Question

When a drop of mercury of radius R is split into n similar drops, What is the change in surface energy? Assume σ as surface tension of mercury.

Answer 1

Volume of initial mercury drop = \(\frac{4}{3}\)πR³

r = radius of smaller drops,

volume conservation

⇒ \(\frac{4}{3}\) πR³ = n\(\frac{4}{3}\) πr³

r = Rn^-1/3

Initial Surface Energy = σ × surface area

= σ × 4πR²

Final Surface Energy = n × σ × 4πr²

= nσ 4π[Rn^-1/3]²

= nσ 4πR²n^-2/3

= n^1/34σ πR²

Change in surface energy

= 4σ πR² [1 – n^1/3]