A rod of mass m and length l is lying on a horizontal table. The work done in making it stand vertically on one end will be

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Solution

The correct option is B

When the rod is lying on a horizontal table, its potential energy = 0

But when we make it stand vertical its centre of mass rises upto height $\frac{l}{2}$ . So it's potential energy $= \frac{m g l}{2}$

$∴$ Work done = change in potential energy $= m g \frac{l}{2} - 0 = \frac{m g l}{2}$ .

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A rod of mass m and length l is lying on a horizontal table. The work done in making it stand vertically on one end will be

The correct option is B When the rod is lying on a horizontal table, its potential energy = 0 But when we make it stand vertical its centre of mass rises upto height l2 . So it's potential energy =mgl2 ∴ Work done = change in potential energy =mgl2−0=mgl2 .