Question

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

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

Answer 1

Answer: (B)

$\frac { m g l } { 2 }$

Answer 2

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

energy = 0

But when we make its stand vertical its centre of mass rises upto high $= \frac { m g l } { 2 }$

∴ Work done = charge in potential energy

$= m g \frac { l } { 2 } - 0 = \frac { m g l } { 2 }$.