A drop of liquid of diameter 2.8 mm breaks up into 125 identical drops. The change in energy is nearly ______ erg. ($S = 75 dyne c{m}^{-1}$)

here,$R=2.8/2=1.4 mm=0.14 cm$

Now $\frac{4}{3}\pi R^3=125\times \frac{4}{3}\pi r^3$

(equality of volume)

Or $r=\frac{R}{5}=\frac{0.14}{5}=0.028\mathrm{cm}$

$\therefore$ change in energy = surface tension  increase in area

$=75\times (125\times 4\times 4\pi r^2-4\pi R^2)=74 \mathrm{erg}$