Question

If the radius and height of a cone both are increased by l0%, then the volume of the cone is increased by:

• A. $10\%$
• B. $21\%$
• C. $33.1\%$
• D. $100\%$

Answer 1

The correct option is C $33.1\%$

Let initial radius of cone $=r_1$

and initial height of cone $=h_1$.

Initial volume of cone, $V_1=\frac{1}{3}\pi r_1^2{h}_1$.

After increasing, radius of cone $=r_2$

and height of cone $=h_2$.

Then, $r_2=r_1+\frac{10}{100}{r}_1=\frac{11}{10}{r}_1$

and $h_2=h_1+\frac{10}{100}{h}_1=\frac{11}{10}{h}_1$.

Therefore, New volume, $V_2=\frac{1}{3}\pi r_2^2{h}_2=\frac{1}{3}\pi \times {\left(\frac{11}{10}\right)}^2{r}_1^2\times {\left(\frac{11}{10}\right)}^2{h}_1$

$⟹$ $V_2=\frac{1}{3}\pi r_1{h}_1\times \frac{1331}{1000}$

$⟹$$V_2=\frac{1331}{1000}{V}_1$.

Percentage increase in volume $=\frac{V_2-V_1}{V_1}\times 100\%$

$=\frac{\frac{1331V_1}{1000}-V_1}{V_1}\times 100=\frac{331V_1}{1000V_1}\times 100=33.1\%$.

Therefore, option $C$ is correct.