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Question

# A chain is held on a frictionless table with (1/n)th of its length hanging over the edge. If the chain has a length L and mass M, how much work is required to pull the hanging part back on the table ?

A
MgL2n2
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B
MgLn2
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C
MgL3n2
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D
MgL4n2
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Solution

## The correct option is A MgL2n2The chain is pulled slowly without acceleration. Let λ is the mass per unit length of the chain. λ=ML Now considering a differential element of the chain of length dy at a depth y from the surface of the table. The force acting on the element due to gravity is equal to λ(dy)g. dF=λ(dy)g So the work done in pulling this element of the chain on the table is dW=dF ycos180∘=−ydF dW=−y(λgdy) For total workdone, integrating both side W=−λg−L/n∫0ydy=−λg[y22]−L/n0 W=−λgL22n2=−MLgL22n2=−MgL2n2 So, magnitude of work is MgL2n2. Hence, option (a) is correct. Alternative approach (using center of mass): Let us take the tabletop as our reference level where potential energy is zero. Initially Ln length is hanging whose mass is MLLn=Mn and center of mass is L2n below the table. Initial PE=−MngL2n Finally, no length is hanging, and the center of mass is on the table. So, Final PE=0 As we know that Wext=Δ(U+K) Looking at initial and final situation, kinetic energy is zero in both cases. ∴Wext=PEf−PEi Wext=0−(−MngL2n)=MgL2n2

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