If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the new cylinder to that of the original one?

1. 1 : 2

2. 1 : 4

3. 1 : 8

4. 4 : 1

For old cylinder,

Let height = h and Radius = r

So, for new cylinder,

Height = h and radius = $\dfrac{\text{r}}{2}$

We know that volume of cylinder = π × radius² × height

∴ Volume of old cylinder = πr²h

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

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

Hence, option 2 is the correct option.