Question

A right circular cone has base radius $5cm$. If the radius is increased by $20\%$ without any change in height of the cone, then the percentage increase in volume is:

• A. $44\%$
• B. $40\%$
• C. $25\%$
• D. $22\%$

Answer 1

The correct option is A $44\%$

$\frac{V_3}{V_1}=\frac{{\left(\frac{6}{5}r\right)}^{2}h}{r^{2}h}=\frac{36}{25}=1.44$

Increase $=44\%$

Let initial volume$V_1=\frac{1}{3}\pi r^{2}h$.

Given, radius is increased by $20\%$.

That is, new radius $=r+20\%r=r+\frac{20}{100}r=\frac{6}{5}r$.

Then, new volume $V_2=\frac{1}{3}\pi (\frac{6}{5}r{)}^{2}h$.

Then, $\frac{V_2}{V_1}=\frac{{\left(\frac{6}{5}r\right)}^{2}h}{r^{2}h}=\frac{36}{25}=1.44$

$⟹$ $V_2=1.44V_1$

$⟹$ $V_2=V_1+0.44V_1$

$⟹$ $V_2=V_1+\frac{0.44}{100}{V}_1$

$⟹$ $V_2=V_1+44\%V_1$.
Thus, increase in volume $=44\%$.

Hence, option $A$ is correct.