Question

A spherical balloon is being inflated at the rate of $35cc/min$. The rate of increase of the surface area of the balloon, when its diameter is $14cm$, is

• A. $7sqcm/min$
• B. $10sqcm/min$
• C. $17.5sqcm/min$
• D. $28sqcm/min$

Answer 1

The correct option is B

$10sqcm/min$

Explanation for the correct option:

As we know, The volume of square; $\mathit{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathbf{\pi }\mathbf{.}{\mathbf{r}}^{\mathbf{3}}$

Step 1. Differentiate it with respect to $r$:

$\Rightarrow \frac{dV}{dt}=\frac{4}{3}\mathrm{\pi }.3{\mathrm{r}}^2\frac{d\mathrm{r}}{d\mathrm{t}}$ ……(1)

Given, $\frac{dV}{dt}=35$

and diameter of the balloon $=14$

so, the radius will be $r=7$

Step 2. Substituting the values of $\frac{dV}{dt}$and $r$ in equation (1)

$\Rightarrow 35=\frac{4}{3}\mathrm{\pi }.3{\left(7\right)}^2\frac{d\mathrm{r}}{d\mathrm{t}}$

$\Rightarrow 35=4\mathrm{\pi }\left(49\right)\frac{d\mathrm{r}}{d\mathrm{t}}$

$\Rightarrow \frac{dr}{dt}=\frac{5}{28\mathrm{\pi }}$

Also, we know that, The surface area; $\mathit{S}\mathbf{=}\mathbf{4}\mathbf{\pi }\mathbf{.}{\mathbf{r}}^{\mathbf{2}}$

Step 3. Differentiate it with respect to $r$:

$\Rightarrow \frac{dS}{dt}=4\mathrm{\pi }\left(2\mathrm{r}\right)\frac{d\mathrm{r}}{d\mathrm{t}}$

Step 4. substitute the value of $\frac{dr}{dt}$:

$\Rightarrow \frac{dS}{dt}=4\mathrm{\pi }\left(2\times 7\right)\left(\frac{5}{28\mathrm{\pi }}\right)$

$\Rightarrow \frac{dS}{dt}=\mathbf{10}$

Thus, The rate of increase of the surface area of the balloon is $\mathbf{10}\mathbf{}\mathit{s}\mathit{q}\mathbf{}\mathit{c}\mathit{m}\mathbf{/}\mathit{m}\mathit{i}\mathit{n}$

Hence, Option ‘C’ is Correct.