Question

A mercury drop of radius 0.5 cm falls from a height on a glass plate and breaks into one million droplets, all of the same size. Find the height from which the drop fell. [Density of mercury = 13600 kg/m³, surface tension of mercury = 0.465 N/m]

Answer 1

Data : R = 0.5 cm = 0.5 × 10^-2 m, n = 10⁶, ρ = 13600 kg m³ , T = 0.465 N/m, g = 9.8 m/s²

\(\frac43\)πR³ = n × \(\frac43\)πR³

as the volume of the mercury remains the same

This gives the radius of a droplet. 

By energy conservation, if h is the height from which the drop of mass m falls,

This gives the required height.