A balloon in the form of a right circular cone surmounted by a hemisphere,

Question:

A balloon in the form of a right circular cone surmounted by a hemisphere, having a diametre equal to the height of the cone, is being inflated. How fast is its volume changing with respect to its total height h, when h = 9 cm.

Solution:

Let $r$ be the radius of the hemisphere, $h$ be the height and $V$ be the volume of the cone.

Then,

$H=h+r$

$\Rightarrow H=3 r$   $[\because h=2 r]$

$\Rightarrow \frac{d H}{d t}=3 \frac{d r}{d t}$

When $H=9 \mathrm{~cm}, r=3 \mathrm{~cm}$

Volume $=\frac{1}{3} \pi \mathrm{r}^{2} h+\frac{2}{3} \pi r^{3}$

Substituting $h=2 r$

$\Rightarrow V=\frac{2}{3} \pi r^{3}+\frac{2}{3} \pi r^{3}$

$\Rightarrow V=\frac{4}{3} \pi r^{3}$

$\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}$

$\Rightarrow \frac{d V}{d t}=\frac{4 \pi r^{2}}{3} \frac{d H}{d t}$

$\Rightarrow \frac{d V}{d H}=\frac{4 \pi(3)^{2}}{3}$

$\Rightarrow \frac{d V}{d H}=12 \pi \mathrm{cm}^{3} / \mathrm{sec}$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now