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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 113 marksmedium

System of Particles and Rotational Motion

State the theorem of perpendicular axes and use it to find the moment of inertia of a ring (mass M, radius R) about a diameter.

Reveal model answer + marking points

Perpendicular axes theorem (for a planar body): the moment of inertia about an axis perpendicular to the plane equals the sum of the moments of inertia about two mutually perpendicular axes in the plane meeting at the same point: Iz = Ix + Iy. For a ring, Iz (about central axis) = M R^2, and by symmetry Ix = Iy = I(diameter). So M R^2 = 2 I(diameter), giving I(diameter) = M R^2 / 2.

Iz = Ix + Iy

Marking-scheme points

  • Iz = Ix + Iy (planar body)
  • Ring: Iz = MR^2, Ix = Iy by symmetry
  • I(diameter) = MR^2/2
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PhysicsClass 112 markseasy

System of Particles and Rotational Motion

What is the radius of gyration of a body? How is it related to the moment of inertia?

Reveal model answer + marking points

The radius of gyration K is the distance from the axis at which the whole mass of the body can be assumed to be concentrated so as to give the same moment of inertia: I = M K^2, hence K = sqrt(I / M). Its SI unit is the metre.

K = sqrt(I / M)

Marking-scheme points

  • I = M K^2
  • K = sqrt(I/M)
  • Unit metre
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PhysicsClass 112 markseasy

System of Particles and Rotational Motion

Find the moment of inertia of a ring of mass 2 kg and radius 0.5 m about an axis passing through its centre and perpendicular to its plane.

Reveal model answer + marking points

For a ring about its central axis, I = M R^2 = 2 * (0.5)^2 = 2 * 0.25 = 0.5 kg m^2.

I = M R^2

Marking-scheme points

  • Ring: I = M R^2
  • = 2 * 0.25
  • I = 0.5 kg m^2
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PhysicsClass 113 marksmedium

Gravitation

State Kepler's three laws of planetary motion.

Reveal model answer + marking points

1) Law of orbits: every planet moves in an ellipse with the Sun at one focus. 2) Law of areas: the line joining a planet to the Sun sweeps out equal areas in equal intervals of time (so a planet moves faster when nearer the Sun); this follows from conservation of angular momentum. 3) Law of periods: the square of the orbital period is proportional to the cube of the semi-major axis, T^2 is proportional to a^3.

T^2 = k a^3

Marking-scheme points

  • Orbits: ellipse, Sun at a focus
  • Areas: equal areas in equal times
  • Periods: T^2 proportional to a^3
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PhysicsClass 113 marksmedium

Gravitation

A satellite revolves in a circular orbit very close to the Earth's surface. Find its orbital speed. Take g = 9.8 m/s^2 and R = 6.4 x 10^6 m.

Reveal model answer + marking points

For an orbit close to the surface the required centripetal force is provided by gravity, so v_o = sqrt(g R) = sqrt(9.8 * 6.4 x 10^6) = sqrt(6.272 x 10^7) = 7.92 x 10^3 m/s = 7.92 km/s (approximately 8 km/s).

v_o = sqrt(g R)

Marking-scheme points

  • v_o = sqrt(g R) for a near-surface orbit
  • Substitute g and R
  • v_o ~ 7.9 km/s
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PhysicsClass 112 marksmedium

Gravitation

What is a geostationary satellite? State any two conditions for a satellite to be geostationary.

Reveal model answer + marking points

A geostationary satellite appears stationary relative to the Earth because it revolves in step with the Earth's rotation. Conditions: (i) its orbital period must equal 24 hours (equal to Earth's rotation period); (ii) it must orbit in the equatorial plane, in the same (west-to-east) sense as the Earth, at a height of about 36000 km.

Marking-scheme points

  • Period = 24 h (matches Earth)
  • Equatorial plane, west to east
  • Height ~ 36000 km
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PhysicsClass 113 marksmedium

Gravitation

Show that the escape velocity from the Earth's surface is sqrt(2) times the orbital velocity of a satellite orbiting close to the surface.

