(C) Given that the rate of decay is proportional to the mass: $\frac{dm}{dt} = -Km$. Integrating both sides: $\int \frac{dm}{m} = \int -K dt$. This gi

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$\left(\frac{1}{K}\right) \log \left(\frac{m_{0}}{m_{1}}\right)$

(C) Given that the rate of decay is proportional to the mass: $\frac{dm}{dt} = -Km$. Integrating both sides: $\int \frac{dm}{m} = \int -K dt$. This gives $\log m = -Kt + c$. At $t = 0$,$m = m_{0}$,so $\log m_{0} = c$. Substituting $c$ back: $\log m = -Kt + \log m_{0}$. Rearranging: $\log m - \log m_{0} = -Kt$,which implies $\log \left(\frac{m}{m_{0}}\right) = -Kt$. When $m = m_{1}$,we have $\log \left(\frac{m_{1}}{m_{0}}\right) = -Kt$. Solving for $t$: $t = -\frac{1}{K} \log \left(\frac{m_{1}}{m_{0}}\right) = \frac{1}{K} \log \left(\frac{m_{0}}{m_{1}}\right)$.

Class 12 Mathematics Differential Equations Application of differential equations

Medium

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 $m_{0} \text{ gm}$ will be reduced to $m_{1} \text{ gm}$ is (where $K$ is the constant of proportionality).

MHT CET 2020 All MHT CET

A
$K \log \left(\frac{m_{1}}{m_{0}}\right)$
B
$\left(\frac{1}{K}\right) \log \left(\frac{m_{1}}{m_{0}}\right)$
C
$\left(\frac{1}{K}\right) \log \left(\frac{m_{0}}{m_{1}}\right)$
D
$K \log \left(\frac{m_{0}}{m_{1}}\right)$

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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 time during which the original mass of $m_{0} \text{ gm}$ will be reduced to $m_{1} \text{ gm}$ is (where $K$ is the constant of proportionality).

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