Question

The amount of energy evolved when eight droplets of mercury (surface tension 0.55 Nm^–1) of radius mm each combine into one drop is.......... 

(A) 18 μJ 

(B) 24 μJ 

(C) 28 μJ 

(D) 16 μJ

Answer 1

The correct option is (C) 28 μJ.

Explanation:

T = 0.55 N/m 

Amount of energy evolved = W = T ∙ ΔA 

∴ W = 055 × ΔA   (1)

ΔA = (n × 4πr²) – (1 × 4πR²) where r is radius of small drop. R is radius of big drop.

= (4π × 8 × r²) – 4πR²

ΔA = 4π(8r² – R²)    (2)

Also as mass remains same hence ρ ∙ (4/3)πr³ × 8 = ρ × (4/3)πR³