(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}

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

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

Class 12 Mathematics Differential Equations Application of differential equations

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

MHT CET 2020 All MHT CET

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

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

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Let $f:[0,2] \rightarrow R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1$. Let $F(x)=\int_0^{x^2} f(\sqrt{t}) dt$ for $x \in [0,2]$. If $F'(x)=f'(x)$ for all $x \in (0,2)$,then $F(2)$ equals

If the function $y = e^{4x} + 2e^{-x}$ is a solution of the differential equation $\frac{\frac{d^3y}{dx^3} - 13\frac{dy}{dx}}{y} = K$,then the value of $K$ is:

Water flows from the base of a rectangular tank of depth $16 \ m$. The rate of flow of the water is proportional to the square root of the depth at any time $t$. If the depth is $4 \ m$ when $t = 2 \ hours$,then after $3.5 \ hours$,the depth (in meters) is:

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