Q132 / 150 · MHT CET 2024 (03 May Shift 2) · Physics

The pressure inside a soap bubble A is 1.01 atmosphere and that in a soap bubble B is 1.02 atmosphere. The ratio of volume of bubble A to that of $B$ is [Surrounding pressure $=1$ atmosphere $]$

  • A
    $101: 102$
  • B
    $102: 101$
  • Correct answer
    C
    $8: 1$
  • D
    $2: 1$

Solution

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 $\Delta \mathrm{P}_1=1.01-1=0.01 \mathrm{~atm}$ and $\Delta \mathrm{P}_2=1.02-1=0.02 \mathrm{~atm}$ $\begin{array}{ll} & \text { Now, } \Delta \mathrm{P} \propto \frac{1}{\mathrm{r}} \Rightarrow \mathrm{r} \propto \frac{1}{\Delta \mathrm{P}} \\ \therefore \quad & \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{\Delta \mathrm{P}_2}{\Delta \mathrm{P}_1}=\frac{0.02}{0.01}=\frac{2}{1} \\ & \text { Now, } \mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3 \\ & \Rightarrow \mathrm{~V} \propto \mathrm{r}^3 \\ \therefore \quad & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^3=\left(\frac{2}{1}\right)^3=\frac{8}{1}\end{array}$

Hence, the correct answer is option: 'C'.

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