Question

If gas is being pumped into a spherical balloon at the rate of $30\mathrm{cubic}\mathrm{feet}/\min .$. Then, the rate at which the radius increases, when it reaches the value $15\mathrm{ft}$ is

• A. $\frac{1}{15\mathrm{\pi }}\mathrm{ft}/\min$
• B. $\frac{1}{30\mathrm{\pi }}\mathrm{ft}/\min$
• C. $\frac{1}{20}\mathrm{ft}/\min$
• D. $\frac{1}{25}\mathrm{ft}/\min$

Answer 1

The correct option is B

$\frac{1}{30\mathrm{\pi }}\mathrm{ft}/\min$

Explanation for the correct option:

Finding the rate of radius $\left(\frac{dr}{dt}\right)$:

Given that,

$\frac{\mathrm{dV}}{\mathrm{dt}}=30\mathrm{cubic}\mathrm{feet}/\min$

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

Differentiate the volume with respect to $t$.

$\begin{array}{rcl}\frac{dV}{dt}& =& \frac{4}{3}\mathrm{\pi }\left(3{\mathrm{r}}^2\frac{\mathrm{dr}}{\mathrm{dt}}\right)\\ & \Rightarrow & \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{30}{4{\mathrm{\pi r}}^2}\\ & \Rightarrow & {\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)}_{\mathrm{r}=15}=\frac{30}{4{\mathrm{\pi r}}^2}\\ & =& \frac{30}{4\mathrm{\pi }\left(15\times 15\right)}\\ & =& \frac{1}{30\mathrm{\pi }}\text{ft/min}\end{array}$

Hence, the correct option is B.