Question

A rod of mass `m` and length `l` is lying on a horizontal table. Work done in making it stand on one end will be
A. `mgl`
B. `mgl//2`
C. `mgl//4`
D. `2mgl`

Answer 1

(b) Work done = final PE. - initial PE.
= `mg (1//2) - 0 = mgl//2`
(as when the rod is standing on one end, its centre of gravity is at a distance `1//2` from the table and when it is lying on the table, its CG is on the table, i.e., at zero distance from the table)