Question

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

• A. $mgl$
• B. $\frac{mgl}{2}$
• C. $\frac{mgl}{4}$
• D. $2mgl$

Answer 1

The correct option is B

$\frac{mgl}{2}$

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 $=\frac{mgl}{2}$

$\therefore$ Work done = change in potential energy $=mg\frac{l}{2}-0=\frac{mgl}{2}$ .