If the rate of change in the volume of sphere is equals to the rate of change in its radius then the radius is

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$\frac{1}{2 \sqrt{π}}$

$\frac{2}{π}$

$\frac{1}{\sqrt{2 π}}$

$\frac{1}{2 π}$

answer is B.

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Detailed Solution

$\frac{dv}{dt} = \frac{dr}{dt} v = \frac{4}{3} {πr}^{3} \frac{dv}{dt} = 4 {πr}^{2} . \frac{dr}{dt} \Rightarrow 4 {πr}^{2} = 1 \Rightarrow r = \frac{1}{2 \sqrt{π}}$

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