A wet substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the open air loses half its moisture during the first hour, then the time t , in which $99 \%$ of the moisture will be lost, is

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Aicharya · Accepted Answer

Let $y$ be the amount of moisture at time t . $\begin{aligned} \therefore \quad & \frac{\mathrm{d} y}{\mathrm{dt}}=-\alpha y \\ & \Rightarrow \frac{\mathrm{~d} y}{y}=-\alpha \mathrm{dt} \end{aligned}$
Integrating on both sides, we get $\int \frac{\mathrm{d} y}{y}=-\alpha \int \mathrm{dt}$ $\therefore \quad \log y=-\alpha \mathrm{t}+\mathrm{c}$ when $\mathrm{t}=0, y=1$ $\therefore \quad$ From (i), we get $\mathrm{c}=0$...(i) when $\mathrm{t}=1, y=0.5$ $\therefore \quad$ From (i) and (ii), we get $\begin{array}{ll} & \log (0.5)=-\alpha \\ \therefore \quad & \alpha=\log 2 \end{array}$ $\therefore \quad$ From (i), (ii) and (iii), we get $\log y=-(\log 2) t$ $\therefore \quad$ When $99 \%$ of the moisture will be lost, $\begin{array}{ll} & y=0.01 \\ \therefore \quad & \log (0.01)=-(\log 2) t \\ \therefore \quad & t=\frac{2 \log 10}{\log 2} \end{array}$

A wet substance in the open air loses its moisture at a rate proportional to the moisture content. If a…

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