Q91 / 125 · BITSAT 2009 · Physics
A chain of mass $\mathrm{M}$ is placed on a smooth table with $1 / \mathrm{n}$ of its length $\mathrm{L}$ hanging over the edge. The work done in pulling the hanging portion of the chain back to the surface of the table is
- A$\mathrm{MgL} / \mathrm{n}$
- B$\mathrm{MgL} / 2 \mathrm{n}$
- C$\mathrm{MgL} / \mathrm{n}^{2}$
- Correct answerD$\mathrm{MgL} / 2 \mathrm{n}^{2}$
Solution
$\quad \mathrm{W}=$ change in $\mathrm{PE}$ of $\mathrm{COM}$ of hanging part
$$
=\frac{\mathrm{M}}{\mathrm{n}} \mathrm{g} \frac{\mathrm{L}}{2 \mathrm{n}}=\frac{\mathrm{MgL}}{2 \mathrm{n}^{2}}
$$
Hence, the correct answer is option: 'D'.