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JEE Main 2019 (9 Apr, Shift 1) · Conservation of Energy

Conservation of Energy question from JEE Main 2019 (9 Apr, Shift 1)

Q30JEE Main 2019 · 9 APR · SHIFT 1Easy

Conservation of Energy

A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its (1/n)th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be:

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Answer: (c)

Hanging part: mass $\frac{M}{n}$ , length $\frac{L}{n}$ , centre of mass $\frac{L}{2n}$ below the edge.
Work = rise in PE of this part: $W = \frac{M}{n} g \cdot \frac{L}{2n}$
$W = \frac{MgL}{2n^2}$
So the answer is (c) $\frac{MgL}{2n^2}$ .

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