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  • Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 77 textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 77 textbook solution.

Answers (1)

Answer : \frac{\log 2}{\lambda}

Given:    The decay rate of radium at any time t is proportional to its mass at that time.

Hint:       Using variable separable method

Explanation:   Acc to given,

\begin{aligned} &\frac{d m}{d t} \alpha m \\ &\Rightarrow \frac{d m}{d t}=-d m \end{aligned}

\begin{aligned} &\text { where } \lambda>0 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \text { [-ve sign } \therefore \text { of decay rate }] \\ &\Rightarrow \frac{d m}{m}=-d t \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{\mathrm{dm}}{\mathrm{m}}=-\int \lambda \mathrm{dt}\\ &\Rightarrow \log \mathrm{m}=-\lambda+\mathrm{c}-(1)\\ &\text { Let } \mathrm{m}_{0} \text { be the initial mass at } \mathrm{t}=0\\ &\log m_{0}=-\lambda \cdot 0+c\\ &\Rightarrow \mathrm{c}=\log \mathrm{m}_{0} \end{aligned}

Put in (1)

\begin{aligned} &\Rightarrow \log \mathrm{m}=-\lambda \mathrm{t}+\log \mathrm{m}_{0} \\ &\Rightarrow \log \mathrm{m}-\log \mathrm{m}_{0} \\ &=-\lambda \mathrm{t} \\ &\Rightarrow \log \left(\frac{\mathrm{m}}{\mathrm{m}_{0}}\right)=-\lambda \mathrm{t}-2 \end{aligned}                                                 \left[\because \log \mathbf{m}-\log \mathbf{x}=\log \mathbf{m}_{\mathrm{h}}\right]

We have to find the time when m=\frac{1}{2}m_{0}

\Rightarrow 2m=m_{0}

Put in (2)

\begin{aligned} &\Rightarrow \log \frac{\mathrm{m}}{2 \mathrm{~m}}=-\lambda \mathrm{t} \\ &\Rightarrow \log \frac{1}{2}=-\lambda \mathrm{t} \\ &\Rightarrow \log 1-\log 2=-\lambda \mathrm{t} \\ &\Rightarrow 0-\log 2=-\lambda \mathrm{t} \end{aligned}                                        [\because \log 1=0]

\begin{aligned} &\Rightarrow \lambda t=\log 2 \\ &\Rightarrow \mathrm{t}=\frac{\log 2}{\lambda} \end{aligned}

 

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