Q109 / 180 · NEET 2014 · Physics

A certain number of spherical drops of a liquid of radius $r$ coalesce to form a single drop of radius $R$ and volume $V .$ If $T$ is the surface tension of the liquid then:

A
Energy $4 V T (\frac{1}{r} - \frac{1}{R})$ is released
B
Energy $= 3 V T (\frac{1}{r} + \frac{1}{R})$ is released
Correct answer
C
Energy $= 3 V T (\frac{1}{r} - \frac{1}{R})$ is released
D
Energy is neither released nor absorbed

Solution

∆ U = (S T) (∆ A)

A (i n i t i a l) = (4 π r^{2}) n

A (f i n a l) = 4 π R^{2}

∆ A = (4 π r^{2}) n - 4 π R^{2}

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

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

∆ A = 4 π [\frac{R^{3}}{r^{3}} . r^{2} - R^{2}] = 4 π [\frac{R^{3}}{r} - \frac{R^{3}}{R}] = (\frac{4 π R^{3}}{3}) 3 [\frac{1}{r} - \frac{1}{R}]

∆ A = 3 V [\frac{1}{r} - \frac{1}{R}]

∆ U = 3 V T [\frac{1}{r} - \frac{1}{R}]

Hence, the correct answer is option: 'C'.

Relevant Questions