Answer:
Weight of length L of the chain \[=Mg\] Weight of length \[\frac{L}{3}\] of the chain\[=\frac{1}{3}Mg\] As the centre of gravity of the hanging part lies at its midpoint i.e. at a distance equal to \[\text{L}/\text{6}\]below the edge of the table, so the work required to pull the hanging part on the table is \[W\text{ }=\]Force \[\times \] distance \[=\frac{1}{3}Mg\times \frac{L}{6}=\frac{MgL}{18}.\]
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