Question

A fixed number of spherical drops of a liquid of radius \(r\) coalesce to form a large drop of radius \(R\) and volume \(V\). If \(T\) is the surface tension, then the energy

• A. is neither released nor absorbed.
• B. \(3VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)\) is released.
• C. \(4VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)\) is released.
• D. \(3VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)\) is absorbed.

Hint:

When liquid drops coalesce, surface area decreases and surface energy is released.

Answer 1

The Correct Option is B

Step 1: Surface energy of a spherical drop.
Surface energy of a drop is:
\[ E = 4\pi r^{2}T \]
Step 2: Number of small drops.
Let the number of drops be \(n\).
Using volume conservation:
\[ n \cdot \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi R^{3} \Rightarrow n = \left(\frac{R}{r}\right)^{3} \]
Step 3: Initial surface energy.
\[ E_i = n \cdot 4\pi r^{2}T = 4\pi T \frac{R^{3}}{r} \]
Step 4: Final surface energy.
\[ E_f = 4\pi R^{2}T \]
Step 5: Energy released.
\[ \Delta E = E_i - E_f = 4\pi T R^{2}\left(\frac{R}{r}-1\right) \] Using \(V=\frac{4}{3}\pi R^{3}\):
\[ \Delta E = 3VT\left(\frac{1}{r}-\frac{1}{R}\right) \]
Step 6: Conclusion.
Energy \(3VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)\) is released.