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

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

Answers (1)

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}

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