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Question

The ratio of time taken by a body to cover an equal distance while sliding down a smooth to rough identical inclined plane of angle θ is n. Find the coefficient of friction.

A
tanθ(1n2)
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B
cotθ(1n2)
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C
sinθ(1n2)
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D
tanθ(1n2)
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Solution

The correct option is D tanθ(1n2)
For a smooth inclined plane friction is zero and,
F=mamgsinθ=ma
a1=gsinθ
Using the second equation of motion,
s1=12g(sinθ)t21


For rough inclined plane:
F=ma2
mgsinθf=ma2
Since f=μN=μ(mgcosθ)
mgsinθμ(mgcosθ)=ma2
a2=g(sinθμcosθ)
Hence, s2=12g(sinθμcosθ)t22; Given: s1=s2 and t1t2=n(1n2)sinθ=μcosθ μ=tanθ(1n2)

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