Excess pressure in a coalesced bigger drop – Rankers Physics

Surface Tension and Viscosity: Practice Problem & Solution

n identical small drops of water having radius \( r \) coalesce to form a bigger drop. If surface tension of water is \( T \) then excess pressure in bigger drop will be
\[ \frac{n^3 4T}{r} \]
\[ \frac{2T}{n^{\frac{1}{3}} r} \]
\[ \frac{4T}{nr} \]
\[ \frac{2T}{n^3 r} \]

Solution Explained:

To solve this problem, we apply the core principles of Surface Tension and Viscosity. Understanding the underlying formula is key to arriving at the correct answer below:

By conserving volume, \( \frac{4}{3} \pi R^3 = n \left(\frac{4}{3} \pi r^3\right) ⇒ R = n^{1/3} r \). The excess pressure in a single-surface liquid drop of radius \( R \) is given by \( \Delta P = \frac{2T}{R} = \frac{2T}{n^{1/3} r} \).

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