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Question: The count rate from \(100\text{ c}{{\text{m}}^{3}}\) of a radioactive liquid is \(c\). Some of this ...

The count rate from 100 cm3100\text{ c}{{\text{m}}^{3}} of a radioactive liquid is cc. Some of this liquid is now discarded. The count rate of the remaining liquid is found to be c/10c/10 after three half-lives. The volume of the remaining liquid in cm3c{{m}^{3}}, is
A.2020
B.4040
C.6060
D.8080

Explanation

Solution

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 88 eventually.

Complete answer:
It is given that the count rate from 100 cm3100\text{ c}{{\text{m}}^{3}} of a radioactive liquid is cc, then count rate of c/10c/10 will be for a volume of 10 cm310\text{ c}{{\text{m}}^{3}}.
As it is given that some of the liquid is discarded from a total volume of 100 cm3100\text{ c}{{\text{m}}^{3}}, then let us assume that the volume of discarded liquid is x cm3x\text{ c}{{\text{m}}^{3}}. If an amount xx is withdrawn, then the remaining amount will be:
100x cm3100-x\text{ c}{{\text{m}}^{3}}
As it is given that the count rate of the remaining liquid is found to be c/10c/10 after three half-lives, i.e., the volume of the remaining liquid becomes 10 cm310\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 cm310\text{ c}{{\text{m}}^{3}}, therefore equating these, we get:
100x2×2×2=10 100x8=10 100x=8×10 100x=80 \begin{aligned} & \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{aligned}
Here 100x100-x is the volume of the remaining liquid when x cm3x\text{ c}{{\text{m}}^{3}} of the liquid is withdrawn from the initial volume. Therefore, the volume of the remaining liquid would be 80 cm380\text{ c}{{\text{m}}^{3}}.

Hence the correct option is DD out of the given options.

Note:
One more way to solve this question was first find out the count rate for 1 cm31\text{ c}{{\text{m}}^{3}} of the liquid, which will be c/100c/100. After three half lives the count rate would become 18×c100\dfrac{1}{8}\times \dfrac{c}{100} and now if the volume of the remaining liquid is taken as VV, then we can write the equation as V×18×c100=c10V\times \dfrac{1}{8}\times \dfrac{c}{100}=\dfrac{c}{10}.