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JEE Main 2025 · Physics · Kinematics and Laws of Motion · medium

A small block slides from rest down a rough incline of angle 30^° and takes 1.5 times as long to reach the bottom as it would on a perfectly smooth…

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A small block slides from rest down a rough incline of angle 30∘30^\circ and takes 1.51.5 times as long to reach the bottom as it would on a perfectly smooth incline of the same angle and length. The coefficient of kinetic friction between the block and the incline is
  1. A. 133\dfrac{1}{3\sqrt{3}}
  2. B. 593\dfrac{5}{9\sqrt{3}}
  3. C. 59\dfrac{5}{9}
  4. D. 123\dfrac{1}{2\sqrt{3}}
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Answer: (B) 593\dfrac{5}{9\sqrt{3}}

For the same distance ss from rest, s=12at2s=\tfrac12 a t^2, so t∝a−1/2t\propto a^{-1/2} and arough=asmooth/n2a_{\text{rough}} = a_{\text{smooth}}/n^2 with n=1.5n = 1.5. Smooth: as=gsin⁡θa_s = g\sin\theta. Rough: ar=gsin⁡θ−μgcos⁡θa_r = g\sin\theta - \mu g\cos\theta. Then gsin⁡θ−μgcos⁡θ=gsin⁡θ/2.25g\sin\theta-\mu g\cos\theta = g\sin\theta/2.25, giving μcos⁡θ=sin⁡θ(1−49)\mu\cos\theta = \sin\theta\left(1-\tfrac{4}{9}\right), i.e. μ=59tan⁡30∘=593≈0.32\mu = \tfrac59\tan 30^\circ = \dfrac{5}{9\sqrt3}\approx 0.32. Using nn instead of n2n^2 gives the wrong value 133\tfrac{1}{3\sqrt3}.
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