\(N_{\mathrm{1}}=\) Remaining nuclei after \(40 \%\) decay

\(=(1-0.4) N_{0}=0.6 N_{0}\)

\(N_{2}=\) Remaining nuclei after \(85 \%\) decay

\(=(1-0.85) N_{0}=0.15 N_{0}\)

\(\therefore\)  \(\frac{N_{2}}{N_{1}}=\frac{0.15 N_{0}}{0.6 N_{0}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{2}\)

Hence, two half life is required between \(40 \%\) decay and \(85 \%\) decay of a radioactive substance.

\(\therefore\) Time taken \(=2 \tau_{1 / 2}=2 \times 30 \mathrm{min}=60 \mathrm{min}\)