A smooth chain of length 2 ,m is kept on a ta | vedclass

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(B) Mass per unit length of the chain, $\lambda = \frac{M}{L} = \frac{4 \,kg}{2 \,m} = 2 \,kg/m$.Let the length of the chain hanging from the table be

Vedclass · Accepted Answer

$3.6$

Vedclass · Answer

(B) Mass per unit length of the chain, $\lambda = \frac{M}{L} = \frac{4 \,kg}{2 \,m} = 2 \,kg/m$.Let the length of the chain hanging from the table be $l = 0.6 \,m$.The mass of the hanging part is $m = \lambda 	imes l = 2 	imes 0.6 = 1.2 \,kg$.The center of mass of the hanging part is at a distance $x = \frac{l}{2} = \frac{0.6}{2} = 0.3 \,m$ below the edge of the table.The work done to pull the chain onto the table is equal to the change in potential energy of the hanging part, which is equivalent to lifting its center of mass to the level of the table.$W = m 	imes g 	imes x = 1.2 \,kg 	imes 10 \,m/s^2 	imes 0.3 \,m = 3.6 \,J$.

$A$ smooth chain of length $2 \,m$ is kept on a table such that its length of $60 \,cm$ hangs freely from the edge of the table. The total mass of the chain is $4 \,kg$. The work done in pulling the entire chain on the table is (Take, $g=10 \,m/s^2$) (in $\,J$)

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