Question

A drop of mercury of radius 2mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is 0.465 J/m²)

• A. 23.4μJ
• B. 18.5μJ
• C. 26.8μJ
• D. 16.8μJ

Answer 1

Answer: (A)

23.4μJ

Answer 2

Increase in surface energy $= 4 \pi R ^ { 2 } T \left( n ^ { 1 / 3 } - 1 \right)$

$= 4\pi(2 \times 10^{- 3})^{2}(0.465)(8^{1/3} - 1)$ =$23.4 \times 10^{- 6}J$ =$23.4\mu J$