Reveal model answer + marking points

Orbital velocity near the surface: v_o = sqrt(g R). Escape velocity: v_e = sqrt(2 g R). Dividing, v_e / v_o = sqrt(2 g R) / sqrt(g R) = sqrt(2). Hence v_e = sqrt(2) * v_o (about 1.41 times).

v_e = sqrt(2) * v_o

Marking-scheme points

  • v_o = sqrt(gR), v_e = sqrt(2gR)
  • Ratio v_e/v_o = sqrt(2)
  • v_e = 1.41 v_o
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PhysicsClass 112 markseasy

Mechanical Properties of Solids

State Hooke's law and define the elastic limit.

Reveal model answer + marking points

Hooke's law: within the elastic limit, the stress developed in a body is directly proportional to the strain produced, i.e. stress / strain = a constant (the modulus of elasticity). The elastic limit is the maximum stress up to which a body returns to its original shape and size on removing the deforming force; beyond it, permanent (plastic) deformation occurs.

stress = E * strain

Marking-scheme points

  • Stress proportional to strain (within elastic limit)
  • stress/strain = modulus of elasticity
  • Elastic limit: max stress for full recovery
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PhysicsClass 113 marksmedium

Mechanical Properties of Solids

A wire of length 2 m and cross-sectional area 1 mm^2 stretches by 1 mm when a load of 10 N is applied. Calculate Young's modulus of the material.

Reveal model answer + marking points

Y = (F L) / (A * delta L). Here F = 10 N, L = 2 m, A = 1 mm^2 = 1 x 10^-6 m^2, delta L = 1 mm = 1 x 10^-3 m. Y = (10 * 2) / (1 x 10^-6 * 1 x 10^-3) = 20 / (1 x 10^-9) = 2 x 10^10 N/m^2.

Y = (F L) / (A * delta L)

Marking-scheme points

  • Y = F L / (A delta L)
  • Convert mm^2 and mm to SI
  • Y = 2 x 10^10 N/m^2
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PhysicsClass 112 markseasy

Mechanical Properties of Fluids

State Pascal's law and give one application of it.

Reveal model answer + marking points

Pascal's law: a pressure applied to an enclosed incompressible fluid is transmitted equally and undiminished to every part of the fluid and to the walls of the container. Application: the hydraulic lift/brake, where a small force on a small piston produces a large force on a large piston because pressure is the same throughout.

F1 / A1 = F2 / A2

Marking-scheme points

  • Pressure transmitted equally in an enclosed fluid
  • Application: hydraulic lift / brakes
  • Small force -> large force via area ratio
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PhysicsClass 112 marksmedium

Mechanical Properties of Fluids

State the equation of continuity for fluid flow and what it represents.

Reveal model answer + marking points

For the steady flow of an incompressible fluid, the product of area of cross-section and speed is constant along the tube: A1 v1 = A2 v2 (A v = constant). It is a statement of conservation of mass - where a pipe is narrow the fluid flows faster, and where it is wide it flows slower.

A1 v1 = A2 v2

Marking-scheme points

  • A v = constant
  • Based on conservation of mass
  • Narrow pipe -> higher speed
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PhysicsClass 112 markseasy

Mechanical Properties of Fluids

Define surface tension. Give one everyday example that demonstrates it.

Reveal model answer + marking points

Surface tension is the property by which the free surface of a liquid behaves like a stretched elastic membrane; it is the force per unit length acting along the surface (SI unit: N/m). Example: small water droplets and mercury beads become spherical (minimum surface area), and some insects can walk on water.

T = F / L

Marking-scheme points

  • Force per unit length on liquid surface
  • Unit N/m
  • Example: spherical droplets, insects on water
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PhysicsClass 113 marksmedium

Mechanical Properties of Fluids

What is viscosity? State Stokes' law and define terminal velocity.

Reveal model answer + marking points

Viscosity is the internal friction between adjacent layers of a fluid moving with different velocities. Stokes' law: the viscous drag on a small sphere of radius r moving with speed v through a fluid of viscosity eta is F = 6 pi eta r v. Terminal velocity is the constant maximum velocity attained by a body falling through a fluid when the net force (weight minus buoyancy and viscous drag) becomes zero.

