If the height of a cone is doubled, then its volume is increased by

(a) 100 %

Suppose that height of the cone becomes 2h and let its radius be r.

Then new volume of the cone $=\frac{1}{3} \pi r^{2}(2 h)=\frac{2}{3} \pi r^{2} h=2 \times$ volume of the cone

Increase in volume $=\frac{2}{3} \pi r^{2} h-\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi r^{2} h$

$\therefore$ Percentage increase $=\frac{\text { increase in volume }}{\text { initial volume }} \times 100 \%=\frac{\frac{1}{3} \pi r^{2} h}{\frac{1}{3} \pi r^{2} h} \times 100 \%=100 \%$

Hence, the volume increases by 100%.