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 the acceleration due to gravity the work required to pull the hanging part of the chain on the table is:

A) \[MgL\]

B) \[\frac{1}{3}MgL\]

C)  \[\frac{1}{9}MgL\]

D)  \[\frac{1}{18}MgL\]

Correct Answer: D

Solution :

 If m is the mass per unit length of the chain, the mass of length y will be \[ym\]and the force acting on it due to gravity will be \[mgy\](assuming that y is the length of the chain hanging over the edge). So, the workdone in pulling the dy length of the chain on the table \[dW=F(-dy)\] (as\[y\]is decreasing) i.e., \[dW=(mgy)(-dy)\] (as \[F=mgy\]) So, the work done in pulling the hanging portion on the table \[W=-\int_{L/g}^{0}{mgy\,dy=\frac{mg{{L}^{2}}}{2{{(3)}^{2}}}}\] \[=\frac{mgL}{18}\] \[(as\,M=mL)\]