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PhysicsJEE Main 23 Jan 2026, Shift 2·Q3·MCQ·Moderate

Newton's Laws of Motion · Friction

A block is sliding down with constant velocity on an inclined plane of slope θ\theta and at an instant t=0t = 0 this block is given an upward momentum so that it starts moving up on the inclined surface with velocity uu. The distance (SS) travelled by the block before its velocity becomes zero, is .
(gg = gravitational acceleration)
  1. (1)u24gsin⁡θ\dfrac{u^2}{4g\sin\theta}
  2. (2)2u2gcos⁡θ\dfrac{2u^2}{g\cos\theta}
  3. (3)u22 gcos⁡θ\dfrac{u^2}{\sqrt2\,g\cos\theta}
  4. (4)u22gcos⁡θ\dfrac{u^2}{2g\cos\theta}
We corrected a printing error in the original paper

Assumption “with constant velocity” added (the source only says “sliding down”; uniform sliding gives μ=tan⁡θ\mu = \tan\theta, retardation 2gsin⁡θ2g\sin\theta); answer corrected; the source key is (4) — u2/(2gcos⁡θ)u^2/(2g\cos\theta) follows from neither reading (frictionless gives u2/(2gsin⁡θ)u^2/(2g\sin\theta), no option); source printed “velocity become zero”.

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