Correct Answer - b
As the number of undecayed nuclei decreases from 25% to 12.5% in 10 s, it shows that the half life of the sample is 10 s, i.e. `T_(1//2)` =10 s
Decay constant ,`lambda=0.06931/T_(1//2)=0.6931/"10 s"`
Mean life , `tau = 1/lambda= "10 s"/0.6931 =14.43 s`