#### Please solve RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 21 maths textbook solution

Answer: $6 \mathrm{~cm}^{3} / \mathrm{sec}$

Hint: The surface area of the bubble will be, $S A=4 \pi r^{2}$ .

Given: The surface area of a spherical bubble is increasing at the rate of 2 cm2/sec

Solution: To find rate of increase of its volume, when the radius is 6 cm.

Let the radius of the given spherical bubble be $r$ cm at any instant time.

It is given that the surface area of a spherical bubble is increasing at the rate of $2 \mathrm{~cm}^{2} / \sec \frac{d s}{d t}=2$

\begin{aligned} &S=4 \pi r^{2} \\\\ &\frac{d S}{d t}=8 \pi r \times \frac{d r}{d t} \end{aligned}

\begin{aligned} &\frac{d r}{d t}=\frac{1}{8 \pi r} \times \frac{d S}{d t} \\\\ &\frac{d r}{d t}=\frac{1}{8 \pi r} \times 2 \quad\left(\frac{d S}{d t}=2(\text { given })\right) \end{aligned}

\begin{aligned} &\frac{d r}{d t}=\frac{1}{8 \pi \times 6} \times 2 \quad(r=6(\text { given })) \\\\ &\frac{d r}{d t}=\frac{1}{24 \pi} \mathrm{cm} / \mathrm{sec} \end{aligned}

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

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

Let’s put value of $r =6$ and  $\frac{d r}{d t}=\frac{1}{24 \pi} \mathrm{cm} / \mathrm{sec}$

$\frac{d V}{d t}=4 \pi \times 6^{2} \times \frac{1}{24 \pi}=6 \mathrm{~cm}^{3} / \mathrm{sec}$