^6\) revolutions.
Q. 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.
Given: \(\tau = 2000\) Nm, \(\theta = 4 \text{ rev} = 4 \times 2\pi = 8\pi\) rad.
Work \(W = \tau \theta = 2000 \times 8\pi = 16000\pi\) Joules.
\(W \approx 16000 \times 3.142 = 50272\) J.
Q. 24. A galvanometer has a resistance of 50 \(\Omega\) and a current of 2 mA is needed for its full scale deflection. Calculate resistance required to convert it (i) into an ammeter of 0.5 A range. (ii) into a voltmeter of 10 V range.
Given: \(G = 50 \Omega, I_g = 0.002 A\).
(i) Ammeter (Shunt): \(S = \frac{I_g G}{I - I_g} = \frac{0.002 \times 50}{0.5 - 0.002} = \frac{0.1}{0.498} \approx 0.2 \Omega\).
(ii) Voltmeter (Series R): \(R = \frac{V}{I_g} - G = \frac{10}{0.002} - 50 = 5000 - 50 = 4950 \Omega\).
Q. 25. Diameter of a water drop is 0.6 mm. Calculate the pressure inside a liquid drop. (T = 72 dyne/cm, atmospheric pressure = \(1.013 \times 10^5 \text{ N/m}^2\))
Radius \(r = 0.3 \text{ mm} = 3 \times 10^{-4}\) m.
\(T = 72 \text{ dyne/cm} = 0.072\) N/m.
Excess pressure \(P_{ex} = 2T/r = \frac{2 \times 0.072}{3 \times 10^{-4}} = 480\) Pa.
Total Pressure \(P = P_{atm} + P_{ex} = 101300 + 480 = 101780\) Pa.
Q. 26. A solenoid of length \(\pi\) m and 5 cm in diameter has a winding of 1000 turns and carries a current of 5A. Calculate the magnetic field at its centre along the axis.
\(n = N/L = 1000/\pi\).
\(B = \mu_0 n I = 4\pi \times 10^{-7} \times (1000/\pi) \times 5\).
\(B = 4 \times 10^{-7} \times 1000 \times 5 = 20 \times 10^{-4} = 2 \times 10^{-3}\) Tesla.

SECTION – D

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

Q. 27. What is Ferromagnetism? Explain it on the basis of domain theory.
Ferromagnetism: The property of substances to be strongly attracted by magnets and can be permanently magnetized.
Domain Theory:
  • Materials contain small regions called domains.
  • In each domain, dipole moments align in the same direction.
  • In unmagnetized state, domains are randomly oriented (net M = 0).
  • In external field, domains parallel to field grow or rotate, causing strong magnetization.
  • Removal of field leaves some alignment (Retentivity).
Q. 28. Obtain an expression for average power dissipated in a series LCR circuit.
Power \(P = V_{rms} I_{rms} \cos \phi\).
Derivation involves integrating instantaneous power \(p = vi\) over one cycle.
\(v = V_m \sin \omega t\), \(i = I_m \sin(\omega t \pm \phi)\).
Average of \(\sin^2 \omega t\) is 1/2. Average of \(\sin \omega t \cos \omega t\) is 0.
Final result: \(P_{avg} = \frac{V_m I_m}{2} \cos \phi = V_{rms} I_{rms} \cos \phi\).
Q. 29. Distinguish between interference and diffraction of light. A double slit arrangement produces interference fringes for sodium light of wavelength 589 nm, that are 0.20 degree apart. What is the angular fringe separation if the entire arrangement is immersed in water? (R.I. of water = 1.33)
Distinction: 1. Interference is superposition of waves from two coherent sources; Diffraction is bending around corners from secondary wavelets of same wavefront.
2. Interference fringes are usually equal width; Diffraction fringes vary.

Numerical:
Angular width in air \(\theta_a = \lambda / d = 0.20^{\circ}\).
In water, wavelength becomes \(\lambda' = \lambda / \mu\).
New angular width \(\theta_w = \lambda' / d = (\lambda / \mu) / d = \theta_a / \mu\).
\(\theta_w = 0.20 / 1.33 \approx 0.15^{\circ}\).
Q. 30. State Einstein’s photoelectric equation and mention physical significance of each term involved in it. The wavelength of incident light is 4000Å. Calculate the energy of incident photon.
Equation: \(h\nu = \phi_0 + K_{max}\) or \(K_{max} = h\nu - \phi_0\).
\(h\nu\) = Energy of incident photon.
\(\phi_0\) = Work function (min energy to escape).
\(K_{max}\) = Max kinetic energy of emitted electron.

Calculation:
\(E = hc/\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4000 \times 10^{-10}}\).
\(E = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}} = 4.97 \times 10^{-19}\) J.
(In eV: \(4.97 \times 10^{-19} / 1.6 \times 10^{-19} \approx 3.1\) eV).
Q. 31. State any four uses of Van de Graaff generator. In a parallel plate air capacitor, intensity of electric field is changing at the rate of \(2 \times 10^{11}\) V/ms. If area of each plate is 20 cm², calculate the displacement current.
Uses: 1. Produce high voltage/potential. 2. Accelerate charged particles. 3. Nuclear physics experiments. 4. Cancer treatment (radiotherapy).

Numerical:
Given: \(dE/dt = 2 \times 10^{11}\) V/m-s. Area \(A = 20 \times 10^{-4}\) m².
Displacement Current \(I_d = \epsilon_0 A (dE/dt)\).
\(I_d = 8.85 \times 10^{-12} \times 20 \times 10^{-4} \times 2 \times 10^{11}\).
\(I_d = 8.85 \times 40 \times 10^{-5} = 354 \times 10^{-5} = 3.54 \times 10^{-3}\) A (or 3.54 mA).

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