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Q.
A hole is drilled half way to the centre of the earth. A body weighs $300\, N$ on the surface of the earth. How much will, it weigh at the bottom of the hole ?
Given, distance of bottom of hole from the surface of earth $(d)=$ half the radius of earth $=\frac{R_{e}}{2}$ If $g$ be the value of gravitational acceleration on the surface of earth, then weight of body
$m g=300 N$
If $g$ 'be the gravitational acceleration at the bottom of hole, then
$g'=g\left(1-\frac{d}{R_{e}}\right)=g\left(1-\frac{\frac{R_{\theta}}{2}}{R_{e}}\right)$
$\Rightarrow g'=\frac{g}{2}$
$\therefore $ Weight of the body on the bottom of hole,
$m g'=\frac{m g}{2}=\frac{300}{2}=150 N$