The decay rate of radio active material at any time $t$ is proportional to its mass at that time. The mass…
The decay rate of radio active material at any time $t$ is proportional to its mass at that time. The mass is 27 grams when $t=0$. After three hours it was found that 8 grams are left. Then the substance left after one more hour is
$\frac{27}{8}$ grams
$\frac{81}{4}$ grams
$\frac{16}{3}$ grams
$\frac{16}{9}$ grams
Solution
Let ' $x$ ' be the mass of the material at time ' $t$ '.
$\therefore \quad \frac{\mathrm{d} x}{\mathrm{dt}}=-\mathrm{k} x,(-\mathrm{ve}$ sign indicates decay. $)$
$\therefore \quad \int \frac{\mathrm{d} x}{x}=-\mathrm{k} \int \mathrm{dt}$
$\therefore \quad \log |x|=-\mathrm{kt}+\mathrm{c}$
When $\mathrm{t}=0, x=27$
$\therefore \quad \mathrm{c}=\log 27$
$\therefore \quad \log |x|=-\mathrm{kt}+\log 27$
When $\mathrm{t}=3, x=8$
$\therefore \quad \mathrm{k}=\log \left(\frac{3}{2}\right)$
When $\mathrm{t}=4$, we get
$\begin{array}{ll}
& \log |x|=-4 \log \left(\frac{3}{2}\right)+\log 27 \\
\therefore \quad & \log |x|=\log \left(\frac{16}{3}\right) \\
\therefore \quad & x=\frac{16}{3} \text { grams }
\end{array}$