Question: The height at which the acceleration due to gravity becomes $\frac{\mathrm{g}}{9}$ (where $\mathrm{g}=$ the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :-
$\frac{R}{2}$
$\sqrt{2} \mathrm{R}$
$2 \mathrm{R}$
$\frac{R}{\sqrt{2}}$
Correct Option: , 3
Solution:
