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 its moisture during the first hour, then the time when it would have lost $99.9\%$ of its moisture is: (weather condition remaining same) [Take ${\log }_{10}2=0.301$ ]

• A. More then 100
• B. More than 10 hours
• C. Approximately 10 hours
• D. Approximately 9 hours

Answer 1

The correct option is C

Approximately 10 hours

Let 'm' be the moisture content at time 't'.

$\therefore \frac{dm}{dt}=kt$; where 'k' is constant of proportionality.

$\therefore logm=kt+c$

At t = 0, m = M - maximum moisture content.

$\therefore c=logM$

$\therefore logm=kt+logM$

$log\left(\frac{m}{M}\right)=kt$

$\therefore m=M{e}^{kt}$

At t = 1 hr; $m=\frac{M}{2}$

$\therefore \frac{M}{2}=M{e}^{kt}$

$\therefore k=\ln \left(\frac{1}{2}\right)=-\ln 2$

$\therefore m=M{e}^{-\left(\ln 2\right)t}$

Let at t = x hr; m = 0.1% of $M=\frac{0.1}{100}M$.

$\therefore \frac{0.1M}{100}=M{e}^{-\left(\ln 2\right)x}$

$\ln \left(\frac{1}{1000}\right)=-\left(\ln 2\right)x$

$\therefore x=\frac{\ln \left(1000\right)}{\ln \left(2\right)}$

$=\frac{{\log }_{10}\left(1000\right)}{{\log }_{10}2}=\frac{3}{{\log }_{10}2}\simeq 10hrs.$