#### provide solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise  12.2 question 18

Answer: $12 \pi \mathrm{cm}^{3} / \mathrm{sec}$

Hint: Use $V=\frac{4}{3} \pi r^{3}$

Given: A balloon in the form of a right circular cone surrounded by hemisphere having a diameter equal to the height of the cone is being inflated.

Solution: Let, the total height of the balloon, then

then

\begin{aligned} &H=h+r \\\\ &H=2 r+r \quad(h=2 r) \\\\ &\frac{d H}{d t}=\frac{3 d r}{d t} \end{aligned}...........(i)

So, volume of the cone volume of the hemisphere

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

Substituting $h =2r$

So, we get

\begin{aligned} &V=\frac{2}{3} \pi r^{3}+\frac{2}{3} \pi r^{3} \\\\ &V=\frac{4}{3} \pi r^{3} \end{aligned}

Let’s take derivative w.r.t to time both side

\begin{aligned} &\frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t} \\\\ &\frac{d V}{d t}=4 \pi r^{2} \times \frac{d H}{3 \times d t} \end{aligned}

Let’s put value of $r = 3$

\begin{aligned} &\frac{d V}{d t}=4 \pi \times 3 \times 3 \times \frac{d H}{3 \times d t} \\\\ &\frac{d V}{d H}=12 \pi \mathrm{cm}^{3} / \mathrm{sec} \end{aligned}