Question

A large number of water drops each of radius r combine to have a drop of radius R. If the surface tension is T and the mechanical equivalent at heat is J then the rise in temperature will be 

(A) (2T / rJ) 

(B) (3T / RJ) 

(C) (3T / J){(1/r) – (1/R)} 

(D) (2T / J){(1/r) – (1/R)}

Answer 1

Correct option: (C) (3T / J){(1/r) – (1/R)}  

Explanation:

n drops of radius r forms a drop of radius R

As mass is constant,

(4/3)πR³ = (4/3)πnr³ 

 ∴ R³ = nr³          (1)

Decrease in surface energy = surface Tension × change in surface area

W = T(n4πr² – 4πR²)

H = (W/J) -------------- J = heat equivalent

∴ m ∙ s ∙ Δt = (W/J)     (2)

Here s = 1 -------- for water

m = density × volume = d × (4/3)πR³ 

∴ msΔt = [{T(n ∙ 4πr² – 4πR²)} / J] from (2)

Δt = [{4πT(n ∙ r² – R²)} / (J × m × s)]

= [{4πT {(R³ / r³) × r² –R²}} / {J × d × (4πR³ / 3) × 1}] ----- from(1),

n = (R³ / r³)

= [4πT × 3 × {(R³ / r) – R²} / (J × d × 4πR³)]

= {(3T) / (JdR³)} × {R²(R/r) – 1}

= {(3T) / (JdR)} {(R/r) – 1}

Δt = (3T / Jd) {(1/r) – (1/R)} 

For water d = density = 1 

Δt = (3T / J) {(1/r) – (1/R)}