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Q.

A homogeneous chain lies in limiting equilibrium on a horizontal table of coefficient of friction 0.5 with part of it hanging over the edge of the table. The fractional length of the chain hanging down the edge of the table is

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a

$1/2$

b

$1/5$

c

$1/3$

d

$2/3$

answer is C.

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Detailed Solution

We know that $m g μ (L - l) = m g l$

$μ L - μ l = l$

$μ L = l (μ + 1)$

$\frac{l}{L} = \frac{μ}{1 + μ} = \frac{0.5}{1 + 0.5} = \frac{0.5}{1.5} = \frac{5}{15} = \frac{1}{3}$

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