(B) Given,the volume of a ball $V = \frac{4}{3} \pi r^3$.When $V = 288 \pi$,we have:$288 \pi = \frac{4}{3} \pi r^3$$r^3 = 288 \times \frac{3}{4} = 216

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(B) Given,the volume of a ball $V = \frac{4}{3} \pi r^3$.When $V = 288 \pi$,we have:$288 \pi = \frac{4}{3} \pi r^3$$r^3 = 288 \times \frac{3}{4} = 216$$r = 6 \text{ cm}$.Now,differentiating $V$ with respect to $t$:$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$Given $\frac{dV}{dt} = 4 \pi \text{ cc/sec}$,substituting the values:$4 \pi = 4 \pi (6)^2 \frac{dr}{dt}$$1 = 36 \frac{dr}{dt}$$\frac{dr}{dt} = \frac{1}{36} \text{ cm/sec}$.

Class 12 Mathematics Application of Derivatives Rate of Change of Quantities

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The volume of a ball is increasing at the rate of $4 \pi \text{ cc/sec}$. The rate of increase of the radius,when the volume is $288 \pi \text{ cc}$,is

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A
$\frac{1}{6} \text{ cm/sec}$
B
$\frac{1}{36} \text{ cm/sec}$
C
$6 \text{ cm/sec}$
D
$36 \text{ cm/sec}$

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