A wet substance in the open air loses its moi | vedclass

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(C) Let $M(t)$ be the moisture content at time $t$. The rate of change is given by $\frac{dM}{dt} = -kM$,where $k > 0$.Solving this differential equat

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$\log _2 10$

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(C) Let $M(t)$ be the moisture content at time $t$. The rate of change is given by $\frac{dM}{dt} = -kM$,where $k > 0$.Solving this differential equation,we get $M(t) = M_0 e^{-kt}$,where $M_0$ is the initial moisture.Given that at $t = 1$ hour,the moisture lost is $50 \%$,so $M(1) = 0.5 M_0$.$0.5 M_0 = M_0 e^{-k} \implies e^{-k} = 0.5 = \frac{1}{2}$.Taking natural logarithms,$-k = \ln(1/2) = -\ln 2$,so $k = \ln 2$.We want to find $t$ such that $90 \%$ of the moisture is lost,meaning $10 \%$ remains.$M(t) = 0.1 M_0 = M_0 e^{-kt}$.$0.1 = e^{-kt} \implies \ln(0.1) = -kt$.$-\ln 10 = -(\ln 2)t$.$t = \frac{\ln 10}{\ln 2} = \log_2 10$ hours.

$A$ 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 $90 \%$ of the moisture will be lost in ...... hours.

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