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:
Show Hint
\(4\pi r^2 T (2 - 2^{2/3})\)
- \( 4\pi r^2 T \left[ 2^{-1} - 2^3 \right] \)
- \( 4\pi r^2 T \left[ 1 + \sqrt{2} \right] \)
- \( 4\pi r^2 T \left[ \sqrt{2} - 1 \right] \)
The Correct Option is A
Solution and Explanation
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)
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