Question

If the radius of a cylinder is reduced by 50%, then the volume of the cylinder will be reduced by_____.

$\left[3\text{Marks}\right]$

Answer 1

Let the radius of the old cylinder be $R$ and that of the new cylinder be $r$.
$\text{Then,}r=\frac{R}{2}$

$\text{Volume of old cylinder}=\pi R^{2}h$

$\text{Volume of new cylinde}r=\pi r^{2}h$

$=\pi \times (\frac{R}{2}{)}^2\times h$

$=\frac{\pi \times R^2\times h}{4}$

$\left[1\text{Mark}\right]$


Hence, reduction in volume

$=\text{Volume of old cylinder - volume of new cylinder}$

$=\pi R^{2}h-\frac{\pi \times R^2\times h}{4}=\frac{3}{4}\pi R^{2}h$

$\left[1\text{Mark}\right]$


$\therefore$ Percentage reduction

$=\frac{\text{Reduction in volume}}{\text{Volume of old cylinder}}\times 100\%$

$=\frac{\frac{3}{4}\pi R^{2}h}{\pi R^{2}h}\times 100\%$

$=75\%$

$\left[1\text{Mark}\right]$