Question

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 same inclination θ, it takes times nth to reach the bottom where n number greater than 1. The coefficient of friction μ is given by:

A
μ=tanθ(11/n2)
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B
μ=cotθ(11/n2)
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C
μ=tanθ(11/n2)1/2
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D
μ=cotθ(11/n2)1/2
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Solution

The correct option is B μ=tanθ(1−1/n2)Van=√2Lg(sinθ−μcosθ)⟶(1)Va=√2Lgsinθ⟶(2)By putting equ(2) in eq(1) we get√2Lgsinθn=√2Lg(sinθ−μcosθ)2Lgsinθn2=2Lg(sinθ−μcosθ)sinθn2=sinθ−μcosθμcosθ=sinθ−sinθn2μcosθ=sinθ(1−1n2)μ=tanθ(1−1n2)

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