Question

If the height of a cone is increased by $50\%$, the what is the percentage increase in the volume of the cone?

• A. $\frac{100}{3}$
• B. $40$
• C. $50$
• D. $\frac{200}{3}$

Answer 1

Answer: (C)

$50$

Let $h$ be the initial height and $h_1$ be the increased height.

Then, $h_1=h+\left(\frac{50}{100}\right)h=1.5h$.

Then, increased volume $V_1=\pi r^2{h}_1=1.5\pi r^{2}h=1.5V$.

$\Rightarrow$ Increase in volume $\left(\frac{V_1-V}{V}\right)100$

$\Rightarrow \left(\frac{1.5V-V}{V}\right)100$ $=50\%$ .

Hence, option $C$ is correct.