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The questions your board exam loves to ask

800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.

800 board questionsModel answersMarking-scheme pointsEvery chapterCBSE · ISC · State boards

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PhysicsClass 112 markseasy

Thermodynamics

State the zeroth law of thermodynamics. What does it define?

Reveal model answer + marking points

The zeroth law states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. It leads to the concept of temperature - a property that is the same for all bodies in thermal equilibrium.

Marking-scheme points

  • A eq C and B eq C => A eq B
  • Defines temperature
  • Basis of thermometry
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PhysicsClass 113 marksmedium

Thermodynamics

Why is the molar specific heat at constant pressure (Cp) greater than that at constant volume (Cv)? State the relation between them.

Reveal model answer + marking points

At constant volume all the heat supplied goes to increase the internal energy (no work is done). At constant pressure the gas also expands and does external work, so extra heat is needed for the same temperature rise; hence Cp > Cv. The relation (Mayer's relation) is Cp - Cv = R, where R is the universal gas constant.

Cp - Cv = R

Marking-scheme points

  • Const V: heat only raises internal energy
  • Const P: heat also does work of expansion
  • Cp - Cv = R
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PhysicsClass 112 marksmedium

Thermodynamics

State the second law of thermodynamics (any one statement).

Reveal model answer + marking points

Kelvin-Planck statement: it is impossible to construct an engine that, working in a cycle, converts all the heat absorbed from a source completely into work with no other effect. (Equivalently, Clausius statement: heat cannot flow of its own accord from a colder body to a hotter body.)

Marking-scheme points

  • Kelvin-Planck: no 100% heat-to-work engine
  • Clausius: heat won't flow cold -> hot on its own
  • Sets a direction for natural processes
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PhysicsClass 113 markshard

Thermodynamics

One mole of an ideal gas expands isothermally at 300 K to twice its original volume. Find the work done by the gas. Take R = 8.31 J/mol/K and ln 2 = 0.693.

Reveal model answer + marking points

For an isothermal process, W = n R T ln(V2/V1) = 1 * 8.31 * 300 * ln(2) = 8.31 * 300 * 0.693 = 1727 J (approximately 1.73 kJ).

W = n R T ln(V2/V1)

Marking-scheme points

  • W = n R T ln(V2/V1)
  • V2/V1 = 2, ln 2 = 0.693
  • W ~ 1727 J
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PhysicsClass 113 marksmedium

Kinetic Theory

Write the expression for the pressure exerted by an ideal gas in terms of density and mean-square speed, and hence relate pressure to the average kinetic energy per unit volume.

Reveal model answer + marking points

From kinetic theory, P = (1/3) rho <v^2>, where rho is the density and <v^2> is the mean-square speed of the molecules. Since the kinetic energy per unit volume is (1/2) rho <v^2>, we get P = (2/3) * (kinetic energy per unit volume). Thus pressure is two-thirds of the translational KE per unit volume.

P = (1/3) rho <v^2>

Marking-scheme points

  • P = (1/3) rho <v^2>
  • KE per volume = (1/2) rho <v^2>
  • P = (2/3) * KE per unit volume
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PhysicsClass 112 marksmedium

Kinetic Theory

State the law of equipartition of energy and give the degrees of freedom of a monatomic and a diatomic gas molecule.

Reveal model answer + marking points

Law of equipartition: in thermal equilibrium, the total energy is shared equally among all degrees of freedom, each contributing (1/2) k T of energy per molecule. A monatomic molecule has 3 degrees of freedom (translational only); a diatomic molecule at ordinary temperatures has 5 (3 translational + 2 rotational).

Energy per degree of freedom = (1/2) k T

Marking-scheme points

  • Each degree of freedom gets (1/2) kT
  • Monatomic: 3 (translational)
  • Diatomic: 5 (3 trans + 2 rot)
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PhysicsClass 113 marksmedium

Kinetic Theory

Calculate the average translational kinetic energy of a gas molecule at 300 K. Take Boltzmann constant k = 1.38 x 10^-23 J/K.

