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