The rate of decay of mass of a certain substa | vedclass

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(D) Let $m$ be the mass of the substance at time $t$.The rate of decay is given by $\frac{dm}{dt} = -km$.Separating the variables,we get $\frac{dm}{m}

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $m$ be the mass of the substance at time $t$.The rate of decay is given by $\frac{dm}{dt} = -km$.Separating the variables,we get $\frac{dm}{m} = -k dt$.Integrating both sides,we get $\log m = -kt + C$.At $t = 0$,$m = m_{0}$,so $\log m_{0} = -k(0) + C$,which gives $C = \log m_{0}$.Substituting $C$ back into the equation: $\log m = -kt + \log m_{0}$.Rearranging gives $\log m - \log m_{0} = -kt$,or $\log \left(\frac{m}{m_{0}}\right) = -kt$.Thus,$t = -\frac{1}{k} \log \left(\frac{m}{m_{0}}\right) = \frac{1}{k} \log \left(\frac{m_{0}}{m}\right)$.When $m = m_{1}$,the time $t$ is $\frac{1}{k} \log \left(\frac{m_{0}}{m_{1}}\right)$.

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 $m_{0}$ grams will be reduced to $m_{1}$ grams is (where $k$ is the constant of proportionality):

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