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The rate of decay of mass of certain substance at time tis proportional to the tt that instant. The time…
The rate of decay of mass of certain substance at time tis proportional to the
tt that instant. The time during which the original mass of $m_{0}$ gram will be leftht
$m_{\text {, gram is }} \quad(k$ is constant of proportionality
$\frac{1}{k} \log \left(\frac{m_{1}}{m_{0}}\right)$ $k \log \left(\frac{m_{0}}{m_{1}}\right)$ $k \log \left(\frac{m_{1}}{m_{0}}\right)$ $\frac{1}{k} \log \left(\frac{m_{0}}{m_{1}}\right)$ Solution
Let $m$ be the initial mass of substance at time $t$.
$\begin{array}{l}
\therefore \frac{\mathrm{dm}}{\mathrm{dt}} \propto \mathrm{km} \Rightarrow \frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{km} \quad \Rightarrow \int \frac{\mathrm{dm}}{\mathrm{m}}=\int-\mathrm{k} \mathrm{dt} \\
\therefore \log \mathrm{m}=-\mathrm{kt}+\mathrm{c}....(1) \\
\text { When } \mathrm{t}=0, \mathrm{~m}=\mathrm{m}_{0} \text { we get } \\
\log \mathrm{m}_{0}=0+\mathrm{c} \Rightarrow \mathrm{c}=\log \mathrm{m}_{0} \\
\text { From }(1), \log \left(\frac{\mathrm{m}}{\mathrm{m}_{0}}\right)=-\mathrm{kt} \Rightarrow \mathrm{t}=\frac{-1}{\mathrm{k}} \log \left(\frac{\mathrm{m}}{\mathrm{m}_{0}}\right)
\end{array}$
$\begin{array}{l}
\text { When } m=m_{1} \text {, we get } \\
\qquad \begin{aligned}
t &=\frac{-1}{k} \log \left(\frac{m_{1}}{m_{0}}\right) \Rightarrow t=\frac{1}{k} \log \left(\frac{m_{0}}{m_{1}}\right)
\end{aligned}
\end{array}$
Asked in: MHT CET 2020 (14 Oct Shift 2)
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