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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

$2\pi sqcm/s$

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B

$\pi sqcm/s$

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C

$\frac{3\pi }{2}sqcm/s$

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D

$\frac{\pi }{2}sqcm/s$

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Solution

## 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$$⇒\frac{dV}{dt}=\frac{4}{3}\pi ×3{r}^{2}\frac{dr}{dt}$ $=4\pi {r}^{2}\frac{dr}{dt}$$⇒$ $\pi =4\pi {r}^{2}\frac{dr}{dt}$ … [from Equation (i)]$⇒$ $\frac{dr}{dt}=\frac{1}{4{r}^{2}}$ …...(ii)Also, we know that, surface of sphere; $S=4\pi {r}^{2}$Step 3. Differentiate it with respect to $r$$⇒\frac{dS}{dt}=8\pi r\frac{dr}{dt}$ $=8\pi r\cdot \frac{1}{4{r}^{2}}$ …​ [from Equation (ii)]$⇒\frac{dS}{dt}=\frac{2\pi }{r}$Step 4. Put the value $r=1$:$⇒\frac{dS}{dt}=\mathbf{2}\mathbit{\pi }$$\therefore$the rate of increase in surface area $=\mathbf{2}\mathbit{\pi }\mathbf{}\mathbit{s}\mathbit{q}\mathbf{}\mathbit{c}\mathbit{m}\mathbf{/}\mathbit{s}$Hence, Option ‘A’ is Correct.

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