JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Fluid Mechanics

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 mechanical equivalent of heat is J , the rise in heat energy per unit volume will be:

Answer: (A)

Solution

When n drops of radius r combine to form a drop of radius R, volume conservation gives

n \cdot \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}

, so

n = \frac{R ^{3}}{r ^{3}}

. The change in surface energy is

Δ E = T (4 π n r^{2} - 4 π R^{2}) = 4 π T (\frac{R ^{3}}{r} - R^{2}) = 4 π T R^{2} (\frac{R}{r} - 1)

. The final volume is

V = \frac{4}{3} π R^{3}

. Heat energy per unit volume is

\frac{Δ E}{J V} = \frac{4 π T R ^{2} ( \frac{R}{r} - 1 )}{J \cdot \frac{4}{3} π R ^{3}} = \frac{3 T}{J} (\frac{1}{r} - \frac{1}{R})

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option