Question

The count rate for \[100c{{m}^{3}}\]of radioactive liquid is c. Some of this liquid is now discarded. The count rate of remaining liquid is found to be c/10 after three half-lives. The volume of the remaining liquid in \[c{{m}^{3}}\] is
A) 20
B) 40
C) 60
D) 80

Answer 1

Hint: Here we have given radioactive liquid and its count rate is given as c for the volume \[100c{{m}^{3}}\]and after three half-lives the count rate is found to be c/10. We have to find the remaining volume after three half-life. Now activity for the radioactive liquid is given as count rate per volume. And by using the formula for activity of remaining volume we can solve the given question.
Formula used:
\[\begin{align}
  & A=\dfrac{c}{V} \\
 & A={{A}_{0}}{{\left( \dfrac{1}{2} \right)}^{n}} \\
\end{align}\]

Complete step by step answer:
Here for a radioactive liquid the count rate is given as c initially when its volume is \[100c{{m}^{3}}\]and then due to radioactivity it decreases and the count rate for the decreased volume is c/10.
Now for a radioactive liquid the activity is given as the count rate per volume for any given instant. Mathematical equation for activity can be given as
\[A=\dfrac{c}{V}\]
Where c is count rate and V is the volume of the radioactive liquid.
For a radioactive substance, if initially the activity was \[{{A}_{0}}\] and after n half-lives the activity A will be given as
\[A={{A}_{0}}{{\left( \dfrac{1}{2} \right)}^{n}}\text{ }......\text{(i)}\]
Now according to the question initially volume was \[100c{{m}^{3}}\]and the count rate was c, therefore initial activity \[{{A}_{0}}\] will be given as
\[{{A}_{0}}=\dfrac{c}{100}\]
Let us say, the volume of radioactive liquid decreased to V after three half-lives and the count rate at this volume was c/10 , then equation (i) can be rewritten as
\[\begin{align}
  & A={{A}_{0}}{{\left( \dfrac{1}{2} \right)}^{3}} \\
 & \Rightarrow \left( \dfrac{\dfrac{c}{10}}{V} \right)=\dfrac{c}{100}\left( \dfrac{1}{8} \right) \\
 & \Rightarrow \left( \dfrac{c}{10V} \right)=\dfrac{c}{800} \\
 & \Rightarrow V=\dfrac{c}{10}\times \dfrac{800}{c} \\
 & \Rightarrow V=80c{{m}^{3}} \\
\end{align}\]

Hence option D is correct.

Note:
“A” is the activity for the remaining volume and not for the volume which is discarded. Also note that here c is count rate and not speed of light as we consider in optics and mostly in other numerical. Half-lives is the time period in which half of the substance is discarded or decreased or we can say it is the time period at which only half of the substance is remaining.