MHT CET · Physics · Mechanical Properties of Fluids
Two soap bubbles having radii ' \(r_1\) ' and ' \(r_2\) ' has inside pressure ' \(\mathrm{P}_1\) ' and ' \(\mathrm{P}_2\) ' respectively. If \(\mathrm{P}_0\) is external pressure then ratio of their volume is
- A \(\frac{\left(\mathrm{P}_1-\mathrm{P}_0\right)}{\left(\mathrm{P}_2-\mathrm{P}_0\right)}\)
- B \(\frac{\left(\mathrm{P}_2-\mathrm{P}_0\right)}{\left(\mathrm{P}_1-\mathrm{P}_0\right)}\)
- C \(\frac{\left(\mathrm{P}_2-\mathrm{P}_0\right)^3}{\left(\mathrm{P}_1-\mathrm{P}_0\right)^3}\)
- D \(\frac{\left(\mathrm{P}_1-\mathrm{P}_0\right)^3}{\left(\mathrm{P}_2-\mathrm{P}_0\right)^3}\)
Answer & Solution
Correct Answer
(C) \(\frac{\left(\mathrm{P}_2-\mathrm{P}_0\right)^3}{\left(\mathrm{P}_1-\mathrm{P}_0\right)^3}\)
Step-by-step Solution
Detailed explanation
Excess pressure \(=P_i-P_0=\frac{4 T}{R} \Rightarrow P_i-P_0 \propto \frac{1}{R}\)
\(\therefore \quad\) As per the equation,
\(\begin{aligned}
& \frac{P_1-P_0}{P_2-P_0}=\frac{r_2}{r_1} \\
& \text { Volume }=\frac{4}{3} \pi R^3 \Rightarrow \quad \text { Volume } \propto R^3 \\
\therefore \quad & \frac{V_1}{V_2}=\left(\frac{r_1}{r_2}\right)^3=\left(\frac{P_2-P_0}{P_1-P_0}\right)^3
\end{aligned}\)
\(\therefore \quad\) As per the equation,
\(\begin{aligned}
& \frac{P_1-P_0}{P_2-P_0}=\frac{r_2}{r_1} \\
& \text { Volume }=\frac{4}{3} \pi R^3 \Rightarrow \quad \text { Volume } \propto R^3 \\
\therefore \quad & \frac{V_1}{V_2}=\left(\frac{r_1}{r_2}\right)^3=\left(\frac{P_2-P_0}{P_1-P_0}\right)^3
\end{aligned}\)
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