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A metal chain of mass $2 \mathrm{~kg}$ and length $90 \mathrm{~cm}$ over hangs a table with $60 \mathrm{~cm}$ on the table. How much work needs to be done to put the hanging part of the chain back on the table? $\left(\right.$ Let $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)$
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The correct answer is:
$1 \mathrm{~J}$
Given, mass of metal chain, $m=2 \mathrm{~kg}$,
Total length of chain, $l=90 \mathrm{~cm}=0.9 \mathrm{~m}$, Hanging length of chain, $l^{\prime}=(90-60) \mathrm{cm}$ $=30 \mathrm{~cm}=0.3 \mathrm{~m}$
The distance of centre of gravity of hanging length from the table,
$l_c=\frac{l^{\prime}}{2}=\frac{0.3}{2}=0.15 \mathrm{~m}$
Mass of the hanging part of chain,
$\begin{aligned} & m^{\prime}=\frac{0.30}{0.90} \times 2 \\ & m^{\prime}=\frac{2}{3} \mathrm{~kg}\end{aligned}$
$\therefore$ Work needed to put hanging part of the chain back on the table,
$\begin{aligned} W & =m^{\prime} g l_c \\ & =\frac{2}{3} \times 10 \times 0.15=1 \mathrm{~J}\left(\because g=10 \mathrm{~m} / \mathrm{s}^2\right)\end{aligned}$
Total length of chain, $l=90 \mathrm{~cm}=0.9 \mathrm{~m}$, Hanging length of chain, $l^{\prime}=(90-60) \mathrm{cm}$ $=30 \mathrm{~cm}=0.3 \mathrm{~m}$
The distance of centre of gravity of hanging length from the table,
$l_c=\frac{l^{\prime}}{2}=\frac{0.3}{2}=0.15 \mathrm{~m}$
Mass of the hanging part of chain,
$\begin{aligned} & m^{\prime}=\frac{0.30}{0.90} \times 2 \\ & m^{\prime}=\frac{2}{3} \mathrm{~kg}\end{aligned}$
$\therefore$ Work needed to put hanging part of the chain back on the table,
$\begin{aligned} W & =m^{\prime} g l_c \\ & =\frac{2}{3} \times 10 \times 0.15=1 \mathrm{~J}\left(\because g=10 \mathrm{~m} / \mathrm{s}^2\right)\end{aligned}$
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