# 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 reduce cylinder to that of the original.

Solution : Let r be the radius of the base and $h$ be the height of the given cylinder .

Then , radius of the base and the height of the reduced cylinder are  $\frac{r}{2}$ and $h$ respectively .

Let $V_{1}$ and $V_{2}$ be the volumes of the given cylinder and reduced cylinder respectively . Then,

$V_{1}= \pi r^2h$   cubic units and , $V_{2}= \pi (\frac{r}{2})^2h =\frac{\pi}{4}r^2h$   cubic units

$\Rightarrow$                                 $\frac{V_{1}}{V_{2}}=\frac{\pi r^2h}{\pi (\frac{r^2}{4})h}=4\Rightarrow \frac{V_{2}}{V_{1}}=\frac{1}{4}\Rightarrow V_{2}:V_{1}=1:4$

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