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Question

A uniform chain of mass \(M\) and length \(L\) rests on a rough table such that one of its ends hangs over the edge. The chain starts sliding off the table by itself when overhanging part is equal to \(\dfrac{L}{4}\). The total work done by frictional force, till the moment the other end of chain slips-off the table, is

\( \Leftrightarrow M _{2} L \)

A
MgL
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B
3MgL32
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C
MgL8
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D
3MgL40
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Solution

The correct option is B 3MgL32
μ×(3Mg4)=Mg4
μ=13
Wf=3L40μML×x×g×dx=3MgL32

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