#### Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 11

Answer: $\frac{1}{k} \log 2$   where lambda is thee constant of proportionality.

Given: Decaying rate of radium at any time is proportional to its mass at that time.

To find: We have to find the time when the mass will be half of its initial mass.

Hint: The amount of radium at any time t is proportional to its mass i.e. $\frac{d A}{d t} \propto A$

Solution: Given that

$\frac{d A}{d t} \propto A$

$=>\frac{d A}{d t}=-\lambda A$           [Where ? is the constant of proportionality and minus sign indicates that A decrease with increase in t]

$=>\frac{d A}{A}=-\lambda d t$

Integrating on both sides

\begin{aligned} &=>\int \frac{d A}{A}=-\lambda \int d t \\\\ &=>\log A=-\lambda t+C \ldots(i) \end{aligned}

Since initial amount of radium is A0 then

\begin{aligned} &=>\log A_{0}=-\lambda \times 0+C \\\\ &=>\log A_{0}=C \end{aligned}

Putting value of C in equation (i) we get

\begin{aligned} &=>\log A=-\lambda t+\log A_{0} \\\\ &=>\log A-\log A_{0}=-\lambda t \\\\ &=>\log \frac{A}{A_{0}}=-\lambda t_{1} \end{aligned}

\begin{aligned} &=>\log \frac{1}{2}=-\lambda t_{1} \\\\ &=>-\log 2=-\lambda t_{1} \\\\ &=>t_{1}=\frac{1}{\lambda} \log 2 \end{aligned}

Hence, the time taken $=\frac{1}{\lambda} \log 2$   where lambda is the constant of proportionality.