Gravitational Field question from NEET 2014 (AIPMT)

Answer: (a)

1. Inside a uniform earth: $E = \frac{GM}{R^3}r$, so $|E| \propto r$ (straight line from 0 to maximum at $r = R$).
2. Outside: $E = \frac{GM}{r^2}$, so $|E| \propto \frac{1}{r^2}$, falling towards zero.
3. The field is attractive (towards the centre), so it is drawn negative.
4. Graph (a) shows the linear rise in magnitude up to $R$ and $1/r^2$ fall beyond, with negative sign. So the answer is (a).