d
Let after \(t\,s\) amount of the \(A_{1}\) and \(A_{2}\) will become equal in the mixture.

As \(N=N_{0}\left(\frac{1}{2}\right)^{n}\)

where \(n\) is the number of half-lives

For \(A_{1}, N_{1}=N_{01}\left(\frac{1}{2}\right)^{t / 20}\)

For \(A_{2}, N_{2}=N_{02}\left(\frac{1}{2}\right)^{t / 10}\)

According to question, \(N_{1}=N_{2}\)

\({\frac{40}{2^{t / 20}}=\frac{160}{2^{t / 10}}}\)

\({2^{t/10}} = 4\left( {{2^{t/20}}} \right)\) or \({2^{t/10}} = {2^2}{2^{t/20}}\) \( \Rightarrow {2^{t/10}} = {2^{\left( {\frac{t}{{20}} + 2} \right)}}\)

\({\frac{t}{10}=\frac{t}{20}+2 \text { or } \frac{t}{10}-\frac{t}{20}=2}\)

or \(\quad \frac{t}{20}=2 \quad\) or \(\quad t=40 \mathrm{\,s}\)