Question

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 reduced cylinder to that of the original one?

• A. $1:2$
• B. $1:4$
• C. $1:8$
• D. $8:1$

Answer 1

The correct option is B $1:4$

Let $V_n$ and $V_o$ are the new and original volumes of cylinder, then

$V_o=\pi r^{2}h$

and $V_n=\pi (r/2{)}^{2}h=\frac{V_o}{4}$ ($\because r\to r/2$ )

Thus, $\frac{V_n}{V_o}=\frac{1}{4}$