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The rate of decay of mass of a certain substance at time 't' is proportional to the mass at that instant.…
The rate of decay of mass of a certain substance at time 't' is proportional to the
mass at that instant. The time during which the original mass of $m_{0}$ gm. will be left to $m_{1} \mathrm{gm}$. is
(K is constant of proportionality)
$\mathrm{K} \log \left(\frac{m_{1}}{m_{0}}\right)$ $\left(\frac{1}{\mathrm{~K}}\right) \log \left(\frac{m_{1}}{m_{0}}\right)$ $\left(\frac{1}{\mathrm{~K}}\right) \log \left(\frac{m_{0}}{m_{1}}\right)$ $\mathrm{K} \log \left(\frac{m_{0}}{m_{1}}\right)$ Solution
Given $\frac{\mathrm{dm}}{\mathrm{dt}} \propto \mathrm{m} \Rightarrow \frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{Km} \Rightarrow \int \frac{\mathrm{dm}}{\mathrm{m}}=\int-\mathrm{Kdt}$
$\therefore \log \mathrm{m}=-\mathrm{Kt}+\mathrm{c}$
When $\mathrm{t}=0, \mathrm{~m}=\mathrm{m}_{0}$
$\therefore \log \mathrm{m}_{0}=0+\mathrm{c} \Rightarrow \mathrm{c}=\log \mathrm{m}_{0}$
$\log \mathrm{m}=-\mathrm{Kt}+\log \mathrm{m}_{0}$
$\therefore \log \left(\frac{\mathrm{m}}{\mathrm{m}_{0}}\right)=-\mathrm{Kt}$
When $\mathrm{m}=\mathrm{m}_{1}$, we get
$\log \left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{0}}\right)=-\mathrm{Kt}$
$\quad \mathrm{t}=\frac{-1}{\mathrm{~K}} \log \left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{0}}\right)=\frac{1}{\mathrm{~K}} \log \left(\frac{\mathrm{m}_{0}}{\mathrm{~m}_{1}}\right)$
Asked in: MHT CET 2020 (15 Oct Shift 1)
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