2014 AIPMT Mechanical Properties of FluidsSurface energy Medium Question 33 of 34 in Mechanical Properties of Fluids

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 $= 4VT\left(\frac{1}{r} - \frac{1}{R}\right)$ is released
B energy $= 3VT\left(\frac{1}{r} + \frac{1}{R}\right)$ is absorbed
C energy $= 3VT\left(\frac{1}{r} - \frac{1}{R}\right)$ is released
D energy is neither released nor absorbed

The answer is C.

Solution

The surface area decreases, so energy is released.
Released energy $= 4\pi R^2T[n^{1/3} - 1]$, where $R = n^{1/3}r$
$= 4\pi R^3T\left[\frac{1}{r} - \frac{1}{R}\right] = 3VT\left[\frac{1}{r} - \frac{1}{R}\right]$

Concept Coalescence of drops

Mechanical Properties of Fluids in past papers

34 questions from this chapter have appeared across 14 exam years.