Question

A rod of length $l$ and mass $m$ fixed at one end, it hanging vertically. The other end is now raised so that the rod makes an angle ${30}^o$ with horizontal line. The work done in this process will be:

• A. $3mgl/4$
• B. $mgl/2$
• C. $mgl/3$
• D. $mgl/4$

Answer 1

Answer: (A)

$3mgl/4$

Initial potential energy $=\frac{-mgl}{2}\text{(since we have take top point of rod as origin and axis is vertically upwards)}$

Final potential energy$=\frac{mg\left(lsin{30}^o\right)}{2}$

$=\frac{mgl}{4}$

Work done is change in potential energy

Work done $=\frac{3mgl}{4}$