Question

Height of a cone is increased by $200\%$. Its volume increases by:

• A. $200\%$
• B. $300\%$
• C. $250\%$
• D. $100\%$

Answer 1

Answer: (A)

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

Given, height is increased by $200\%$.

That is, new height $=h+200\%h=h+\frac{200}{100}h=h+2h=3h$.

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

Then, $\frac{V_2}{V_1}=\frac{\pi r^2\left(3h\right)}{\pi r^{2}h}=\frac{3}{1}$.

$⟹$ $V_2=3V_1$

$⟹$ $V_2=V_1+2V_1$

$⟹$ $V_2=V_1+\frac{200}{100}{V}_1$

$⟹$ $V_2=V_1+200\%V_1$.
Thus, increase in volume $=200\%$.

Hence, option $A$ is correct.