If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone?
Let r be the radius of the base and h be the height of the original cone.
Then, $$V_1=\dfrac{1}{3}\pi r^2h$$
Let $$V_2$$ be the volume of the reduced cone.
reduced radius$$=\dfrac r2$$
Then,
$$V_2=\dfrac{1}{3}\pi\left(\dfrac{r}{2}\right)^2\times h$$
$$\therefore \dfrac{V_1}{V_2}=\dfrac{\dfrac{1}{3}\pi r^2h}{\dfrac{1}{3}\pi\left(\dfrac{r}{2}\right)^2\times h}=\dfrac{4}{1}$$
$$\Rightarrow \dfrac{V_1}{V_2}=\dfrac{4}{1}$$.