Question

A wet porous substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the wind loses half of its moisture during the first hour, when will it have lost 95% moisture, weather conditions remaining the same.

Answer 1

Let the original amount of moisture in the porous substance be N and the amount of moisture in the porous substance at any time t be P.

$$\begin{gathered} \mathrm{Given}:\frac{dP}{dt}\alpha P \\ \Rightarrow \frac{dP}{dt}=-aP \\ \Rightarrow \frac{dP}{P}=-adt \\ \Rightarrow \log \left|P\right|=-at+C.....\left(1\right) \\ \mathrm{Now},P=N\mathrm{at}t=0 \\ \mathrm{Putting}P=N\mathrm{and}t=0\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log \left|N\right|=C \\ \mathrm{Putting}C=\log \left|N\right|\mathit{}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log \left|P\right|=-at+\log \left|N\right| \\ \Rightarrow \log \left|\frac{P}{N}\right|=-at.....\left(2\right) \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}, \\ \mathrm{P}=\frac{1}{2}\mathrm{N}\mathrm{a}\mathrm{t}\mathrm{t}=1 \\ \log \left|\frac{N}{2N}\right|=-a \\ \Rightarrow a=\log \left|2\right| \\ \mathrm{Putting}a=\log \left|2\right|\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ \log \left|\frac{P}{N}\right|=-t\log \left|2\right| \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{time}\mathrm{when}\mathrm{it}\mathrm{will}\mathrm{loss}95\%\mathrm{moisture},\mathrm{we}\mathrm{have} \\ P=\left(1-95\%\right)N=\frac{5}{100}N \\ \therefore \log \left|\frac{5N}{100N}\right|=-t\log \left|2\right| \\ \Rightarrow \log 20=t\log 2 \\ \Rightarrow t=\frac{\log 20}{\log 2} \end{gathered}$$