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

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

Answers (1)

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}

Now,

\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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