Question

If the radius of a cylinder is reduced by 50 $\%$, then the volume of the cylinder will get reduced by
( Take $\pi =\frac{22}{7}$ )

• A. 25$\%$
• B. 50$\%$
• C. 75$\%$
• D. 40$\%$

Answer 1

The correct option is C

75$\%$

Let the radius of the old cylinder be $R$ and that of the new cylinder be $r$.
Given that radius is reduced by 50% then $r=\frac{R}{2}$
Volume of old cylinder $=\pi R^{2}h$
Volume of new cylinder $=\pi r^{2}h$
$=\pi \times (\frac{R}{2}{)}^2\times h$

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

$\therefore$ Reduction in volume $=\pi R^{2}h-\frac{\pi \times R^2\times h}{4}=\frac{3}{4}\pi R^{2}h$

$\therefore$ Percentage reduction $=\frac{\frac{3}{4}\pi R^{2}h}{\pi R^{2}h}\times 100$
$=\frac{3}{4}\times 100$
$=75\%$