# A drop of liquid evaporates at a rate proportional to its area of surface. If the radius initially is $4\;mm$ and $5$ minutes later, the radius is reduced to $2\; mm$, find radius of drop as a function of time ?

$(a)\;R=\frac{20}{t+5} \\ (b)\;R= \frac{t+5}{20} \\ (c)\;R=\frac{10}{t+5} \\ (d)\;R=\frac{10}{t+4}$

According to given condition
$\large\frac{dR}{dt}$$=K ' \times 4 \pi R^2 \large\frac{dR}{dt}$$=K' R^2$
$\large\frac{dR}{R^2}$$=K'dt -\large\frac{1}{R}\bigg]_4^2=k't \bigg]_0^5 - \bigg[\large\frac{1}{2} -\frac{1}{4}\bigg]$$=5K^1$
$K=-\large\frac{1}{20}$
$\large\frac{-1}{R}\bigg]_4^R =-\large\frac{1}{20}$$t\bigg]_0^t -\bigg[\large\frac{1}{R} -\frac{1}{4}\bigg]=\large\frac{-1}{20}$$ \times t$
$\large\frac{1}{R} -\frac{1}{4} =\frac{t}{20}$
$\large\frac{1}{R} =\frac{1}{4}+ \frac{t}{20}$
$\large\frac{1}{R} =\frac{t+5}{20}$
$R=\large\frac{20}{t+5}$
Hence a is the correct answer.