Question

If the height of a cone is doubled, then its volume is increased by
(a) 100%
(b) 200%
(c) 300%
(d) 400%

Answer 1

(a) 100 %

Suppose that height of the cone becomes 2h and let its radius be r.
Then new volume of the cone = $\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}\mathit{\left(}2h\right)=\frac{2}{3}\mathrm{\pi }{r}^{\mathit{2}}h\mathit{}=2\times \mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cone}$
Increase in volume = $\frac{2}{3}\mathrm{\pi }{r}^{\mathit{2}}h-\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h=\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h$
$\therefore \mathrm{Percentage}\mathrm{increase}=\frac{\mathrm{increase}\mathrm{in}\mathrm{volume}}{\mathrm{initial}\mathrm{volume}}\times 100\%$=$\frac{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r}^{\mathit{2}}h}{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r}^{\mathit{2}}h}\times 100\%$=100%

Hence, the volume increases by 100%.