MHT CET · Physics · Mechanical Properties of Fluids
The excess pressure inside the first soap bubble of radius ' \(\mathrm{R}_{1}\) ' is two times, that inside the second soap bubble of radius ' \(\mathrm{R}_{2}\) '. The ratio of volumes of the first bubble to that of second bubble is
- A \(1: 4\)
- B \(1: 1\)
- C \(1: 2\)
- D \(1: 8\)
Answer & Solution
Correct Answer
(D) \(1: 8\)
Step-by-step Solution
Detailed explanation
Excess pressure \(\quad P=\frac{4 T}{R}\)
\(\therefore \frac{P_{1}}{P_{2}}=\frac{R_{2}}{R_{1}}\)
\(\frac{P_{1}}{P_{2}}=2\)
\(\frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi R_{1}^{2}}{\frac{4}{3} \pi R_{2}^{3}}=\frac{R_{1}^{3}}{R_{2}^{3}}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}\)
\(\therefore \frac{P_{1}}{P_{2}}=\frac{R_{2}}{R_{1}}\)
\(\frac{P_{1}}{P_{2}}=2\)
\(\frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi R_{1}^{2}}{\frac{4}{3} \pi R_{2}^{3}}=\frac{R_{1}^{3}}{R_{2}^{3}}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}\)
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