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A smooth chain of length $2 \mathrm{~m}$ is kept on a table such that its length of $60 \mathrm{~cm}$ hangs frecly from the edge of the table. The total mass of the chain is $4 \mathrm{~kg}$. The work done in pulling the entire chain on the table is (Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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Verified Answer
The correct answer is:
$3.6 \mathrm{~J}$
Mass per unit length of chain,
$$
\frac{M}{L}=\frac{4}{2}=2 \mathrm{~kg} / \mathrm{m}
$$
Work done in pulling the chain of small length $d x$.
$$
\begin{aligned}
d W & =\text { mass of length } d x \times g \times x \\
& =\frac{M}{L} \times d x \times g \times x \\
& =2 \times d x \times 10 \times x=20 x d x
\end{aligned}
$$
Work done, $W=\int_0^{0.6} d W d x=\int_0^{0.6} 20 x d x$
$$
=\left[20 \frac{x^2}{2}\right]_0^{0.6}=\left[10 x^2\right]_0^{0.6}=10 \times(0.6)^2=36 \mathrm{~J}
$$
$$
\frac{M}{L}=\frac{4}{2}=2 \mathrm{~kg} / \mathrm{m}
$$
Work done in pulling the chain of small length $d x$.
$$
\begin{aligned}
d W & =\text { mass of length } d x \times g \times x \\
& =\frac{M}{L} \times d x \times g \times x \\
& =2 \times d x \times 10 \times x=20 x d x
\end{aligned}
$$
Work done, $W=\int_0^{0.6} d W d x=\int_0^{0.6} 20 x d x$
$$
=\left[20 \frac{x^2}{2}\right]_0^{0.6}=\left[10 x^2\right]_0^{0.6}=10 \times(0.6)^2=36 \mathrm{~J}
$$
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