(B) Let $M(t)$ be the mass at time $t$. The rate of disintegration is given by $\frac{dM}{dt} = -kM$.Separating variables, we get $\frac{dM}{M} = -k d

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(B) Let $M(t)$ be the mass at time $t$. The rate of disintegration is given by $\frac{dM}{dt} = -kM$.Separating variables, we get $\frac{dM}{M} = -k dt$.Integrating both sides, $\ln M = -kt + C$.At $t = 0$, $M = M_0 = 1.5 \text{ gm}$, so $C = \ln(1.5)$.Thus, $\ln M = -kt + \ln(1.5)$, which simplifies to $\ln(\frac{M}{1.5}) = -kt$.We want to find $t$ when $M = 0.5 \text{ gm}$.Substituting the values, $\ln(\frac{0.5}{1.5}) = -kt$.$\ln(\frac{1}{3}) = -kt$.$-\ln(3) = -kt$.$t = \frac{1}{k} \ln 3$.

Class 12 Mathematics Differential Equations Application of differential equations

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The rate of disintegration of a radioactive element at any time is proportional to its mass at that time, where $k (k > 0)$ is the constant of proportionality. The time during which an original mass of $1.5 \text{ gm}$ will disintegrate to a mass of $0.5 \text{ gm}$ is ...

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A
$k \log 3$
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$\frac{1}{k} \log 3$
C
$k \log 5$
D
$\frac{1}{k} \log 5$

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The rate of disintegration of a radioactive element at any time is proportional to its mass at that time, where $k (k > 0)$ is the constant of proportionality. The time during which an original mass of $1.5 \text{ gm}$ will disintegrate to a mass of $0.5 \text{ gm}$ is ...

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Let the curve $y = y(x)$ be the solution of the differential equation,$\frac{dy}{dx} = 2(x + 1)$. If the numerical value of the area bounded by the curve $y = y(x)$ and the $x-$axis is $\frac{4\sqrt{8}}{3}$,then the value of $y(1)$ is equal to

The solution of the differential equation $\frac{dy}{dx} = \frac{a + bx}{c + dy}$ represents a family of circles centered at the origin if...

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