Answer:
Option D
Explanation:
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
mg= 300N
If 'g' be the gravitational acceleration at the bottom of hole , then
$g'=g\left(1-\frac{d}{R_{e}}\right)=g\left(1-\frac{\frac{R_{e}}{2}}{R_{e}}\right)\Rightarrow g'=\frac{g}{2}$
$\therefore$ Weight of the body on the bottom of hole,
$mg'=\frac{mg}{2}=\frac{300}{2}=150 N$
