MHT CET · Physics · Mechanical Properties of Fluids
small drops of liquid of same radius coalesce to form a big drop. The ratio of
the total surface energies after and before the change is
- A \(2^{3}: 1\)
- B \(2^{-\frac{1}{3}}: 1\)
- C \(2^{-\frac{2}{3}}: 1\)
- D \(2^{\frac{2}{3}}: 1\)
Answer & Solution
Correct Answer
(B) \(2^{-\frac{1}{3}}: 1\)
Step-by-step Solution
Detailed explanation
\(2 \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3} \quad \therefore \quad R=2^{\frac{1}{3}} r\)
Ratio of energies \(\frac{E_{2}}{E_{1}}=\frac{4 \pi R^{2} T}{2 \times \pi r^{2} \times T}\)
\(\frac{E_{2}}{E_{1}}=\frac{R^{2}}{2 r^{2}}=\frac{2^{\frac{2}{3}} r^{2}}{2 r^{2}}=2^{\frac{2}{3}-1}=2^{-\frac{1}{3}}\)
Ratio of energies \(\frac{E_{2}}{E_{1}}=\frac{4 \pi R^{2} T}{2 \times \pi r^{2} \times T}\)
\(\frac{E_{2}}{E_{1}}=\frac{R^{2}}{2 r^{2}}=\frac{2^{\frac{2}{3}} r^{2}}{2 r^{2}}=2^{\frac{2}{3}-1}=2^{-\frac{1}{3}}\)
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