Here, s = 0.072 N/m
r = 0.5 mm = 0.5 × 10-3 m.
Let R be the radius of big drop formed.
Volume of the big drop = volume of 8 small drops.
\(\frac{4}{3}\)π R3 = 8 × \(\frac{4}{3}\)πr3
or R = 2 r = 2 × 0.5 × 10-3 = 10-3 m
Surface area of big drop :
= 4π R2 = 4π × (10-3 )2
= 4π × 10-6m2
Surface area of 8 small drops:
= 8 × 4π × r2
= 8 × 4π × (0.5 × 10-3 )2
= 8π × 10-6m2
Energy evolved = S.T × decrease in area.
= 0.072 × 4π × 10-6 = 9.05 × 107J.
