CBSE Class 11 Physics Sample Paper 2026 (with Solutions) — Free PDF
This is a full-length CBSE Class 11 Physics paper built to the board's own pattern: 33 questions across five sections for 70 marks in 3 hours, the remainin
TL;DR: This is a full-length CBSE Class 11 Physics paper built to the board's own pattern: 33 questions across five sections for 70 marks in 3 hours, the rem…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
This is a full-length CBSE Class 11 Physics paper built to the board's own pattern: 33 questions across five sections for 70 marks in 3 hours, the remaining 30 marks being the practical component assessed in school. It covers units and measurement, kinematics, laws of motion, work energy and power, rotational motion, gravitation, thermodynamics, kinetic theory and oscillations. Every question is followed by a complete worked solution with the derivation set out step by step.
Board: CBSE · Class: Class 11 · Subject: Physics · Total: 70 marks · Time: 3 hours
Section A — Objective (1 mark each) — 16 marks
Questions 1–14 are multiple choice. Questions 15 and 16 are Assertion–Reason: mark (a) if both A and R are true and R explains A, (b) if both are true but R does not explain A, (c) if A is true and R is false, (d) if A is false and R is true.
Q1. The dimensional formula of pressure is (1 mark)
Answer: [ML⁻¹T⁻²]. Pressure is force per unit area, so [MLT⁻²]/[L²] = [ML⁻¹T⁻²].
Q2. The number of significant figures in 0.00230 is (1 mark)
Answer: Three — 2, 3 and the final 0. Leading zeros only fix the decimal point, while the trailing zero after a decimal is significant.
Q3. A body thrown vertically upward returns to the thrower. Its displacement is (1 mark)
Answer: Zero, since it ends where it began, although the distance covered is twice the maximum height.
Q4. The area under a velocity–time graph gives (1 mark)
Answer: The displacement of the body over that interval; the slope of the same graph gives the acceleration.
Q5. For a projectile launched at angle θ with speed u, the range is maximum when θ equals (1 mark)
Answer: 45°. The range is (u² sin 2θ)/g, which is greatest when sin 2θ = 1, that is 2θ = 90°.
Q6. The centripetal acceleration of a body moving in a circle of radius r with speed v is (1 mark)
Answer: v²/r, directed towards the centre of the circle.
Q7. Newton's first law of motion defines (1 mark)
Answer: Inertia, and thereby the concept of force — a body changes its state of rest or uniform motion only when an external force acts.
Q8. The impulse of a force equals the change in (1 mark)
Answer: Momentum. Impulse = F·Δt = Δp, which is the impulse–momentum theorem.
Q9. A body of mass 2 kg moving at 3 m/s has kinetic energy (1 mark)
Answer: 9 J. KE = ½mv² = ½(2)(9) = 9 joules.
Q10. The work done by a centripetal force on a body in uniform circular motion is (1 mark)
Answer: Zero, because the force is always perpendicular to the displacement, and W = Fs cos 90° = 0.
Q11. The moment of inertia of a solid sphere of mass M and radius R about its diameter is (1 mark)
Answer: (2/5)MR². For a hollow sphere the corresponding value is (2/3)MR².
Q12. The escape velocity from the surface of the Earth is approximately (1 mark)
Answer: 11.2 km/s, obtained from v = √(2GM/R) — independent of the mass of the escaping body.
Q13. In an isothermal process the quantity that stays constant is (1 mark)
Answer: Temperature, so for an ideal gas the internal energy is unchanged and the heat supplied equals the work done.
Q14. The time period of a simple pendulum depends on (1 mark)
Answer: Its length and the acceleration due to gravity, T = 2π√(l/g) — it does not depend on the mass of the bob or on the amplitude, provided the swing is small.
Q15. Assertion (A): A body can have zero velocity and still be accelerating. Reason (R): Acceleration is the rate of change of velocity. (1 mark)
Answer: (a) Both true and R explains A. At the top of its flight a body thrown upward has zero velocity but its velocity is still changing at g, so it is accelerating — exactly what the reason describes.
Q16. Assertion (A): The internal energy of an ideal gas depends only on its temperature. Reason (R): There are no intermolecular forces in an ideal gas. (1 mark)
Answer: (a) Both true and R explains A. With no intermolecular forces there is no potential energy term, so the internal energy is purely the kinetic energy of the molecules, which depends only on temperature.
Section B — Very Short Answer (2 marks each) — 10 marks
Answer all 5 questions. Each carries 2 marks.
Q1. State the principle of homogeneity of dimensions and give one use of it. (2 marks)
Answer: The principle states that every term on both sides of a physically correct equation must have the same dimensions. It is used to check the correctness of an equation and to derive relations between quantities — for example it shows at once that s = ut + ½at² is dimensionally consistent, since every term has the dimension of length.
