A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be :

Complete Solution:

Diameter of the original water drop: d = 2 cm = 0.02 m

Radius of the original drop: r = d / 2 = 0.01 m

Number of smaller droplets: n = 64

Surface tension of water: γ = 0.075 N/m

Volume of the original drop = Volume of one small droplet × n

(4/3)πr³ = n × (4/3)πr_small³

r_small = r / ³√n = 0.01 / 4 = 0.0025 m

A_original = 4πr² = 4π(0.01)² = 4π × 10^-4 m²

A_small = 4πr_small² = 4π(0.0025)² = 25π × 10^-6 m²

A_total-small = n × A_small = 64 × 25π × 10^-6 = 1.6π × 10^-3 m²

ΔA = A_total-small - A_original = 1.6π × 10^-3 - 4π × 10^-4 = 1.2π × 10^-3 m²

ΔE = γ × ΔA = 0.075 × 1.2π × 10^-3 = 0.09π × 10^-3

Approximating π ≈ 3.14:

ΔE = 0.09 × 3.14 × 10^-3 = 0.2826 × 10^-3 = 2.826 × 10^-4 J

Final Answer:

The gain in surface energy is 2.83 × 10^-4 J.