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Q.

A block, released from rest from the top of a smooth inclined plane of inclination θ, has a speed v when it reaches the bottom. The same block, released from the top of a rough inclined plane of the same inclination θ, has a speed v/n on reaching the bottom, where n is a number greater than unity. The coefficient of friction is given by

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a

μ=1−1n2tan⁡θ

b

μ=1−1n2cot⁡θ

c

μ=1−1n21/2tan⁡θ

d

μ=1−1n21/2cot⁡θ

answer is A.

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Detailed Solution

We use the relation v2−u2=2as. Since u = 0, we have v2=2as.Now v12=2a1s or v2=2g sin⁡θ×sand v22=2a2s or v2n2=2(gsin⁡θ−μgcos⁡θ)×sDividing, we get  n2(sin⁡θ−μcos⁡θ)=sin⁡θwhich gives μ=1−1n2tan⁡θ, which is choice (1).
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A block, released from rest from the top of a smooth inclined plane of inclination θ, has a speed v when it reaches the bottom. The same block, released from the top of a rough inclined plane of the same inclination θ, has a speed v/n on reaching the bottom, where n is a number greater than unity. The coefficient of friction is given by