Q2. Distinguish between distance and displacement. (2 marks)
Answer: Distance is the total length of the path actually covered, is a scalar, and can never be negative or zero for a moving body. Displacement is the shortest straight-line vector from the initial to the final position, has both magnitude and direction, and can be zero even when the distance is large — as for a body returning to its starting point.
Q3. Why is it easier to pull a lawn roller than to push it? (2 marks)
Answer: When the roller is pulled, the vertical component of the applied force acts upward and reduces the normal reaction, so friction, which is proportional to the normal reaction, decreases. When it is pushed, that component presses downward, increasing the normal reaction and hence the friction. Pulling therefore needs a smaller effort.
Q4. Define angular momentum and state the condition under which it is conserved. (2 marks)
Answer: Angular momentum is the rotational analogue of linear momentum, L = Iω for a rigid body, or r × p for a particle. It is conserved when the net external torque acting on the system is zero — which is why a spinning skater speeds up on drawing her arms in, reducing I and so raising ω.
Q5. State the first law of thermodynamics and identify each term. (2 marks)
Answer: ΔQ = ΔU + ΔW, where ΔQ is the heat supplied to the system, ΔU is the increase in its internal energy, and ΔW is the work done by the system on its surroundings. It is a statement of conservation of energy applied to thermal processes.
Section C — Short Answer (3 marks each) — 21 marks
Answer all 7 questions. Each carries 3 marks.
Q1. Derive the equation v = u + at using the definition of acceleration. (3 marks)
Answer: Acceleration is the rate of change of velocity, a = (v − u)/t for uniform acceleration, where u is the velocity at t = 0 and v at time t. Multiplying both sides by t gives at = v − u, and rearranging gives v = u + at. The derivation assumes the acceleration is constant throughout the interval.
Q2. A ball is thrown at 20 m/s at 30° to the horizontal. Find its time of flight and maximum height. (g = 10 m/s²) (3 marks)
Answer: The vertical component of the initial velocity is u sin θ = 20 × 0.5 = 10 m/s. Time of flight T = 2u sin θ/g = 2 × 10/10 = 2 s. Maximum height H = (u sin θ)²/2g = 100/20 = 5 m.
Q3. Explain why a passenger falls forward when a moving bus stops suddenly. (3 marks)
Answer: While the bus is moving, the passenger's whole body moves with it at the same speed. When the brakes are applied, the force acts on the bus and, through friction with the floor, on the passenger's feet, which stop with it. The upper part of the body has no such force acting on it and, by the law of inertia, continues to move forward at the original speed. The result is that the upper body is thrown forward relative to the feet.
Q4. A machine gun fires 60 bullets per minute, each of mass 50 g at 400 m/s. Calculate the force needed to hold the gun. (3 marks)
Answer: Sixty bullets per minute is one bullet per second. The momentum carried away by each bullet is p = mv = 0.05 × 400 = 20 kg m/s. The rate of change of momentum is therefore 20 kg m/s per second, and by Newton's second law the force required to hold the gun steady against the recoil is 20 N.
Q5. State and prove the work–energy theorem for a constant force. (3 marks)
Answer: The theorem states that the work done by the net force on a body equals the change in its kinetic energy. For a constant force F producing acceleration a over a displacement s, W = Fs = mas. From the equation of motion v² = u² + 2as, we have as = (v² − u²)/2. Substituting, W = m(v² − u²)/2 = ½mv² − ½mu², which is the final kinetic energy minus the initial kinetic energy.
Q6. Derive an expression for the orbital velocity of a satellite close to the Earth's surface. (3 marks)
Answer: For a satellite of mass m in a circular orbit of radius r about the Earth of mass M, the gravitational attraction supplies the centripetal force: GMm/r² = mv²/r. Cancelling m and one factor of r gives v² = GM/r, so v = √(GM/r). For an orbit close to the surface r ≈ R, and since GM = gR², this becomes v = √(gR) ≈ √(9.8 × 6.4 × 10⁶) ≈ 7.9 km/s.
Q7. Explain why the boiling point of water decreases at high altitudes. (3 marks)
Answer: A liquid boils when its saturated vapour pressure becomes equal to the external pressure on its surface. At high altitude the atmospheric pressure is lower, so the vapour pressure reaches that smaller value at a lower temperature and the water boils below 100 °C. Because the water is cooler, food takes longer to cook, which is why a pressure cooker — which raises the pressure and hence the boiling point — is used in the hills.
Section D — Case-based questions (4 marks each) — 8 marks
Answer both case-based questions. Each has sub-parts totalling 4 marks.
Q1. CASE STUDY 1 — A car of mass 1000 kg travelling at 20 m/s is brought to rest in 5 s by braking. (i) Find the retardation. (ii) Find the braking force. (iii) Find the distance travelled while stopping. (iv) State the energy conversion that takes place. (4 marks)
Answer: (i) a = (v − u)/t = (0 − 20)/5 = −4 m/s², a retardation of 4 m/s². (ii) F = ma = 1000 × 4 = 4000 N, opposing the motion. (iii) s = ut + ½at² = 20(5) + ½(−4)(25) = 100 − 50 = 50 m. (iv) The kinetic energy of the car, ½(1000)(400) = 2 × 10⁵ J, is converted almost entirely into heat in the brake pads and tyres, with a little into sound.
