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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

a
μ=tan⁡θ1−1n2
b
μ=cot⁡θ1−1n2
c
μ=tan⁡θ1−1n21/2
d
μ=cot⁡θ1−1n21/2

detailed solution

Correct option is A

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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Similar Questions

A skier starts from rest at point A and slides down the hill without turning or breaking. The friction coefficient is μ. When he stops at point B, his horizontal displacement is s. What is the height difference between points A and B

(The velocity of the skier is small, so that the additional pressure on the snow due to the curvature
can be neglected. Neglect also the friction of air and the dependence of μ on the velocity of the skier.) 


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