Reveal model answer + marking points

Average translational KE per molecule = (3/2) k T = (3/2)(1.38 x 10^-23)(300) = 1.5 * 1.38 x 10^-23 * 300 = 6.21 x 10^-21 J.

KE = (3/2) k T

Marking-scheme points

  • KE = (3/2) k T
  • Independent of the type of gas
  • KE = 6.21 x 10^-21 J
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PhysicsClass 112 markseasy

Oscillations

Define simple harmonic motion (SHM) and write its defining equation.

Reveal model answer + marking points

Simple harmonic motion is an oscillation in which the restoring force (or acceleration) is directly proportional to the displacement from the mean position and is always directed towards it. Defining equation: a = - omega^2 x, where omega is the angular frequency and x the displacement.

a = - omega^2 x

Marking-scheme points

  • Restoring force proportional to -x
  • Directed towards mean position
  • a = - omega^2 x
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PhysicsClass 113 marksmedium

Oscillations

Show that the total mechanical energy of a particle in SHM is constant and independent of the displacement.

Reveal model answer + marking points

For SHM of amplitude A, kinetic energy KE = (1/2) m omega^2 (A^2 - x^2) and potential energy PE = (1/2) m omega^2 x^2. Adding, total energy E = KE + PE = (1/2) m omega^2 A^2. This is constant - it does not depend on x - and equals (1/2) m omega^2 A^2. Energy shuttles between KE (maximum at the mean position) and PE (maximum at the extremes).

E = (1/2) m omega^2 A^2

Marking-scheme points

  • KE = (1/2) m omega^2 (A^2 - x^2)
  • PE = (1/2) m omega^2 x^2
  • E = (1/2) m omega^2 A^2 = constant
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PhysicsClass 112 markseasy

Oscillations

Write the expression for the time period of a mass m attached to a spring of force constant k, and state how it changes if the mass is quadrupled.

Reveal model answer + marking points

T = 2 pi sqrt(m / k). Since T is proportional to sqrt(m), quadrupling the mass (m -> 4m) makes sqrt(4m) = 2 sqrt(m), so the time period doubles.

T = 2 pi sqrt(m / k)

Marking-scheme points

  • T = 2 pi sqrt(m/k)
  • T proportional to sqrt(m)
  • 4x mass => 2x period
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PhysicsClass 113 marksmedium

Oscillations

A particle executes SHM with amplitude 5 cm and time period 2 s. Find its maximum velocity. Take pi = 3.14.

Reveal model answer + marking points

Angular frequency omega = 2 pi / T = 2 pi / 2 = pi rad/s. Maximum velocity v_max = A omega = 0.05 * pi = 0.05 * 3.14 = 0.157 m/s (about 15.7 cm/s).

v_max = A omega

Marking-scheme points

  • omega = 2 pi / T = pi rad/s
  • v_max = A omega
  • v_max = 0.157 m/s
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PhysicsClass 112 marksmedium

Oscillations

What are forced oscillations and resonance?

Reveal model answer + marking points

Forced oscillations occur when a body is made to oscillate under an external periodic force with the frequency of that force. Resonance is the special case when the driving frequency equals the body's natural frequency; the amplitude of oscillation then becomes maximum. Example: a child's swing pushed at its natural frequency.

Marking-scheme points

  • Forced: oscillation at the driving frequency
  • Resonance: driving frequency = natural frequency
  • Amplitude becomes maximum
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PhysicsClass 112 markseasy

Waves

Derive the relation between wave speed, frequency and wavelength.

Reveal model answer + marking points

In one time period T, a wave advances by one wavelength lambda. So wave speed v = distance / time = lambda / T. Since frequency f = 1/T, we get v = f lambda. This holds for all progressive waves.

v = f lambda

Marking-scheme points

  • In time T, wave moves one wavelength
  • v = lambda / T
  • f = 1/T => v = f lambda
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PhysicsClass 112 marksmedium

Waves

Distinguish between progressive (travelling) and stationary (standing) waves.

Reveal model answer + marking points

A progressive wave transfers energy continuously in one direction, and every particle has the same amplitude but oscillates with a phase lag. A stationary wave is formed by two identical waves travelling in opposite directions; it does not transfer net energy, has fixed nodes (zero amplitude) and antinodes (maximum amplitude), and the amplitude varies from point to point.

