Question

A liquid drop of radius 4 mm breaks into 1000 identical drops. Find the change in surface energy. S = 0.007 Nm to the power of -1.

Answer 1

Hi

Here is your answer,

Volume of 1000 shapes drops = Volume of a large drop

                  1000 × 4/3 πr³ = 4/3 πR³

                       r = R/10

Surface area of large drop = 4πR³

Surface area of 1000 drop = 4π × 1000 r² = 40πR²

∴ Increase in surface area = (40 - 4)πR² = 36 πR²

The increase in surface area 

     = Surface tension × Increase in surface area 

  = 36πR² × 0.07 = 36 × 3.14 × ( 4 × 10⁻³) × 0.07

 = 1.26 × 10⁻⁴ J.



Hope it helps you !