A liquid drop of radius R = 10⁻² m and surface tension S = 0.1/(4π) N m⁻¹ divides into K identical drops. The total increase in surface energy is ΔU = 10⁻³ J. If K = 10^α, find α.
Physics · Properties of solid and fluid
JEE Advanced 2017

Question

A liquid drop of radius R = 10⁻² m and surface tension S = 0.1/(4π) N m⁻¹ divides into K identical drops. The total increase in surface energy is ΔU = 10⁻³ J. If K = 10^α, find α.

G1 Physics · Accepted Answer

Step 1: Volume conservation gives K r³ = R³, so new droplet radius r=R/K^(1/3). Step 2: Increase in surface area is 4πR²(K^(1/3)−1), and ΔU=SΔA. With the given numbers, K=10⁶, so α=6. Final answer: 6.