PhysicsMedium148×since 2002Q8022A mercury drop of radius 10−3 m10^{-3}~\mathrm{m}10−3 m is broken into 125 equal size droplets. Surface tension of mercury is 0.45 Nm−10.45~\mathrm{Nm}^{-1}0.45 Nm−1. The gain in surface energy is :A28×10−5 J28\times10^{-5}~\mathrm{J}28×10−5 JB17.5×10−5 J17.5\times10^{-5}~\mathrm{J}17.5×10−5 JC5×10−5 J5\times10^{-5}~\mathrm{J}5×10−5 JD2.26×10−5 J2.26\times10^{-5}~\mathrm{J}2.26×10−5 JCheck answerSkip