A) \[2mgl\]
B) \[\frac{mgl}{5}\]
C) \[\frac{mgl}{9}\]
D) \[\frac{mgl}{18}\]
Correct Answer: D
Solution :
The mass hanging over the edge is\[\frac{mg}{6}\], so initial force required to pull the hanging mass on the tablet. The final force required to pull the remaining mass\[=0\]. So, Average force\[=\frac{mg}{6}\]. Displacement in the chain\[=\frac{l}{3}\]. Thus, work done by the variable force\[=\]force\[\times \]distance\[=\frac{mgl}{2}{{x}^{2}}\]where \[\frac{1}{x}\] is the fraction of the length of the chain which is initially hanging. Here,\[x=3\] So, work done\[=mg\frac{l}{2\times {{3}^{2}}}=mg\frac{l}{18}\]
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