question_answer

A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is      [IIT 1985; MNR 1990; AIEEE 2002; MP PMT 1994, 97, 2000; JIPMER 2000]

A)             \[MgL\]

B)             \[MgL/3\]

C)             \[MgL/9\]

D)               \[MgL/18\]

Correct Answer: D

Solution :

    \[W=\frac{MgL}{2{{n}^{2}}}=\frac{MgL}{2{{(3)}^{2}}}=\frac{MgL}{18}\]          (n = 3 given)