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(C) Given that the rate of change of the radius is $\frac{dr}{dt} = 5 	ext{ inch/min}$.When the radius $r = 10 	ext{ inches}$.The volume $V$ of a sp

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$2000$

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(C) Given that the rate of change of the radius is $\frac{dr}{dt} = 5 	ext{ inch/min}$.When the radius $r = 10 	ext{ inches}$.The volume $V$ of a spherical balloon is given by $V = \frac{4}{3} \pi r^3$.Differentiating both sides with respect to time $t$, we get:$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 ight) = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Substituting the given values $r = 10$ and $\frac{dr}{dt} = 5$:$\frac{dV}{dt} = 4 \pi (10)^2 (5) = 4 \pi (100) (5) = 2000 \pi 	ext{ cubic inches/min}$.

If the radius of a spherical balloon is increasing at the rate of $5 \text{ inch/min}$, then the rate at which the volume increases (in $\text{cubic inches/min}$) when the radius is $10 \text{ inches}$ is: (in $\pi$)

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