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Q.
A stone weighs $100\, N$ on the surface of the Earth. The ratio of its weight at a height of half the radius of the Earth to a depth of half the radius of the earth will be approximately
Weight of a mass depends on the acceleration due to gravity $(g)$.
Acceleration due to gravity at height $h = \bigg( = \frac{R}{2} \bigg)$ from the surface of earth
$g_{h}=\frac{g}{\left(1+\frac{\frac{R}{2}}{R}\right)^{2}} = \frac{4}{9} g$ ....(i)
Acceleration due.to gravity at depth $ d \bigg( = \frac{R}{2} \bigg)$ from the surface of earth
$g_{d} = g \left(1 - \frac{\frac{R}{2}}{R}\right) - g \left(1 -\frac{1}{2}\right) = \frac{g}{2} $
.....(ii)
Required ratio = $\frac{W_{h}}{W_{d}} =\frac{mg_{h}}{mg_{d}} = \frac{\frac{4}{9}g}{\frac{g}{2}} = \frac{8}{9} =0.9$