Q2. CASE STUDY 2 — A simple pendulum of length 1 m oscillates with small amplitude at a place where g = 9.8 m/s². (i) Write the formula for its time period. (ii) Calculate the time period. (iii) What happens to the period if the bob's mass is doubled? (iv) What happens if the pendulum is taken to the Moon? (4 marks)
Answer: (i) T = 2π√(l/g). (ii) T = 2π√(1/9.8) = 2π × 0.3194 ≈ 2.0 s. (iii) Nothing — the period is independent of the mass of the bob, since the restoring force and the inertia both scale with mass and the effect cancels. (iv) On the Moon g is about one sixth of its value on Earth, and since T is inversely proportional to √g, the period becomes about √6 ≈ 2.45 times longer, roughly 4.9 s — the pendulum swings much more slowly.
Section E — Long Answer (5 marks each) — 15 marks
Answer all 3 questions. Each carries 5 marks.
Q1. (a) State the law of conservation of linear momentum and derive it from Newton's third law. (b) Explain the recoil of a gun using it. (c) A 4 kg gun fires a 20 g bullet at 200 m/s — find the recoil velocity. (5 marks)
Answer: (a) In the absence of an external force, the total linear momentum of a system of bodies remains constant. Consider two bodies A and B colliding. By the third law the force A exerts on B is equal and opposite to that B exerts on A, so F_AB = −F_BA. Since force is the rate of change of momentum, the momentum gained by one is exactly the momentum lost by the other over the same interval of contact, and the total change is zero. Hence the total momentum before the collision equals that after it. (b) Before firing, gun and bullet are at rest and the total momentum is zero. After firing, the bullet carries forward momentum, so the gun must carry an equal momentum backwards for the total to remain zero — this backward motion is the recoil. (c) 0 = m₁v₁ + m₂v₂, so (0.02)(200) + (4)v = 0, giving 4v = −4 and v = −1 m/s. The gun recoils at 1 m/s, opposite to the bullet.
Q2. (a) Define the coefficient of static and kinetic friction. (b) Why is kinetic friction less than static friction? (c) A block of mass 5 kg rests on a horizontal surface with μs = 0.4 — find the least force needed to start it moving. (g = 10 m/s²) (5 marks)
Answer: (a) The coefficient of static friction μs is the ratio of the maximum force of static friction to the normal reaction, μs = f_max/N — it governs the force needed to start motion. The coefficient of kinetic friction μk is the ratio of the force of kinetic friction to the normal reaction while the body slides, μk = f_k/N. (b) When two surfaces are at rest in contact, the microscopic irregularities interlock and cold welds form at the points of contact, and these must be broken to begin motion. Once sliding begins the surfaces are in contact at any given point for too short a time for such bonds to form fully, so the resisting force is smaller — hence μk < μs. (c) N = mg = 5 × 10 = 50 N. The limiting friction is f_max = μs N = 0.4 × 50 = 20 N, so a force just exceeding 20 N is needed to start the block moving.
Q3. (a) State the assumptions of the kinetic theory of gases. (b) Derive the relation between the pressure of an ideal gas and the mean square speed of its molecules. (c) Hence show that the average kinetic energy per molecule is proportional to the absolute temperature. (5 marks)
Answer: (a) A gas consists of a very large number of identical molecules in constant random motion; their own volume is negligible compared with the volume of the container; there are no intermolecular forces except during collisions; all collisions between molecules and with the walls are perfectly elastic; and the time of a collision is negligible compared with the time between collisions. (b) Consider N molecules each of mass m in a cube of side L. A molecule moving along x with velocity vₓ rebounds elastically from a wall, changing its momentum by 2mvₓ, and returns to the same wall after a time 2L/vₓ. The force it exerts is therefore 2mvₓ/(2L/vₓ) = mvₓ²/L. Summing over all molecules and dividing by the wall area L² gives P = m(Σvₓ²)/L³. Since the motion is random, the mean square speed satisfies v² = vₓ² + v_y² + v_z² = 3vₓ², so P = (1/3)(Nm/V)v̄², that is PV = (1/3)Nmv̄². (c) Comparing with the ideal gas equation PV = NkT gives (1/3)Nmv̄² = NkT, so ½mv̄² = (3/2)kT. The average kinetic energy per molecule is thus (3/2)kT — directly proportional to the absolute temperature and independent of the nature of the gas.
FAQs
What is the total marks for Class 11 Physics?
70 marks for theory and 30 marks for practicals; total 100 marks.
Which topics are most important?
Laws of motion, energy conservation, thermodynamics, and waves carry high weightage.
Are derivations important in physics?
Yes, derivations of major equations are essential and often asked as part of long-answer questions.
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