359696
The height at which the weight of a body becomes \(1 / 16\) th, its weight on the surface of the earth (radius \(R\)), is
1 \(3R\)
2 \(15 R\)
3 \(5R\)
4 \(4R\)
Explanation:
According to the question, \(\dfrac{G M m}{(R+h)^{2}}=\dfrac{1}{16} \dfrac{G M m}{R^{2}}\) where, \(m=\) mass of the body and \(\dfrac{G M}{R^{2}}=\) gravitational acceleration \(\begin{aligned}& \dfrac{1}{(R+h)^{2}}=\dfrac{1}{16 R^{2}} \\& \text { or } \dfrac{R}{R+h}=\dfrac{1}{4} \text { or } \dfrac{R+h}{R}=4 \\& h=3 R\end{aligned}\)
