Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picksup moisture at a rate proportional to its surface area. Under these circumstances, what will bethe e‚ect on the drop’s radius.

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Solution:

V=43πr3dVdt=43.3πr2drdt=4πr2drdt …(i)V=dfrac43pi r^3
dfrac{dV}{dt}=dfrac43.3pi r^2 dfrac{dr}{dt}
=4pi r^2 dfrac{dr}{dt} …(i)V=34​πr3dtdV​=34​.3πr2dtdr​=4πr2dtdr​ …(i)

A=4πr2…(ii)A=4pi r^2 …(ii)A=4πr2…(ii)

It is given that dVdt=k.(Surface area)dfrac{dV}{dt}=k.(Surface area)dtdV​=k.(Surface area)

⇒4πr2drdt=k.(4πr2)Rightarrow 4pi r^2 dfrac{dr}{dt}=k.(4pi r^2)⇒4πr2dtdr​=k.(4πr2) [using (i) and (ii)]

⇒drdt=kRightarrow dfrac{dr}{dt}=k⇒dtdr​=k

It shows radius increases at a constant rate.