Question

A spherical balloon is expanding. If the radius is increasing at the rate of $2cm/min$, then the rate at which the volume increases $(incm/min)$, when the radius is $5cm$, is

• A. $10\mathrm{\pi }$
• B. $100\mathrm{\pi }$
• C. $200\mathrm{\pi }$
• D. $50\mathrm{\pi }$

Answer 1

The correct option is C

$200\mathrm{\pi }$

Explanation for the correct option:

Step 1. Let $r$ be the radius of the spherical balloon

As we know,

The volume of sphere; $V=\frac{4}{3}{\mathrm{\pi r}}^3$

Step 2. Differentiate it with respect to $r$:

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

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

$\Rightarrow \frac{dV}{dt}=4{\mathrm{\pi r}}^2\times 2$ ….$\left[\because \frac{dr}{dt}=2\right]$

$\Rightarrow \frac{dV}{dt}=8{\mathrm{\pi r}}^2{\mathrm{cm}}^3/\min$

Step 3. Put the given value of $r=5$:

$\Rightarrow \frac{dV}{dt}=8\mathrm{\pi }{\left(5\right)}^2$

$=\mathbf{200}\mathbf{\pi }\mathbf{}{\mathbf{cm}}^{\mathbf{3}}\mathbf{/}\mathbf{min}$

Hence, Option ‘C’ is Correct.