Physics · Laws of motion · chapter 4 of 14

Class 11 Physics Chapter 4: Laws of Motion MCQs with answers

10 multiple-choice questions on Laws of Motion, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
A hockey puck slides across a smooth, level sheet of ice in a straight line at a steady 4 m s⁻¹. Neglecting air resistance, the net force on the puck is
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Answer: (a) zero

The puck moves with constant velocity, so its acceleration is zero; by the first law the net external force on it is zero. Its weight is cancelled by the normal force of the ice.

Question 2
Which pair of quantities has the same SI unit?
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Answer: (d) impulse and momentum

Impulse = force × time = change in momentum, so both are measured in N s = kg m s⁻¹, with dimensions [MLT⁻¹]. Force is in N, [MLT⁻²].

Question 3
For the action and reaction forces between two bodies, which statement is correct?
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Answer: (c) They act at the same instant, on different bodies.

The two mutual forces are simultaneous, equal and opposite, and act on different bodies; there is no cause and effect between them, and they cannot cancel because they do not act on the same body.

Question 4
A bar magnet lying on a table pulls a small iron nail held near it with a force of 0.2 N. The force exerted by the nail on the magnet is
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Answer: (b) 0.2 N, directed towards the nail

Force on the magnet by the nail is equal and opposite to the force on the nail by the magnet. The magnet pulls the nail towards itself, so the nail pulls the magnet towards the nail with 0.2 N. The forces act at a distance; contact is not needed.

Question 5
A steady force of 6 N acts for 2 s on a 1 kg cart and, separately, for 2 s on a 4 kg cart, each starting from rest on a smooth floor. The ratio of their final momenta, p₁/p₂ (1 kg cart to 4 kg cart), is
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Answer: (b) 1

Change in momentum = impulse = F × t = 6 × 2 = 12 N s for each cart, whatever its mass. The lighter cart moves faster (12 m s⁻¹ against 3 m s⁻¹) but the momenta are equal, so p₁/p₂ = 1.

Question 6
A ball of mass 0.2 kg moving at 15 m s⁻¹ strikes a wall head-on and bounces straight back. The wall gives it an impulse of magnitude 5 N s. The speed of the ball after the bounce is
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Answer: (a) 10 m s⁻¹

Take the rebound direction as positive. Impulse = pfinal − pinitial: 5 = 0.2v − (−0.2 × 15) = 0.2v + 3, so v = 10 m s⁻¹. Forgetting the reversal gives 25 m s⁻¹.

Question 7
A 5 kg crate rests on a floor; the coefficient of static friction is 0.4 (g = 10 m s⁻²). A horizontal push of 12 N is applied and the crate does not move. The frictional force on the crate is
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Answer: (d) 12 N

The limiting friction is μₛN = 0.4 × 50 = 20 N. The push of 12 N is less, so static friction self-adjusts to balance it: fₛ = 12 N. Writing fₛ = μₛN gives 20 N, which is wrong here.

Question 8
An artificial satellite moves round the earth in a circular orbit. The centripetal force on it is
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Answer: (c) the gravitational pull of the earth on it

Centripetal force is not a new kind of force; it is the name for whatever material force gives the inward acceleration v²/R. For a satellite that is the earth's gravitational pull, just as the sun's pull does for a planet.

Question 9
A spring of force constant 200 N m⁻¹ is stretched by 3 cm from its natural length. The magnitude of the restoring force is
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Answer: (c) 6 N

F = kx = 200 × 0.03 = 6 N, directed opposite to the extension. Using x in centimetres gives 600 N.

Question 10
Ball bearings are fitted between the moving parts of a machine mainly because
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Answer: (b) rolling friction is much smaller than sliding friction for the same load

Rolling friction is smaller than sliding friction, even by two or three orders of magnitude, so less power is lost as heat. It is not zero, and friction does not depend on the area of contact.

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