KCET · Physics · Work Power Energy
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, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(6.3 \mathrm{~J}\)
- B \(3.6 \mathrm{~J}\)
- C \(2.0 \mathrm{~J}\)
- D \(12.9 \mathrm{~J}\)
Answer & Solution
Correct Answer
(B) \(3.6 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
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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