F = 6 pi eta r v

Marking-scheme points

  • Viscosity: internal friction between fluid layers
  • Stokes: F = 6 pi eta r v
  • Terminal velocity: net force zero, constant speed
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PhysicsClass 113 markseasy

Mechanical Properties of Fluids

Calculate the total pressure at a depth of 10 m in a lake. Take atmospheric pressure = 1.0 x 10^5 Pa, density of water = 1000 kg/m^3 and g = 10 m/s^2.

Reveal model answer + marking points

Total pressure P = P_atm + rho g h = 1.0 x 10^5 + (1000)(10)(10) = 1.0 x 10^5 + 1.0 x 10^5 = 2.0 x 10^5 Pa.

P = P_atm + rho g h

Marking-scheme points

  • P = P_atm + rho g h
  • rho g h = 1.0 x 10^5 Pa
  • P = 2.0 x 10^5 Pa
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PhysicsClass 112 markseasy

Thermal Properties of Matter

Define the coefficient of linear expansion. Write the relation for the increase in length of a rod on heating.

Reveal model answer + marking points

The coefficient of linear expansion (alpha) is the fractional increase in length per degree rise in temperature: alpha = (delta L) / (L * delta T). The increase in length is delta L = L alpha (delta T), and the new length is L' = L (1 + alpha delta T). SI unit of alpha: per kelvin (K^-1).

delta L = L alpha delta T

Marking-scheme points

  • alpha = delta L / (L delta T)
  • delta L = L alpha delta T
  • Unit K^-1
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PhysicsClass 113 marksmedium

Thermal Properties of Matter

A metal rod of length 1 m is heated through 100 K. If its coefficient of linear expansion is 1.2 x 10^-5 K^-1, find the increase in its length.

Reveal model answer + marking points

delta L = L alpha (delta T) = 1 * (1.2 x 10^-5) * 100 = 1.2 x 10^-3 m = 1.2 mm.

delta L = L alpha delta T

Marking-scheme points

  • delta L = L alpha delta T
  • = 1 * 1.2e-5 * 100
  • delta L = 1.2 mm
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PhysicsClass 112 markseasy

Thermal Properties of Matter

Distinguish between specific heat capacity and latent heat.

Reveal model answer + marking points

Specific heat capacity is the heat required to raise the temperature of 1 kg of a substance by 1 K (Q = m c delta T; unit J/kg/K) - the temperature changes. Latent heat is the heat required to change the state of 1 kg of a substance at constant temperature (Q = m L; unit J/kg) - the temperature stays constant during the phase change.

Q = m c delta T ; Q = m L

Marking-scheme points

  • Specific heat: Q = m c delta T, temperature changes
  • Latent heat: Q = m L, temperature constant (phase change)
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PhysicsClass 113 marksmedium

Thermal Properties of Matter

How much heat is required to melt 100 g of ice at 0 degrees C into water at 0 degrees C? The latent heat of fusion of ice is 3.36 x 10^5 J/kg.

Reveal model answer + marking points

Q = m L = 0.1 kg * 3.36 x 10^5 J/kg = 3.36 x 10^4 J = 33600 J. The temperature stays at 0 degrees C throughout the melting.

Q = m L

Marking-scheme points

  • Q = m L (phase change)
  • m = 0.1 kg
  • Q = 33600 J
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PhysicsClass 113 markseasy

Thermal Properties of Matter

Name and briefly describe the three modes of heat transfer.

Reveal model answer + marking points

Conduction: heat flows through a material from the hotter to the cooler region by molecular vibrations, without bulk movement of the material (e.g. a metal spoon heating up). Convection: heat is carried by the actual movement of the heated fluid (e.g. boiling water, sea breeze). Radiation: heat travels as electromagnetic waves and needs no medium (e.g. heat from the Sun).

Marking-scheme points

  • Conduction: through solids, no bulk motion
  • Convection: bulk movement of fluid
  • Radiation: EM waves, no medium needed
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PhysicsClass 112 marksmedium

Thermal Properties of Matter

State Newton's law of cooling.

Reveal model answer + marking points

Newton's law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings, provided this difference is small: (- dQ/dt) is proportional to (T - T_surroundings).

dQ/dt = -k (T - T0)

Marking-scheme points

  • Rate of cooling proportional to temperature difference
  • Valid for small differences
  • - dQ/dt proportional to (T - T0)
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