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ExamsJEE MainPhysics

A great many tiny water droplets, each of radius r, coalesce into one larger drop of radius R. If T is the surface tension and J is the mechanical equivalent of heat, find the rise in heat energy per unit volume of the resulting drop.

  1. 2T/J (1/r - 1/R)
  2. 2T/(rJ)
  3. 3T/(rJ)
  4. 3T/J (1/r - 1/R)

Correct answer: 3T/J (1/r - 1/R)

Solution

Volume conserved: N r³ = R³, so N = (R/r)³. Surface energy released = T(N*4pi r² - 4pi R²) = 4pi T(N r² - R²) = 4pi T(R³/r - R²). Heat per unit volume = released energy / [(4/3)pi R³ * J] = 4pi T(R³/r - R²) / ((4/3)pi R³ J) = 3T(1/r - 1/R)/J.

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