Graph is drawn between rate of decay $\frac{\mathrm{dN}}{\mathrm{dt}}$ and time for radioactive samples A and B. The sample having shorter mean life is

From the given graph, slope of A is greater than B so rate of decay of A is greater than of B.

$\mathrm{slope} \mathrm{of} \mathrm{the} \mathrm{graph} = \frac{\mathrm{d}\left({\displaystyle \raisebox{1ex}{$\mathrm{dN}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{dt}$}\right.}\right)}{\mathrm{dt}}={\mathrm{\lambda }}^2\mathrm{N}$

$\mathrm{m}= \frac{\mathrm{d}\left({\displaystyle \raisebox{1ex}{$\mathrm{dN}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{dt}$}\right.}\right)}{\mathrm{dt}}\propto \frac{1}{{\tau }^2}$

${\mathrm{m}}_{\mathrm{A}}<{\mathrm{m}}_{\mathrm{B}}\Rightarrow {\tau }_{\mathrm{A}}>{\tau }_{\mathrm{B}}$