Question

The volume of a spherical balloon is increasing at the rate of $40$ $\raisebox{1ex}{$c{m}^3$}\!\left/ \!\raisebox{-1ex}{$min$}\right.$.

The rate of change of the surface area of the balloon at the instant when its radius is $8$ $cm$, is

• A. $\frac{5}{2}\frac{sq.cm}{min}$
• B. $5\frac{sq.cm}{min}$
• C. $10\frac{sq.cm}{min}$
• D. $20\frac{sq.cm}{min}$

Answer 1

The correct option is C

$10\frac{sq.cm}{min}$

Explanation for the correct option:

Step-1: Solve for rate of change of radius:

Given that,

• The volume of a spherical balloon is increasing at the rate of $40$ $\raisebox{1ex}{$c{m}^3$}\!\left/ \!\raisebox{-1ex}{$min$}\right.$
• Radius is $8$ $cm$

The volume of the spherical balloon is given as $V=\frac{4}{3}{\mathrm{\pi r}}^3$where $r$ is the radius of the balloon

Differentiating the volume with respect to time we get

$\frac{dV}{dt}=\frac{4}{3}\times \mathrm{\pi }\times 3{\mathrm{r}}^2\times \frac{\mathrm{dr}}{\mathrm{dt}}$

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

where, $\frac{dV}{dt}$ is the rate of change of volume and $\frac{dr}{dt}$ is the rate of change of radius

It is given that $\frac{dV}{dt}$$=40$ $\raisebox{1ex}{$c{m}^3$}\!\left/ \!\raisebox{-1ex}{$min$}\right.$ and $r$$=8cm$

Substituting these values we get

$\Rightarrow$ $40=4\pi \times 8^2\frac{\mathrm{dr}}{\mathrm{dt}}$

$\Rightarrow$$\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{5}{32\mathrm{\pi }}\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$min$}\right.$$...\left(i\right)$

Step-2: Solve for rate of change of surface area:

The surface area of the spherical balloon is given as $S=4{\mathrm{\pi r}}^2$

Differentiating the surface area with respect to time we get

$\Rightarrow$$\frac{dS}{dt}=4\mathrm{\pi }\times 2\mathrm{r}\frac{\mathrm{dr}}{\mathrm{dt}}$

$\Rightarrow$$\frac{dS}{dt}=8\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}}$$...\left(ii\right)$

Substituting the value of $r$$=8cm$ and $\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{5}{32\mathrm{\pi }}\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$min$}\right.$ we get

$\Rightarrow$$\frac{dS}{dt}=8\pi \times 8\times \frac{5}{32\mathrm{\pi }}$

$\Rightarrow$$\frac{dS}{dt}=10\raisebox{1ex}{$c{m}^2$}\!\left/ \!\raisebox{-1ex}{$min$}\right.$

Thus the surface area of the balloon is changing at a rate of $10\frac{c{m}^2}{min}$ when the radius is $8cm$.

Hence option $\left(C\right)$ i.e. $10\frac{sq.cm}{min}$ is the correct answer.