#### provide solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise  12.2 question 6

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

Hint: We know that the volume of the spherical balloon is $V=\frac{4}{3}\pi r^{3}$ .

Given: The spherical balloon inflated by pumping in 900 cubic cm of gas per second.

Solution: Suppose the radius of the given spherical balloon be r cm and let V be the volume of the spherical balloon at any instant time.

The balloon is inflated by pumping 900 cubic cm of gas per second hence the rate of volume of spherical balloon is increased by,

$\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=900 \mathrm{~cm}^{3} / \mathrm{sec}$

$\frac{d V}{d t}=\frac{d\left(\frac{4}{3} \pi r^{3}\right)}{d t}$                                   $\therefore\left(V=\frac{4}{3} \pi r^{3}\right)$

lets put value of $\frac{d V}{d t}=900$(given)

Let’s differentiate $\left(\frac{4}{3} \pi r^{3}\right)$  w.r.t time

$\begin{gathered} 900=\frac{4 \times 3}{3} \times \pi r^{2} \frac{d r}{d t} \\\\ 900=4 \pi r^{2} \frac{d r}{d t} \end{gathered}$

So, lets put value of $r= 15$ (given)

\begin{aligned} &\Rightarrow 900=4 \pi \times 15 \times 15 \times \frac{d r}{d t} \\\\ &\Rightarrow 900=900 \pi \times \frac{d r}{d t} \\\\ &\Rightarrow \frac{900}{900 \pi}=\frac{d r}{d t} \end{aligned}

\begin{aligned} &\Rightarrow \frac{1}{\pi}=\frac{d r}{d t} \\\\ &\text { So, } \frac{d r}{d t}=\frac{1}{\pi} \end{aligned}