The surface area of a spherical bubble


The surface area of a spherical bubble is increasing at the rate of 2 cm2/s. When the radius of the bubble is 6 cm, at what rate is the volume of the bubble increasing?


Let $r$ be the radius, $S$ be the surface area and $V$ be the volume of the sphere at any time $t$. Then,

$S=4 \pi r^{2}$

$\Rightarrow \frac{d S}{d t}=8 \pi r \frac{d r}{d t}$

$\Rightarrow \frac{d r}{d t}=\frac{1}{8 \pi r} \frac{d S}{d t}$

$\Rightarrow \frac{d r}{d t}=\frac{2}{8 \pi \times 6}$

$\Rightarrow \frac{d r}{d t}=\frac{1}{24 \pi} \mathrm{cm} / \mathrm{sec}$


Volume of sphere $=\frac{4}{3} \pi r^{3}$

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

$\Rightarrow \frac{d V}{d t}=\frac{4 \pi(6)^{2}}{24 \pi}$

$\Rightarrow \frac{d V}{d t}=6 \mathrm{~cm}^{3} / \mathrm{sec}$

Leave a comment


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