If the surface area of a sphere of radius r is increasing uniformly at the rate $8 {cm}^{2} / s$ then the rate of change of its volume is :

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proportional to $\sqrt{r}$

proportional to $r^{2}$

proportional to r

constant

answer is D.

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Detailed Solution

$V = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d t} = 4 π r^{2} \cdot \frac{d r}{d t}$

$S = 4 π r^{2} \Rightarrow \frac{d S}{d t} = 8 π r \cdot \frac{d r}{d t} \Rightarrow 8 = 8 π r \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{1}{π r}$

we get $\frac{d V}{d t} = 4 π r^{2} \times \frac{1}{π r} = 4 r \Rightarrow \frac{d V}{d t}$ is proportional to r

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