A body of mass ' $\mathrm{m}$ ' is raised through a height above the earth's surface so that the increase in…
A body of mass ' $\mathrm{m}$ ' is raised through a height above the earth's surface so that the increase in potential energy is $\frac{\mathrm{mgR}}{5}$. The height to which the body is raised is ( $\mathrm{R}=$ radius of earth, $\mathrm{g}=$ acceleration due to gravity)
$\mathrm{R}$
$\frac{\mathrm{R}}{2}$
$\frac{\mathrm{R}}{4}$
$\frac{\mathrm{R}}{8}$
Solution
When a particle of mass $m$ is taken from the Earth's surface to a height $h=n R$, then the change in P.E. can be calculated as,
$\begin{array}{ll}
& \Delta \mathrm{U}=\operatorname{mgR}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right) \\
\therefore & \frac{\mathrm{mgR}}{5}=\mathrm{mgR}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right) \\
\therefore \quad \mathrm{n}+1=5 \mathrm{n} & \mathrm{n}=\frac{1}{4} \\
\therefore \quad & \mathrm{h}=\frac{\mathrm{R}}{4}
\end{array}$