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 pan 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 \[m\]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 work done 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                 \[W=-\int_{L/g}^{0}{mgy\,\,dy=\frac{mg{{L}^{2}}}{2{{(3)}^{2}}}}\]