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The upper portion of an inclined plane of inclination $α$ is smooth and the lower portion is rough. A particle slide down from rest from the top and just comes to rest at the foot. If the ratio of the smooth length to rough length is $m : n$ , the coefficient of friction is

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[\frac{m + n}{n}] \cot α

\frac{1}{2}

[\frac{m - n}{n}] \cot α

[\frac{m + n}{n}] \tan α

answer is A.

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Detailed Solution

$s_{1} : s_{2} = m : n$

$v^{2} - u^{2} = 2 as$

$v^{|^{2}} = 2 g \sin α s_{1}$

$o^{2} - {(v^{|})}^{2} = 2 [g (\sin α - μ \cos α)] s_{2}$

$= \frac{v^{|^{2}}}{- v^{|^{2}}} = \frac{2 g \sin α s_{1}}{2 g (\sin α - μ \cos α) s_{2}}$

$\sin α s_{1} = - (\sin α - μ \cos α) s_{2}$

$\sin α m = - (\sin α - μ \cos α) n$

$sin α (m + n) = μ \cos α n$

$μ = \frac{\sin α}{\cos α} (\frac{m + n}{n})$

$μ = \tan α (\frac{m + n}{n})$

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