MHT CET · Physics · Mechanical Properties of Fluids
The pressure inside two soap bubbles, (A) is 1.01 and that of (B) is 1.02 atmosphere respectively. The ratio of their respective radii (A to B) is (outside pressure \(=1 \mathrm{~atm}\).)
- A \(2: 1\)
- B \(4: 1\)
- C \(6: 1\)
- D \(8: 1\)
Answer & Solution
Correct Answer
(A) \(2: 1\)
Step-by-step Solution
Detailed explanation
Outside pressure \(=1 \mathrm{~atm}\)
Pressure inside first bubble \(=1.01 \mathrm{~atm}\)
Pressure inside second bubble \(=1.02 \mathrm{~atm}\)
\(\therefore \quad\) Excess pressures will be \(\Delta \mathrm{P}_{\mathrm{A}}=1.01-1=0.01 \mathrm{~atm}\) and \(\Delta \mathrm{P}_{\mathrm{B}}=1.02-1=0.02 \mathrm{~atm}\) Now, \(\Delta \mathrm{P} \propto \frac{1}{\mathrm{r}} \Rightarrow \mathrm{r} \propto \frac{1}{\Delta \mathrm{P}}\) \(\therefore \quad \frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=\frac{\Delta \mathrm{P}_{\mathrm{B}}}{\Delta \mathrm{P}_{\mathrm{A}}}=\frac{0.02}{0.01}=\frac{2}{1}\)
Pressure inside first bubble \(=1.01 \mathrm{~atm}\)
Pressure inside second bubble \(=1.02 \mathrm{~atm}\)
\(\therefore \quad\) Excess pressures will be \(\Delta \mathrm{P}_{\mathrm{A}}=1.01-1=0.01 \mathrm{~atm}\) and \(\Delta \mathrm{P}_{\mathrm{B}}=1.02-1=0.02 \mathrm{~atm}\) Now, \(\Delta \mathrm{P} \propto \frac{1}{\mathrm{r}} \Rightarrow \mathrm{r} \propto \frac{1}{\Delta \mathrm{P}}\) \(\therefore \quad \frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=\frac{\Delta \mathrm{P}_{\mathrm{B}}}{\Delta \mathrm{P}_{\mathrm{A}}}=\frac{0.02}{0.01}=\frac{2}{1}\)
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