Question

A sphere increases its volume at the rate of $\pi c{m}^3/s$. The rate at which its surface area increases, when the radius is $1cm$ is

• A. $2\pi sqcm/s$
• B. $\pi sqcm/s$
• C. $\frac{3\pi }{2}sqcm/s$
• D. $\frac{\pi }{2}sqcm/s$

Answer 1

The correct option is A

$2\pi sqcm/s$

Explanation for the correct option:

Step 1. Given rate of volume increase of sphere is

$\frac{dV}{dt}=\pi$ …..(i)

As we know that,
Volume of sphere; $V=\frac{4}{3}\pi r^3$

Step 2. Differentiate it with respect to $r$

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

$=4\pi r^2\frac{dr}{dt}$
$\Rightarrow$ $\pi =4\pi r^2\frac{dr}{dt}$ … [from Equation (i)]
$\Rightarrow$ $\frac{dr}{dt}=\frac{1}{4r^2}$ …...(ii)

Also, we know that, surface of sphere; $S=4\pi r^2$

Step 3. Differentiate it with respect to $r$

$\Rightarrow \frac{dS}{dt}=8\pi r\frac{dr}{dt}$

$=8\pi r\cdot \frac{1}{4r^2}$ …​ [from Equation (ii)]
$\Rightarrow \frac{dS}{dt}=\frac{2\pi }{r}$

Step 4. Put the value $r=1$:

$\Rightarrow \frac{dS}{dt}=\mathbf{2}\mathit{\pi }$

$\therefore$the rate of increase in surface area $=\mathbf{2}\mathit{\pi }\mathbf{}\mathit{s}\mathit{q}\mathbf{}\mathit{c}\mathit{m}\mathbf{/}\mathit{s}$

Hence, Option ‘A’ is Correct.