question_answer

A homogeneous chain of length L lies on a table. If the coefficient of friction is\[\mu \], the maximum length of the chain that can hang over the table is

A)  \[\frac{\mu L}{1-\mu }\]

B)  \[\frac{\mu L}{1+\mu }\]

C)  \[\frac{L}{\mu }\]

D)  \[\frac{L}{1+\mu }\]

Correct Answer: B

Solution :

: Let\[m=\]mass of unit length. The hanging part\[(l)\]pulls the chain\[(L-l)\]to right by a force \[=(wl)g\] Force of friction\[=\mu \times R=\mu m(L-l)g\] Equate the two force for equilibrium \[\therefore \] \[m\lg =\mu m(L-l)g\] Or \[l=\mu (L-l)\] or \[l=\mu L-\mu l\] Or \[l=\frac{\mu L}{(1+\mu )}\]