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(D) Let the time after which the amounts become equal be $t$.For radioactive material $A_1$,the amount remaining is given by $N_1 = N_{01} \left(\frac

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$40$

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(D) Let the time after which the amounts become equal be $t$.For radioactive material $A_1$,the amount remaining is given by $N_1 = N_{01} \left(\frac{1}{2}\right)^{t/T_1} = 40 \left(\frac{1}{2}\right)^{t/20}$.For radioactive material $A_2$,the amount remaining is given by $N_2 = N_{02} \left(\frac{1}{2}\right)^{t/T_2} = 160 \left(\frac{1}{2}\right)^{t/10}$.Setting $N_1 = N_2$,we get:$40 \left(\frac{1}{2}\right)^{t/20} = 160 \left(\frac{1}{2}\right)^{t/10}$.Dividing both sides by $40$:$\left(\frac{1}{2}\right)^{t/20} = 4 \left(\frac{1}{2}\right)^{t/10}$.$\left(\frac{1}{2}\right)^{t/20} = 2^2 \left(\frac{1}{2}\right)^{t/10} = \left(\frac{1}{2}\right)^{-2} \left(\frac{1}{2}\right)^{t/10}$.$\left(\frac{1}{2}\right)^{t/20} = \left(\frac{1}{2}\right)^{t/10 - 2}$.Equating the exponents:$\frac{t}{20} = \frac{t}{10} - 2$.$2 = \frac{t}{10} - \frac{t}{20} = \frac{2t - t}{20} = \frac{t}{20}$.$t = 40 \ s$.

$A$ mixture consists of two radioactive materials $A_1$ and $A_2$ with half-lives of $20 \ s$ and $10 \ s$ respectively. Initially,the mixture has $40 \ g$ of $A_1$ and $160 \ g$ of $A_2$. The amount of the two in the mixture will become equal after: (in $s$)

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