Solution

Acceleration due to gravity at depth $d$: $g'=g\left(1-\frac{d}{R}\right)$ Multiplying by $m$: $mg'=mg\left(1-\frac{d}{R}\right)$ Half way down, $d=\frac{R}{2}$: $mg'=200\left(1-\frac{R}{2R}\right)=100\ N$

Concept weight at half the depth to the centre

Gravitation in past papers

66 questions from this chapter have appeared across 26 exam years.