Question

Two water drops each of radius $ r $ coalesce to form a bigger drop. If $ T $ is the surface tension, the surface energy released in this process is:

• A. \(4\pi r^2 T (2 - 2^{2/3})\)
• B. \( 4\pi r^2 T \left[ 2^{-1} - 2^3 \right] \)
• C. \( 4\pi r^2 T \left[ 1 + \sqrt{2} \right] \)
• D. \( 4\pi r^2 T \left[ \sqrt{2} - 1 \right] \)

Hint:

To solve problems involving surface tension, always use the principle of volume conservation and the fact that energy is related to the surface area of the drop.

Answer 1

The Correct Option is A

The problem is about finding the surface energy released when two identical spherical water drops of radius \( r \) coalesce into a single larger drop. The surface tension of the liquid is \( T \).

Concept Used:

The surface energy \( E \) of a liquid drop is directly proportional to its surface area:

\[ E = 4\pi R^2 T \]

When two drops combine, the total volume remains constant. Since the new radius \( R \) changes, the total surface area decreases, releasing energy equal to the difference in initial and final surface energies.

Step-by-Step Solution:

Step 1: Apply volume conservation:

\[ 2 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \] \[ \Rightarrow R^3 = 2r^3 \quad \Rightarrow \quad R = 2^{1/3}r \]

Step 2: Calculate the total initial surface energy of two small drops:

\[ E_i = 2 \times 4\pi r^2 T = 8\pi r^2 T \]

Step 3: Calculate the final surface energy of the single large drop:

\[ E_f = 4\pi R^2 T = 4\pi (2^{1/3}r)^2 T = 4\pi 2^{2/3} r^2 T \]

Step 4: Energy released is the difference between the initial and final energies:

\[ \Delta E = E_i - E_f = 8\pi r^2 T - 4\pi 2^{2/3} r^2 T = 4\pi r^2 T (2 - 2^{2/3}) \]

Final Computation & Result:

The surface energy released when two identical drops of radius \( r \) coalesce is:

\[ \boxed{\Delta E = 4\pi r^2 T (2 - 2^{2/3})} \]

Correct Option: (A)