Question

A metre stick of mass 400 g is pivoted at one end and displaced through an angle ${60}^0$. The increase in its potential energy is:

• A. 2 J
• B. 3 J
• C. 0 J
• D. 1 J

Answer 1

The correct option is A 2 J

Consider a small element 'dx' of mass 'dm' which gets rained by height h.

change in potential energy (PE) : [Ref. image]

$du=dm.g.h$

from figure,

$dm=\frac{M}{L}dx,$ [where $M=0.4kgL=1m$]

and $h=x\sin \theta$

as $\theta ={60}^{\circ },h=\frac{\sqrt{3}}{2}x$

thus, net change in $PE$

$\Delta u=\int du=\int dm.g.h$

$={\int }_0^L\frac{M}{L}g\frac{\sqrt{3}}{2}xdx=\frac{\sqrt{3}M}{2L}{\left[\frac{x^2}{2}\right]}_0^L$

$=\frac{\sqrt{3}}{4}MgL$

putting values,

$\Delta u=\frac{\sqrt{3}}{4}\times 0.4\times 10\times 1=\sqrt{3}=1.73J$