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Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 78 textbook solution.

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Answer:  0.04%

Given: Radius disintegrates at a rate proportional to the amount of radium present at the moment

Hint: Using variable separable method

Explanation: Let A be the amount radium at that time

According to given,

\begin{aligned} &\Rightarrow \frac{\mathrm{dA}}{\mathrm{dt}} \propto \mathrm{A} \\ &\Rightarrow \frac{\mathrm{dA}}{\mathrm{dt}}=-\lambda \mathrm{A}, \lambda>0 \\ &\Rightarrow \frac{\mathrm{dA}}{\mathrm{A}}=-\lambda \mathrm{dt} \end{aligned}

Integrating both sides,

\begin{aligned} &\int \frac{\mathrm{d} \mathrm{A}}{\mathrm{A}}=-\lambda \int \mathrm{dt} \\ &\Rightarrow \log _{\mathrm{e}} \mathrm{A}=-\lambda \mathrm{t}+\mathrm{c}-(1) \end{aligned}

Let A_{0} be the initial amount at t=0

\begin{aligned} &\Rightarrow \log _{\mathrm{e}} \mathrm{A}_{0}=\mathrm{C} \\ &\therefore \log _{\mathrm{e}} \mathrm{A}=-\lambda \mathrm{t}+\log _{\mathrm{e}} \mathrm{A}_{0} \\ &\Rightarrow \log _{\mathrm{e}} \mathrm{A}-\log _{\mathrm{e}} \mathrm{A}_{0}=-\lambda \mathrm{t} \\ &\Rightarrow \log \mathrm{e} \frac{\mathrm{A}}{\mathrm{A}_{0}}=-\lambda \mathrm{t} \Rightarrow \log \frac{\mathrm{A}_{0}}{\mathrm{~A}}=\lambda \mathrm{t} \end{aligned}


\begin{aligned} &\mathrm{A}=\frac{\mathrm{A}_{\mathrm{o}}}{2} \\ \\&\mathrm{t}=1590 \\ \\&\Rightarrow \lambda \times 1590 \\ \\&=\log \frac{2 \mathrm{~A}}{\mathrm{~A}} \end{aligned}

\begin{aligned} &\Rightarrow \frac{\log 2}{1590}=\lambda \\ \\&\Rightarrow \lambda=\log _{c}\left(\frac{1+0}{A}\right) \\ \\&t=1 \\ \\&\Rightarrow \frac{\log 2}{1590}=\log c \end{aligned}

\begin{aligned} &\frac{A_{0}}{A} \Rightarrow \frac{A_{0}}{A}=e^{\frac{\log _{2}}{1590}} \\ \\&\Rightarrow A=e^{-\frac{l o g 2}{1590}} A_{0} \\ \\&\Rightarrow A=0.9996 A_{0} \end{aligned}                                                               [\because \text {of given}]

\begin{aligned} \therefore \text { Reqd. } \%=& \frac{\mathrm{A}-\mathrm{A}_{0}}{\mathrm{~A}_{0}} \times 100 \\ =& \frac{0.9996 \mathrm{~A}_{0}-\mathrm{A}_{0}}{\mathrm{~A}_{0}} \times 100 \\ &=-0.04 \% \end{aligned}

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