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

1. 1 : 2

2. 1 : 3

3. 1 : 4

4. 2 : 3

For older cone,

⇒ Radius = r

⇒ Height = h

⇒ Volume = v

For new cone,

⇒ Radius = R = $\dfrac{\text{r}}{2}$

⇒ Height = h

⇒ Volume = V

Volume of cone = $\dfrac{1}{3}π \text{r}^2 \text{h}$

$\Rightarrow \dfrac{\text{V}}{\text{v}} = \dfrac{\dfrac{1}{3}π \text{R}^2 \text{h}}{\dfrac{1}{3}π \text{r}^2 \text{h}} \\[1em] = \dfrac{\Big(\dfrac{\text{r}}{2}\Big)^2}{\text{r}^2} \\[1em] = \dfrac{\dfrac{\text{r}^2}{4}}{\text{r}^2} \\[1em] = \dfrac{\text{r}^2}{4\text{r}^2} \\[1em] = \dfrac{1}{4}.$

Hence, option 3 is the correct option.