A uniform chain of mass m and length L hangs on a table with two thirds of its length on the table. The work done by a person to put the hanging part back on the table will be
A
mgL18
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B
mgL9
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C
mgL27
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D
mgL
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Solution
The correct option is AmgL18 On applying, Wext=ΔU=Uf−Ui...(1)
Let us take a small mass dm of length dx at a distance x below reference level (table) , then its potential energy is dU=−dmgx L3 length of the chain is hanging from the table.
On integrating both sides by taking limits of x=0tox=L3 U∫0dU=L/3∫0−dmgx ⇒U=−L/3∫0(mLdx)gx
[ ∵dm=λdx=mLdx] ⇒U=−mgL[x22]L/30 ∴U=−mg2L[L29−0]=−mgL18
When L3 length of the chain is hanging from the table, then potential energy Ui=−mgL18
When the hanging part is pulled back on the table, potential energy of the system is Uf=0 as h=0 from reference level.
Substituting in Eq.(1), Wext=Uf−Ui Wext=0−(−mgL18) ∴Wext=mgL18
Hence work done by the person to put the hanging part back on the table is mgL18