#### Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 79 textbook solution.

Answer : $\mathrm{T}=\frac{\log 20}{\log 2}$

Given: A wet porous substance in the wet air loses its moisture at a rate proportional to the moisture content

Hint: Using variable separable method

Explanation: Let M be the moisture content at any time A

According to given,

\begin{aligned} &\frac{\mathrm{d} \mathrm{M}}{\mathrm{dt}} \alpha \mathrm{M} \\ &\Rightarrow \frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{km} \end{aligned}

when k be any constant and [-ve sign. ? of it losses its moisture]

$\Rightarrow \frac{d m}{m}=-k d t$

Integrating both the sides.

\begin{aligned} &\int \frac{\mathrm{dm}}{\mathrm{m}}=-\mathrm{k} \int \mathrm{dt}\\ &\log \mathrm{m}=-\mathrm{kt}+\mathrm{c}-(1)\\ &\text { Let at } \mathrm{t}=0 \text {, }\\ &\mathrm{M}=\mathrm{M}_{0} \end{aligned}

Put in (1)

\begin{aligned} &\log \mathrm{M}=-\mathrm{kt}+\log \mathrm{M}_{0} \\ &\Rightarrow \log \mathrm{M}-\log \mathrm{M}_{0}=-\mathrm{Kt} \\ &\Rightarrow \log \left|\frac{\mathrm{M}}{\mathrm{M}_{0}}\right|=-\operatorname{lct}-(2) \end{aligned}                                  $\left[\because \log \mathrm{a}-\log \mathrm{b}=\frac{\log \mathrm{a}}{\mathrm{b}}\right]$

Given tht at $t=1,$

$\mathrm{m}=\frac{1}{2} \times \mathrm{m}_{0}$

Put in (2)

\begin{aligned} &\Rightarrow \log \left|\frac{\mathrm{M}}{2 \mathrm{M}}\right|=-\mathrm{k} \times 1 \\ &\Rightarrow \log \frac{1}{2}=-\mathrm{K} \\ &\Rightarrow \log 2^{-1}=-\mathrm{k} \\ &\Rightarrow \log 2^{-1}=-\mathrm{k} \end{aligned}                          $\left[\because \log \mathbf{a}^{\mathbf{b}}=\text { blo g } \mathbf{a}\right]$

\begin{aligned} &\Rightarrow-1 \cdot \log 2=-\mathrm{k} \\ &\mathrm{k}=\log 2 \end{aligned}

Put in (2)

\begin{aligned} &\log \left|\frac{\mathrm{M}}{\mathrm{M}_{0}}\right|=-\log 2 \cdot \mathrm{t} \\ &\Rightarrow \log \left|\frac{\mathrm{M}_{0}}{\mathrm{M}}\right|=\log 2 \cdot \mathrm{t}-(3) \end{aligned}

Let at $t=T,$

moisture = 95%

$\Rightarrow$ Remaining = 5 %

\begin{aligned} &\mathrm{M}=5 \% \text { of } \mathrm{M}_{0} \\ &\Rightarrow \mathrm{M}=\frac{5}{100} \times \mathrm{M}_{0} \\ &=\frac{1}{20} \mathrm{M}_{0} \end{aligned}

Put in (3)

\begin{aligned} &\log \left(\frac{20 \mathrm{M}}{\mathrm{M}}\right)=+\log 2 \cdot \mathrm{T} \\ &\Rightarrow \log 2_{0}=\log 2 \cdot \mathrm{T} \\ &\Rightarrow \mathrm{T}=\frac{\log 20}{\log 2} \end{aligned}