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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

Waves

Distinguish between transverse and longitudinal waves, giving one example of each.

Reveal model answer + marking points

In a transverse wave the particles of the medium vibrate perpendicular to the direction of wave propagation (example: a wave on a stretched string, light waves). In a longitudinal wave the particles vibrate parallel to the direction of propagation, forming compressions and rarefactions (example: sound waves in air).

v = f * lambda

Marking-scheme points

  • Transverse: vibration perpendicular to propagation (string, light)
  • Longitudinal: vibration parallel; compressions and rarefactions (sound)
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PhysicsClass 112 markseasy

Laws of Motion

State Newton's second law of motion and show that it leads to F = m a.

Reveal model answer + marking points

Newton's second law: the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force. F = dp/dt = d(mv)/dt. For constant mass, F = m (dv/dt) = m a.

F = dp/dt = m a

Marking-scheme points

  • F proportional to rate of change of momentum
  • F = dp/dt
  • Constant mass => F = m a
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PhysicsClass 112 markseasy

Units and Measurements

Distinguish between systematic errors and random errors.

Reveal model answer + marking points

Systematic errors have a definite cause and always shift readings in one direction (e.g. a zero error in an instrument, or a mis-calibrated scale); they can be minimised by correcting the instrument. Random errors occur due to unpredictable fluctuations and vary in size and sign; they are reduced by taking many readings and averaging.

Marking-scheme points

  • Systematic: one-directional, known cause, correctable
  • Random: irregular, reduced by averaging many readings
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PhysicsClass 113 marksmedium

Units and Measurements

The percentage errors in measuring quantities A and B are 2% and 3% respectively. Find the maximum percentage error in the quantity P = A * B^2.

Reveal model answer + marking points

For a product/power, percentage errors add with the powers as weights: (dP/P) = (dA/A) + 2 (dB/B) = 2% + 2(3%) = 2% + 6% = 8%.

dP/P = dA/A + 2 dB/B

Marking-scheme points

  • % error in product adds
  • Power multiplies its error
  • Max error = 2 + 2*3 = 8%
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PhysicsClass 112 markseasy

Motion in a Straight Line

Distinguish between distance and displacement.

Reveal model answer + marking points

Distance is the total length of the actual path travelled; it is a scalar and is always positive. Displacement is the shortest straight-line vector from the initial to the final position; it is a vector and can be positive, negative or zero. Displacement magnitude is always less than or equal to the distance.

Marking-scheme points

  • Distance: scalar, total path, always >= 0
  • Displacement: vector, straight line initial->final
  • |displacement| <= distance
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PhysicsClass 113 marksmedium

Motion in a Straight Line

Using a velocity-time graph, derive the equation s = u t + (1/2) a t^2 for uniformly accelerated motion.

Reveal model answer + marking points

On a v-t graph for constant acceleration, the velocity rises linearly from u to v = u + a t. The displacement equals the area under the graph, which is a rectangle (height u, width t) plus a triangle (base t, height a t): s = (u * t) + (1/2)(t)(a t) = u t + (1/2) a t^2.

s = u t + (1/2) a t^2

Marking-scheme points

  • Displacement = area under v-t graph
  • Area = rectangle (u t) + triangle ((1/2) a t^2)
  • s = u t + (1/2) a t^2
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PhysicsClass 113 marksmedium

Motion in a Straight Line

A car moving at 20 m/s is brought to rest with a uniform deceleration of 5 m/s^2. Find the stopping distance.

Reveal model answer + marking points

Using v^2 = u^2 - 2 a s with v = 0: 0 = (20)^2 - 2(5) s, so 10 s = 400, giving s = 40 m.

v^2 = u^2 - 2 a s

Marking-scheme points

  • v^2 = u^2 - 2 a s
  • Set v = 0
  • s = 400 / 10 = 40 m
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PhysicsClass 113 marksmedium

Motion in a Plane

State the parallelogram law of vector addition and write the expression for the magnitude of the resultant of two vectors P and Q inclined at an angle theta.

Reveal model answer + marking points

Parallelogram law: if two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, their resultant is represented by the diagonal from that point. Magnitude: R = sqrt(P^2 + Q^2 + 2 P Q cos(theta)), and the resultant makes an angle alpha with P where tan(alpha) = (Q sin(theta)) / (P + Q cos(theta)).

R = sqrt(P^2 + Q^2 + 2 P Q cos(theta))

Marking-scheme points

  • Adjacent sides -> diagonal is resultant
  • R = sqrt(P^2 + Q^2 + 2PQ cos theta)
  • tan(alpha) = Q sin theta / (P + Q cos theta)
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PhysicsClass 112 marksmedium

Motion in a Plane

What is centripetal acceleration? Write its expression and state its direction.

Reveal model answer + marking points

In uniform circular motion, centripetal acceleration is the acceleration directed towards the centre of the circle that continuously changes the direction of velocity. Its magnitude is a_c = v^2 / r = omega^2 r, and it always points radially inward (towards the centre).

a_c = v^2 / r = omega^2 r

Marking-scheme points

  • Directed towards centre
  • a_c = v^2/r = omega^2 r
  • Changes direction of v, not speed
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PhysicsClass 113 marksmedium

Motion in a Plane

A ball is projected with a speed of 30 m/s at 30 degrees to the horizontal. Find its horizontal range. Take g = 10 m/s^2.

Reveal model answer + marking points

R = u^2 sin(2 theta) / g = (30)^2 * sin(60 deg) / 10 = 900 * 0.866 / 10 = 77.9 m (about 78 m).

R = u^2 sin(2 theta) / g

Marking-scheme points

  • R = u^2 sin(2 theta)/g
  • sin 60 = 0.866
  • R ~ 78 m
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PhysicsClass 112 markseasy

Laws of Motion

State Newton's third law of motion and give one example.

