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Gas is being pumped into a spherical balloon at the rate of $30 {ft}^{3} / \min$ Then the rate at which the radius increases when is reaches the value $15 f t,$ is

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a

\frac{1}{30 π} ft / \min

b

\frac{1}{15 π} ft / \min

c

\frac{1}{20} ft / \min

d

\frac{1}{25} ft / \min

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detailed solution

Correct option is A

Let $r$ be the radius of the spherical balloon and $V$

be the volume at any time $t$ .

It is given that $\frac{d V}{d t} = 30 f t^{3} / \min$

Now, $V = \frac{4}{3} π r^{3}$

$\begin{array}{l} \Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} \\ \Rightarrow {(\frac{d V}{d t})}_{r = 15} = 4 π \times 15^{2} \times {(\frac{d r}{d t})}_{r = 15} \\ \Rightarrow 30 = 900 π {(\frac{d r}{d t})}_{r = 15} \\ \Rightarrow {(\frac{d r}{d t})}_{r = 15} = \frac{1}{30 π} ft / \min \end{array}$

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