(d) $\frac{1}{2}mv^{2}\left( \frac{K^{2}}{R^{2}} \right) = 40\%\ \ \frac{1}{2}mv^{2} \Rightarrow \ \therefore\frac{K^{2}}{R^{2}} = \frac{40}{100} = \frac{2}{5}$ $i.e.$ the body is solid sphere
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.