An object is placed on the surface of a smooth inclined plane of inclination θ. It takes time t, to reach the bottom. If the same object is allowed to slide down a rough inclined plane of inclination q, it takes time n t to reach the bottom where n is a number greater than one. The coefficient of friction μ is given by

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μ=tan⁡θ1−1n2

μ=cot⁡θ1−1n2

μ=tan⁡θ1−1n21/2

μ=cot⁡θ1−1n21/2

answer is A.

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Detailed Solution

For smooth inclined plane, acceleration = g sin qFor rough inclined plane, acceleration =gsin⁡θ−μgcos⁡θLet l be the length of inclined plane, then l=12(gsin⁡θ)t2 …(1)also l=12(gsin⁡θ−μgcos⁡θ)(nt)2 …(2)∴ 12(gsin⁡θ)t2=12(gsin⁡θ−μgcos⁡θ)(nt)2Solving, we getμ=tan⁡θ1−1n2.

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