At a place, value of \( g \) is less by \( 1 \% \) than its value on the surface of the Earth (Radius of Earth \( =6400 \mathrm{~km} \) ). The place is
- \( 64 \mathrm{~km} \) below the surface of the Earth
- \( 64 \mathrm{~km} \) above the surface of the Earth
- \( 30 \mathrm{~km} \) above the surface of the Earth
- \( 32 \mathrm{~km} \) below the surface of the Earth.
Here, $g$ is the gravitational acceleration on the surface and let $g_{d}$ be the $1$ % less gravity at the distance $d$ from the surface.
Radius of the earth $R=6400\ km$
$g_d=g(1-\frac{d}{R})$
Or $0.99g=g(1-\frac{d}{R})$
Or $0.99=1-\frac{d}{R}$
Or $\frac{d}{R}=1-0.99$
Or $\frac{d}{R}=0.01$
Or $d=0.01R$
Or $d=0.01\times 6400\ km$
Or $d=64\ km$
Therefore, at $64\ km$ below the surface of the earth $g$ is $1$ % less.
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