UPSC CDS Math Volume and Surface Area Questions

We have, A cylinder height = 10 cm and radius is 6 cm. Volume of a solid cylinder $\;=\pi r^{2} h$ $\;\; =\pi \times 6^{2} \times 10$ $\;\; =360 \pi$ Flat surface area of cylinder $=2\pi r^{2}=2\pi \times 6^{2}=72\pi \,\mathrm{cm}^{2}$ A solid cylinder is melted into two cones with same height $10 \,\mathrm{cm}$. The ratio of volume of two cone is 1: 2 . Means if we divide $360 \pi$ into 1: 2 then $120 \pi: 240 \pi$ So one cone volume is $120 \pi$ and second has $240 \pi$. $\therefore$ Cone volume $=\;\frac{\pi}{3} r^{2} h$ Then first bigger cone has $R$ radius. then $240 \pi=\;\frac{\pi}{3} R^{2} h \Rightarrow R^{2}=72$ Then flat surface area of first cone $=\pi R^{2}=72 \pi$ Then second smaller cone has $r$ radius. then $120 \pi=\;\frac{\pi}{3} r^{2} h \Rightarrow r^{2}=36$ Then flat surface area of second cone $=\pi r^{2}=36 \pi$ Total surface area of cone $=72 \pi+36 \pi=108 \pi$ So flat surface area from cylinder to cone is increased by $36 \pi$ which is 50 $\%$ increase.