In 10 hours 75% mass of a radio active sample is converted into a stable element. Then mean life of the radio active element is

If $m_0$ be the original mass of radio active element and T be its half life, then after 10 hours mass of remaining radio active element will be $m_0/4$

$\therefore \frac{{\mathrm{m}}_0}{4}=\frac{{\mathrm{m}}_0}{{\left(2\right)}^{\raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$\mathrm{T}$}\right.}}\Rightarrow \frac{1}{{\left(2\right)}^2}=\frac{1}{{\left(2\right)}^{10/\mathrm{T}}}\Rightarrow \frac{10}{\mathrm{T}}=2\Rightarrow \mathrm{T}=5\mathrm{hrs}$

If half life is $\mathrm{T}$, then ${\mathrm{T}}^1=\frac{\mathrm{T}}{\ln 2_{ }}=\frac{5}{0.693}\mathrm{hr}=7.215\mathrm{hr}$