Question

Eight droplets of mercury, each of radius 1 mm, coalesce to form a single drop. Find the change in the surface energy. [Surface tension of mercury = 0.472 J/m²]

Answer 1

Data : r = 1 mm = 1 × 10^-3 m, T = 0.472 J/m²

Let R be the radius of the single drop formed due to the coalescence of 8 droplets of mercury.

Volume of 8 droplets = volume of the single drop as the volume of the liquid remains constant.

∴ 8 × \(\frac43\) πr³ = \(\frac43\) πr³

∴ 8r³ = R³

∴ 2r = R

Surface area of 8 droplets = 8 × 4πr²

Surface area of single drop = 4πR²

∴ Decrease in surface area = 8 × 4πr² – 4πR²

= 4π(8r² – R²) 

= 4π[8r² – (2r)²] 

= 4π × 4r²

∴ The energy released

= surface tension × decrease in surface area

= T × 4π × 4r² 

= 0.472 × 4 × 3.142 × 4 × (1 × 10^-3)² 

= 2.373 × 10^-5 J