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Q.

The coefficient of friction between a chain and a table is $μ$ . If a chain is placed on a horizontal table so that a part of it is hanging from one end, the minimum fraction of length of the chain that can be on the table, so that the chain may not slip off is

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a

zero

b

1

c

$\frac{μ}{μ + 1}$

d

$\frac{1}{μ + 1}$

answer is B.

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Detailed Solution

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$Weight of the hanging part is balanced by fricitional force on the table if mass per unit length of the string is m and x length is on the table then frictional force = xmμg weight of the hanging part = (L - x) mg xmμg = (L - x) mg μx = (L - x) x (μ + 1) = L \frac{x}{L} = \frac{1}{μ + 1}$

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