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Question

# An uniform cable of mass M and length L is placed on a horizontal surface such that its (Ln) th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be

A
2 MgLn2
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B
nMgL
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C
MgLn2
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D
MgL2n2
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Solution

## The correct option is D MgL2n2 Given, mass of the cable is M So, mass of 1n th part of the cable i.e hanged part of the cable is = M/n ...... (i) Now, centre of mass of the hanged part will be its middle point. So, its distance from the top of the table will be L/2n. ∴ Initial potential energy of the hanged part of cable, Ui=(Mn)(−g)(L2n) ⇒Ui=−MgL2n2....(ii) When whole cable is on the table, its potential energy will be zero. ∴Uf=0....(iii) Now, using work-energy theorem, Wnet=ΔU=Uf−Ui ⇒Wnet=0−(−MgL2n2) [Using Eqs. (ii) and (iii)] ⇒Wnet=MgL2n2

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