Question

If the height of a right circular cone is increased by 100% and the radius of the base is reduced by 50%, then the volume of the cone

• A. Decreases by 25%
• B. Decreases by 50%
• C. Increases by 25%
• D. Increases by 15%

Answer 1

The correct option is B Decreases by 50%

Let $r$ and $h$ be the radius and height of the original cone, respectively.
And, $r\text{'}$ and $h\text{'}$ be the radius and height of the new cone, respectively.
Now, height of the cone is increased by 100%.
∴ $h\text{'}=h$ + 100% of $h$
$=h+h$
$\Rightarrow h\text{'}=2h$ …..(i)
And, radius of the cone is decreased by 50%.
∴ $r\text{'}=r$ – 50% of $r$
$=r-\frac{r}{2}$
$\Rightarrow r=\frac{r}{2}$ .....(ii)
$\therefore \text{Volume of new cone}=\frac{1}{3}\pi \left(r^{\prime }{)}^2\right(h^{\prime })$
$\frac{1}{3}\pi \times {\left(\frac{r}{2}\right)}^2\times 2h$ [From (i) and (ii)]
$=\frac{1}{2}\times \frac{1}{3}\pi r^{2}h$
$\Rightarrow \text{Volume of new cone}=\frac{1}{2}\times \text{Volume of original cone}$
∴ Percentage decreased in volume $=\frac{(\frac{1}{3}\pi r^{2}h-\frac{1}{2}\times \frac{1}{3}\pi r^{2}h)}{\frac{1}{3}\pi r^{2}h}\times 100$
$=\frac{1}{2}\times 100$
= 50%
Therefore, the volume of the cone is decreased by 50%.
Hence, the correct answer is option (b).