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MHT CET · Physics · Mechanical Properties of Fluids

Two small drops of mercury each of radius 'R' coalesce to form a large single drop.
The ratio of the total surface energies before and after the change is

  1. A \(\sqrt{2}: 1\)
  2. B \(2^{1 / 3}: 1\)
  3. C \(2: 1\)
  4. D \(2^{2 / 3}: 1\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2^{1 / 3}: 1\)

Step-by-step Solution

Detailed explanation

\(\frac{4}{3} \pi R^{3} \times 2=\frac{4}{3} \pi R^{\prime 3}\)
\(R^{\prime}=2^{1 / 3} R\)
Total surface energy before \(=2 \times 4 \pi R^{2} T \quad(T=\) surface tension \()\)
Total surface energy after \(=4 \pi R^{\prime 2} T\)
\(\therefore \quad\) Ratio \(=\frac{2 \times R^{2}}{R^{\prime 2}}=\frac{2 \times R^{2}}{2^{2 / 3} R^{2}}=\frac{2}{2^{2 / 3}}=2^{1-\frac{2}{3}}=2^{1 / 3}\)
\(\therefore\) Ratio \(=2^{1 / 3}: 1\)