The height at which the weight of the body becomes $\frac{1^{\text {th }}}{16}$ of its weight on the surface…
The height at which the weight of the body becomes $\frac{1^{\text {th }}}{16}$ of its weight on the surface of the earth of radius ' $R$ ' is
2 R
3 R
4 R
5 R
Solution
At height $h=\frac{R}{n}$ the value of acceleration due to gravity is given by,
$\begin{aligned}
\quad \frac{\mathrm{g}_{\mathrm{h}}}{\mathrm{~g}} & =\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)^2=\frac{1}{16} \Rightarrow \frac{\mathrm{n}}{\mathrm{n}+1}=\frac{1}{4} \\
\therefore \quad \mathrm{n} & =\frac{1}{3} \\
\mathrm{~h} & =\frac{\mathrm{R}}{\mathrm{n}}=3 \mathrm{R}
\end{aligned}$
.