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Q.
How much deep inside the earth (radius $R$ ) should a man go, so that his weight becomes one-fourth of that on the earth's surface?
ManipalManipal 2020
Solution:
Suppose depth inside the earth is $h$
As we know that, inside the earth
$g_{i} \propto\left(\frac{R-h}{R^{3}}\right)$ ...(i)
On the surface of the earth
$g_{s} \propto \frac{1}{R^{2}}$ ... (ii)
From Eqs. (i) and (ii), we get
$\frac{g_{i}}{g_{s}}=\frac{(R-h) R^{2}}{R^{3}}=\frac{R-h}{R}$
$\Rightarrow \frac{1}{4}=\frac{R-h}{R}$
$\Rightarrow R=4 R-4 h$
$\Rightarrow 3 R=4 h$
$\Rightarrow h=\frac{3}{4} R$