Activities of three radioactive substance $\text{A, B}$ and $\text{C}$ are represented by the curves $\text{A, B}$ and $\text{C}$, in the figure. Then their half gives $T_{1/2}\left(A\right):T_{1/2}\left(B\right):T_{1/2}\left(C\right)$ are in the ratio

The correct option is D $2:1:3$
We know that,

$\frac{R}{R_0}=e^{-\lambda t}$

Where, $R$ is the activity after time $t$.
$R_0$ is the initial activity.

$\Rightarrow \ln \left(\frac{R}{R_0}\right)=-\lambda t$

$\Rightarrow \ln R-\ln R_0=-\lambda t$

$\Rightarrow \ln R=-\lambda t+\ln R_0$

From the above equation, we get $\lambda =$ slope of the graph.

From graph :

${\lambda }_A=\frac{6}{10}=\frac{\ln 2}{T_{1/2}\left(A\right)}$

${\lambda }_B=\frac{6}{5}=\frac{\ln 2}{T_{1/2}\left(B\right)}$

${\lambda }_C=\frac{2}{5}=\frac{\ln 2}{T_{1/2}\left(C\right)}$

$T_{1/2}\left(A\right):T_{1/2}\left(B\right):T_{1/2}\left(C\right)=\frac{\ln 2}{{\lambda }_A}:\frac{\ln 2}{{\lambda }_B}:\frac{\ln 2}{{\lambda }_C}$

$=\frac{10}{6}:\frac{5}{6}:\frac{5}{2}=10:5:15$

$\therefore T_{1/2}\left(A\right):T_{1/2}\left(B\right):T_{1/2}\left(C\right)=2:1:3$

Hence, $\left(\text{D}\right)$ is the correct answer.

Why this question? This question gives an idea how to solve the question related decay graph.