A uniform cable of mass ' M ' and length ' L ' is placed on a horizontal surface such that its $\left(\frac{1}{n}\right)^{\text {th }}$ 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:

(D) $\frac{\mathrm{MgL}}{\mathrm{n}^{2}}$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Work, Power & Energy

A uniform cable of mass ' M ' and length ' L ' is placed on a horizontal surface such that its

(\frac{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:

(A)nMgL
(B) $\frac{M g L}{2 n ^{2}}$ ✓ Correct
(C) $\frac{2 M g L}{n ^{2}}$
(D) $\frac{MgL}{n ^{2}}$

Answer: (B)

Solution

The hanging piece has mass

\frac{M}{n}

and length

\frac{L}{n}

, so its centre of mass sits

\frac{L}{2 n}

below the edge. Lifting that piece up to the edge raises its centre of mass by

\frac{L}{2 n}

, and the work is

W = \frac{M}{n} g \frac{L}{2 n} = \frac{M g L}{2 n ^{2}}

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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