Reveal model answer + marking points

Newton's third law: to every action there is an equal and opposite reaction, and the two forces act on different bodies. Example: when we walk, the foot pushes the ground backward (action) and the ground pushes the foot forward (reaction), which moves us ahead.

F(AB) = - F(BA)

Marking-scheme points

  • Equal and opposite forces
  • Act on different bodies
  • Example: walking / rocket / gun recoil
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PhysicsClass 113 marksmedium

Laws of Motion

A gun of mass 4 kg fires a bullet of mass 20 g with a muzzle speed of 400 m/s. Find the recoil velocity of the gun.

Reveal model answer + marking points

By conservation of linear momentum (initial total momentum = 0): m_bullet * v_bullet = m_gun * v_gun. So 0.020 * 400 = 4 * v_gun, giving v_gun = 8 / 4 = 2 m/s (opposite to the bullet).

m1 v1 = m2 v2

Marking-scheme points

  • Total initial momentum = 0
  • m1 v1 = m2 v2
  • v_gun = 2 m/s (recoil, opposite direction)
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PhysicsClass 112 markseasy

Laws of Motion

Distinguish between static friction and kinetic friction.

Reveal model answer + marking points

Static friction acts on a body at rest and is self-adjusting up to a maximum value (limiting friction) f_s(max) = mu_s N. Kinetic (sliding) friction acts on a moving body and has a nearly constant value f_k = mu_k N. For the same surfaces, mu_k < mu_s, so it is harder to start motion than to keep it going.

f_s(max) = mu_s N ; f_k = mu_k N

Marking-scheme points

  • Static: on body at rest, self-adjusting, up to mu_s N
  • Kinetic: on moving body, ~constant mu_k N
  • mu_k < mu_s
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PhysicsClass 113 marksmedium

Laws of Motion

A block slides down a smooth inclined plane of inclination 30 degrees. Find its acceleration. Take g = 10 m/s^2.

Reveal model answer + marking points

On a smooth incline the acceleration along the plane is a = g sin(theta) = 10 * sin(30 deg) = 10 * 0.5 = 5 m/s^2, directed down the incline.

a = g sin(theta)

Marking-scheme points

  • Component of g along incline = g sin theta
  • sin 30 = 0.5
  • a = 5 m/s^2
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PhysicsClass 112 markseasy

Work, Energy and Power

Define work done by a constant force. When is the work done (a) zero and (b) negative?

Reveal model answer + marking points

Work done by a constant force is the dot product of force and displacement: W = F s cos(theta), where theta is the angle between them. (a) W = 0 when theta = 90 degrees (force perpendicular to displacement, e.g. centripetal force). (b) W is negative when theta is obtuse (90 to 180 degrees), e.g. friction opposing motion.

W = F s cos(theta)

Marking-scheme points

  • W = F s cos(theta)
  • Zero when theta = 90 deg
  • Negative when 90 < theta <= 180 deg
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PhysicsClass 112 marksmedium

Work, Energy and Power

Distinguish between conservative and non-conservative forces with one example each.

Reveal model answer + marking points

A conservative force does work that depends only on the initial and final positions, not on the path, and the work done in a closed loop is zero (example: gravity, spring force). A non-conservative force does path-dependent work and dissipates energy (example: friction, viscous drag).

Marking-scheme points

  • Conservative: path-independent, zero work in a loop (gravity, spring)
  • Non-conservative: path-dependent, dissipative (friction)
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PhysicsClass 113 marksmedium

Work, Energy and Power

A pump lifts 100 kg of water to a height of 10 m in 5 s. Calculate the power of the pump. Take g = 10 m/s^2.

Reveal model answer + marking points

Work done W = m g h = 100 * 10 * 10 = 10000 J. Power P = W / t = 10000 / 5 = 2000 W = 2 kW.

P = W / t = m g h / t

Marking-scheme points

  • W = m g h = 10000 J
  • P = W / t
  • P = 2000 W = 2 kW
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PhysicsClass 112 marksmedium

Work, Energy and Power

Distinguish between elastic and inelastic collisions.

Reveal model answer + marking points

In an elastic collision both linear momentum and kinetic energy are conserved (e.g. collisions between hard steel balls, or gas molecules). In an inelastic collision momentum is conserved but kinetic energy is not - some is lost as heat, sound or deformation; in a perfectly inelastic collision the bodies stick together.

Marking-scheme points

  • Elastic: momentum AND KE conserved
  • Inelastic: momentum conserved, KE not
  • Perfectly inelastic: bodies stick together
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PhysicsClass 112 markseasy

System of Particles and Rotational Motion

Define torque (moment of a force). Write its expression and SI unit.

Reveal model answer + marking points

Torque is the turning effect of a force about an axis, equal to the product of the force and the perpendicular distance of its line of action from the axis: tau = r F sin(theta) = r x F (cross product). SI unit: newton metre (N m). It is a vector along the axis of rotation.

tau = r F sin(theta)

Marking-scheme points

  • Turning effect of force
  • tau = r F sin(theta)
  • Unit N m; vector quantity
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PhysicsClass 113 marksmedium

System of Particles and Rotational Motion

State the law of conservation of angular momentum and explain why a spinning skater speeds up on pulling in the arms.

Reveal model answer + marking points

When the net external torque on a system is zero, its total angular momentum L = I omega remains constant. A spinning skater experiences almost no external torque, so I omega is constant. On pulling the arms in, the moment of inertia I decreases, so the angular speed omega increases to keep L constant - the skater spins faster.

I1 omega1 = I2 omega2

Marking-scheme points

  • No external torque => L = I omega constant
  • Pull arms in => I decreases
  • omega increases so L stays constant
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