Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Radium disappears at a rate propotional to the amount present .If $5\%$ of the original amount disappears in $50$ years , how much will remain at the end of $100$ years [Take $A_{0}$ as a intial amount.]

Can you answer this question?

1 Answer

0 votes
  • First order,first degree DE variable separatable : Variable of a DE are rearranged to separate there (i.e.) $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
  • It can be written as $\large\frac{g_1(y)}{g_2(y)}$$dy=\large\frac{f_1(x)}{f_2(x)}$$dx$
  • The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
Step 1:
Let $x$ be the quantity of radium at time $t$ (in years)
At $t=0$,the quantity was $A_0$
Step 2:
The rate of change of $x,\large\frac{dx}{dt}\propto x$
$\therefore \large\frac{dx}{dt}$$=kx$
Where $k$ is the constant of proportionality .
Step 3:
Integrating on both sides we get,
$\int\large\frac{dx}{x}$$=\int kdt+\log c$
$\log x=kt+\log c$
$\log \large\frac{x}{c}$$=kt$
$\Rightarrow x=ce^{kt}$
Step 4:
When $t=0,x=A_0$
$\therefore A_0=c$
When $t=50,x=0.95A_0$
$\therefore 0.95A_0=A_0e^{50k}$
$\Rightarrow e^{50k}=0.95$
Step 5:
When $t=100$
$\therefore$ the amount remaining after 100 years =$0.9025A_0$
answered Sep 6, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App