Question

A metal chain of mass $2\, kg$ and length $90\, cm$ over hangs a table with $60\, cm$ 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 $\left.g=10\, m / s ^{2}\right)$

• A. 2 J
• B. 10 J
• C. 1 J
• D. 3 J

Answer 1

Answer: (C)

Given, mass of metal chain, $m=2\, kg$,
Total length of chain, $l=90\, cm =0.9\, m$,
Hanging length of chain, $l^{'}=(90-60)\, cm$
$=30 \,cm =0.3\, m$
The distance of centre of gravity of hanging length from the table,
$l_{c}=\frac{l^{'}}{2}=\frac{0.3}{2}=0.15\, m$
Mass of the hanging part of chain,
$m^{'} =\frac{0.30}{0.90} \times 2$
$m^{'} =\frac{2}{3}\, kg$
$\therefore $ Work needed to put hanging part of the chain back on the table,
$W =m^{'} \, g \, l_{c} $
$=\frac{2}{3} \times 10 \times 0.15=1 \, J \left(\because g=10\, m / s ^{2}\right)$