Question

A spherical balloon is filled with $4500\pi$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72\pi$ cubic meters per minute, the rate in meter per minute at which the radius of the balloon decreases $49$ minutes after the leakage began is

• A. $\frac{9}{7}$
• B. $\frac{7}{9}$
• C. $\frac{2}{9}$
• D. $\frac{9}{2}$

Answer 1

The correct option is C

$\frac{2}{9}$

Explanation for correct option:

Step 1. Find the volume after $49$ minutes

Given, $\frac{dV}{dt}=72\mathrm{\pi }$ m³/ min

Initial Volume $=4500\pi$

$\therefore$Volume after $49$ minutes well be $=4500\mathrm{\pi }-49\left(72\mathrm{\pi }\right)$

$=972\mathrm{\pi }$ …..(1)

Step 2. Find the radius after $49$ minutes,

Volume of spherical balloon is given by $\mathit{V}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{4}}{\mathbf{3}}\mathbf{}\mathit{\pi }{\mathit{r}}^{\mathbf{3}}$ ……(2)

$\begin{array}{rcl}972\pi & =& \left(\frac{4}{3}\right)\pi r^3\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{r}^3& =& 243\times 3\\ & =& 36\end{array}$

$\Rightarrow$ $\begin{array}{rcl}r& =& 9\end{array}$

Step 3.Find the rate of change in radius:

$\frac{dV}{dt}=\frac{4}{3}\mathrm{\pi }3{\mathrm{r}}^2\frac{d\mathrm{r}}{d\mathrm{t}}$

$\Rightarrow$$72\mathrm{\pi }=\frac{4}{3}\mathrm{\pi }3{\left(9\right)}^2\frac{d\mathrm{r}}{d\mathrm{t}}$

$\Rightarrow$ $72=4\times 81\times \frac{d\mathrm{r}}{d\mathrm{t}}$

$\mathbf{\therefore }\frac{\mathbf{d}\mathbf{r}}{\mathbf{d}\mathbf{t}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{2}}{\mathbf{9}}$

Hence, Option ‘C’ is Correct.