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Q.

If the volume of a sphere increases at the rate of $2 π c m^{3} / s e c,$ then the rate of increase of its radius when the volume is $288 π c m^{3}$ is

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a

$\frac{1}{36} c m / \sec$

b

$36 c m / \sec$

c

$\frac{1}{72} c m / \sec$

d

$\frac{1}{6} c m / \sec$

answer is D.

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Detailed Solution

$\begin{array}{l} r \to r a d i u s \\ v \to v o l u m e \\ v = \frac{4}{3} π r^{3} = 288 π \\ \Rightarrow r^{3} = 216 \\ \Rightarrow r = 6 c m \\ v = \frac{4}{3} π r^{3} \\ \frac{d v}{d t} = 4 π r^{2} . \frac{d r}{d t} \\ \Rightarrow \frac{d r}{d t} = \frac{1}{72} c m / \sec \end{array}$

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If the volume of a sphere increases at the rate of 2πcm3/sec, then the rate of increase of its radius when the volume is 288π cm3 is