Question

Find the work done when a drop of mercury of radius 2 mm breaks into 8 equal droplets. [Surface tension of mercury = 0.4855 J/m²].

Answer 1

Let R be the radius of the drop and r be the radius of each droplet.

Data: R = 0.2 cm = 2 × 10⁻³ m, n = 8, T = 0.4855 Jm²

As the volume of the liquid remains constant,

the volume of n droplets = volume of the drop

∴ `n × 4/3 π r^3 = 4/3 π r^3`

∴ `r^3 = R^3/n`

∴ r = `R/(root(3)(n)) = R/(root(3)(8)) = R/2`

Surface area of the drop = `4π R^2`

Surface area of n droplets = n × `4π R^2`

∴ The increase in the surface area = surface area of n droplets − surface area of drop

= 4π (nr² − R²)

= `4π (8 xx (R^2)/4 - R^2)`

=  4π (2 − 1)= R²

=  4πR²

∴ The work done = surface tension × increase in surface area

= T × 4πR2

= 0. 4855 × 4 × 3.142 × (2 × 10⁻³)²

= 1. 942 × 3.142 × 4 × 10⁻⁶

log 1.942   =     0.2882

log 3.142   =  + 0.4972

log 4         =   + 0.6021

                         1.3875

AL(1.3875) = 24.41

= 24.41 × 10⁻⁶ J

= 24.41 μJ