ABC26GN4959 · Volume and Surface Area

Let the original radius and height of the cone be $r$ and $h$ respectively. Then, Original volume $=\frac{1}{3} \pi r^{2} h$. New radius $=\frac{120}{100} r=\frac{6}{5} r$, New height $=\frac{6}{5} h$. New volume $=\frac{1}{3} \pi \times\left(\frac{6}{5} r\right)^{2} \times\left(\frac{6}{5} h\right)=\frac{216}{125} \times \frac{1}{3} \pi r^{2} h$. Increase in volume $=\frac{91}{125} \times \frac{1}{3} \pi r^{2} h$. $$\text { ∴ Increase } \%=\left(\frac{\frac{91}{125} \times \frac{1}{3} \pi r^{2} h}{\frac{1}{3} \pi r^{2} h} \times 100\right) \%=72.8 \% .$$