Question

The count rate from $100\text{ c}{{\text{m}}^{3}}$ of a radioactive liquid is $c$. Some of this liquid is now discarded. The count rate of the 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: The first step should be to find out a relation between the count rate and the volume of the liquid. Then three half-lives mean, the count rate is halved after each half life, which means that the count rate of the remaining liquid decreases by a factor of $8$ eventually.

Complete answer:
It is given that the count rate from $100\text{ c}{{\text{m}}^{3}}$ of a radioactive liquid is $c$, then count rate of $c/10$ will be for a volume of $10\text{ c}{{\text{m}}^{3}}$.
As it is given that some of the liquid is discarded from a total volume of $100\text{ c}{{\text{m}}^{3}}$, then let us assume that the volume of discarded liquid is $x\text{ c}{{\text{m}}^{3}}$. If an amount $x$ is withdrawn, then the remaining amount will be:
$100-x\text{ c}{{\text{m}}^{3}}$
As it is given that the count rate of the remaining liquid is found to be $c/10$ after three half-lives, i.e., the volume of the remaining liquid becomes $10\text{ c}{{\text{m}}^{3}}$ after three half-lives. We also know that after each life the liquid will be halved and then eventually becomes $10\text{ c}{{\text{m}}^{3}}$, therefore equating these, we get:
$\begin{align}
  & \dfrac{100-x}{2\times 2\times 2}=10 \\
 & \Rightarrow \dfrac{100-x}{8}=10 \\
 & \Rightarrow 100-x=8\times 10 \\
 & \therefore 100-x=80 \\
\end{align}$
Here $100-x$ is the volume of the remaining liquid when $x\text{ c}{{\text{m}}^{3}}$ of the liquid is withdrawn from the initial volume. Therefore, the volume of the remaining liquid would be $80\text{ c}{{\text{m}}^{3}}$.

Hence the correct option is $D$ out of the given options.

Note:
One more way to solve this question was first find out the count rate for $1\text{ c}{{\text{m}}^{3}}$ of the liquid, which will be $c/100$. After three half lives the count rate would become $\dfrac{1}{8}\times \dfrac{c}{100}$ and now if the volume of the remaining liquid is taken as $V$, then we can write the equation as $V\times \dfrac{1}{8}\times \dfrac{c}{100}=\dfrac{c}{10}$.