STANDARD Physics Work done by a variable force MCQ Question
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, work required to pull the hanging part on to the table is
MgL
MgL/3
MgL/9
MgL/18
Correct Answer
Detailed Explanation
The weight of the hanging part of the chain is (1/3)Mg. This weight acts at the center of gravity of the hanging part, which is at a distance of L/6 from the table. As work done is force × distance, W = (Mg/3) × (L/6) = MgL/18.
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