(D) The volume $V$ of a spherical snowball is given by $V = \frac{4}{3} \pi r^3$.Differentiating both sides with respect to time $t$,we get $\frac{dV}

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$\frac{1}{2\pi} \text{ cm/sec}$

(D) The volume $V$ of a spherical snowball is given by $V = \frac{4}{3} \pi r^3$.Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Given that $\frac{dV}{dt} = 8 \text{ cm}^3/\text{sec}$ and $r = 2 \text{ cm}$.Substituting these values,we have $8 = 4 \pi (2)^2 \frac{dr}{dt}$.$8 = 16 \pi \frac{dr}{dt}$.Therefore,$\frac{dr}{dt} = \frac{8}{16\pi} = \frac{1}{2\pi} \text{ cm/sec}$.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

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$A$ spherical snowball is forming such that its volume is increasing at the rate of $8 \text{ cm}^3/\text{sec}$. Find the rate of increase of its radius when the radius is $2 \text{ cm}$.

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A
$\pi \text{ cm/sec}$
B
$\frac{1}{8\pi} \text{ cm/sec}$
C
$2\pi \text{ cm/sec}$
D
$\frac{1}{2\pi} \text{ cm/sec}$

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$A$ spherical snowball is forming such that its volume is increasing at the rate of $8 \text{ cm}^3/\text{sec}$. Find the rate of increase of its radius when the radius is $2 \text{ cm}$.

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