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(B) The volume $V$ of a spherical balloon is given by $V = \frac{4}{3} \pi r^3$,where $r$ is the radius.Differentiating both sides with respect to tim

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$\frac{2}{\sqrt{\pi}}$

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(B) The volume $V$ of a spherical balloon is given by $V = \frac{4}{3} \pi r^3$,where $r$ is the radius.Differentiating both sides with respect to time $t$,we get:$\frac{dV}{dt} = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.According to the problem,the rate of increase of volume is $16$ times the rate of increase of the radius:$\frac{dV}{dt} = 16 \frac{dr}{dt}$.Substituting this into the derivative equation:$16 \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Assuming $\frac{dr}{dt} 
eq 0$,we can divide both sides by $\frac{dr}{dt}$:$16 = 4 \pi r^2$.$r^2 = \frac{16}{4 \pi} = \frac{4}{\pi}$.Taking the square root of both sides:$r = \sqrt{\frac{4}{\pi}} = \frac{2}{\sqrt{\pi}}$.

$A$ spherical balloon is expanding. If at any instant the rate of increase of its volume is $16$ times the rate of increase of its radius,then its radius at that instant is:

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