(d) Here, $g = GM/R$ and $g^{'} = \frac{G(M/2)}{(R/2)^{2}} = \frac{2GM}{R^{2}} = 2g$ $\therefore$ % increase in $g = \left( \frac{g' - g}{g} \right) \times 100$ $= \left( \frac{2g - g}{g} \right) \times 100 = 100\%$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.