Marking-scheme points

  • Progressive: transfers energy, same amplitude, phase lag
  • Stationary: no net energy transfer, fixed nodes and antinodes
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PhysicsClass 113 marksmedium

Waves

Write the expression for the fundamental frequency of a string of length L fixed at both ends, and explain what harmonics are.

Reveal model answer + marking points

For a string of length L, linear density mu and tension T, the fundamental frequency (first harmonic) is f1 = (1/2L) sqrt(T/mu), because the fundamental mode fits half a wavelength in the length L. Harmonics (overtones) are the higher allowed frequencies, which are integer multiples of the fundamental: fn = n f1 (n = 1, 2, 3, ...).

fn = (n / 2L) sqrt(T / mu)

Marking-scheme points

  • f1 = (1/2L) sqrt(T/mu)
  • Fundamental: L = lambda/2
  • Harmonics: fn = n f1
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PhysicsClass 113 markseasy

Waves

Define beats. Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together - find the number of beats heard per second.

Reveal model answer + marking points

Beats are the periodic rise and fall in the loudness of sound produced when two waves of slightly different frequencies superpose. The beat frequency equals the difference of the two frequencies: 260 - 256 = 4 beats per second.

f_beat = |f1 - f2|

Marking-scheme points

  • Beats: periodic variation of loudness
  • Beat frequency = |f1 - f2|
  • = 4 beats per second
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PhysicsClass 113 markshard

Waves

A source of sound of frequency 340 Hz moves towards a stationary observer at 34 m/s. Find the apparent frequency heard. Speed of sound = 340 m/s.

Reveal model answer + marking points

For a source approaching a stationary observer, f' = f * v / (v - v_s) = 340 * 340 / (340 - 34) = 340 * 340 / 306 = 377.8 Hz (about 378 Hz). The pitch appears higher.

f' = f v / (v - v_s)

Marking-scheme points

  • Approaching source: f' = f v/(v - v_s)
  • v_s = 34 m/s
  • f' ~ 378 Hz (higher pitch)
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PhysicsClass 112 markseasy

Motion in a Straight Line

Define relative velocity. Two trains move in the same direction at 60 km/h and 40 km/h - what is the velocity of the first relative to the second?

Reveal model answer + marking points

Relative velocity of a body A with respect to B is the velocity of A as seen from B: v_AB = v_A - v_B. For the trains, v = 60 - 40 = 20 km/h, so the faster train appears to move ahead at 20 km/h relative to the slower one.

v_AB = v_A - v_B

Marking-scheme points

  • v_AB = v_A - v_B
  • Same direction: subtract speeds
  • = 20 km/h
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PhysicsClass 113 marksmedium

Work, Energy and Power

Derive the expression for the elastic potential energy stored in a spring stretched by x, with force constant k.

Reveal model answer + marking points

The restoring force at extension x' is F = k x'. The work done in stretching the spring from 0 to x is W = integral of (k x') dx' from 0 to x = (1/2) k x^2. This work is stored as elastic potential energy: U = (1/2) k x^2.

U = (1/2) k x^2

Marking-scheme points

  • F = k x' (Hooke's law)
  • Work = integral k x' dx' from 0 to x
  • U = (1/2) k x^2
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PhysicsClass 113 marksmedium

System of Particles and Rotational Motion

Write the expression for the total kinetic energy of a body rolling without slipping, and explain its two parts.

Reveal model answer + marking points

A rolling body has both translation and rotation. Its total kinetic energy is KE = (1/2) m v^2 + (1/2) I omega^2, where the first term is the translational KE of the centre of mass and the second is the rotational KE about the centre. Using v = R omega and I = m K^2, this becomes KE = (1/2) m v^2 (1 + K^2/R^2).

KE = (1/2) m v^2 + (1/2) I omega^2

Marking-scheme points

  • KE = (1/2) m v^2 + (1/2) I omega^2
  • Translational + rotational parts
  • = (1/2) m v^2 (1 + K^2/R^2)
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