Question

If the radius of the base of a right circular cone is halved, keeping the height same, then the ratio of the volume of the reduced cone to that of the original cone is

• A. $2:1$
• B. $4:1$
• C. $1:4$
• D. $1:2$

Answer 1

The correct option is C $1:4$

Suppose that

Radius of the base is r, and the height of the original cone is $h$.

Then,

Volume of the original cone $\left(V_1\right)=\frac{1}{3}\pi r^{2}h$

When cone reduced then,

$radius=\frac{r}{2}$

$height=h$

Now,

Volume of the reduced cone $\left(V_2\right)=\frac{1}{3}\pi {\left(\frac{r}{2}\right)}^{2}h$

$=\frac{1}{4}\left(\frac{1}{3}\pi r^{2}h\right)$

$=\frac{1}{4}{V}_1$$

$\therefore \frac{V_2}{V_1}=\frac{1}{4}=1:4$

This is the required solution.