The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of the new cylinder to that of the original cylinder ?

For old cylinder,

Let Height = h and Radius = r.

So, for new cylinder,

Height = 2h and Radius = $\dfrac{r}{2}$.

We know that volume of cylinder = π × (radius)² × height.

∴ Volume of old cylinder = πr²h

and Volume of new cylinder = $π.\Big(\dfrac{r}{2}\Big)^2.2h$

Hence,

$\Rightarrow \dfrac{\text{Volume of new cylinder}}{\text{Volume of old cylinder}} = \dfrac{π.\Big(\dfrac{r}{2}\Big)^2.2h}{πr^2h} \\[1em] = \dfrac{π.\Big(\dfrac{r^2}{4}\Big).2h}{πr^2h} \\[1em] = \dfrac{2πr^2h}{4πr^2h} \\[1em] = \dfrac{2}{4} \\[1em] = \dfrac{1}{2}.$

Hence, the ratio of the volume of the new cylinder to that of the original cylinder is 1 : 2.