MHT CET · Physics · Mechanical Properties of Fluids
Pressure inside two soap bubbles is \(1.01 \mathrm{~atm}\) and \(1.03 \mathrm{~atm}\). The ratio between their volumes is (Pressure outside the soap bubble is \(1 \mathrm{~atm}\) )
- A 9:1
- B 27:1
- C 81:1
- D 3:1
Answer & Solution
Correct Answer
(B) 27:1
Step-by-step Solution
Detailed explanation
Excess pressure inside a soap bubble is given by
\(\Delta \mathrm{P}=\mathrm{P}_{\mathrm{i}}-\mathrm{P}_0=\frac{4 \mathrm{~T}}{\mathrm{r}}\)
\(\begin{aligned} & \Delta \mathrm{P}_1=1.01 \mathrm{~atm}-1 \mathrm{~atm}=0.01 \mathrm{~atm} \\ & \Delta \mathrm{P}_2=1.03 \mathrm{~atm}-1 \mathrm{~atm}=0.03 \mathrm{~atm}\end{aligned}\)
\(\begin{aligned} & \therefore \frac{\Delta \mathrm{P}_1}{\Delta \mathrm{P}_2}=\frac{\mathrm{r}_2}{\mathrm{r}_1} \\ & \frac{0.03}{0.01}=3=\frac{\mathrm{r}_2}{\mathrm{r}_1} \\ & \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^3=(3)^3=27\end{aligned}\)
\(\Delta \mathrm{P}=\mathrm{P}_{\mathrm{i}}-\mathrm{P}_0=\frac{4 \mathrm{~T}}{\mathrm{r}}\)
\(\begin{aligned} & \Delta \mathrm{P}_1=1.01 \mathrm{~atm}-1 \mathrm{~atm}=0.01 \mathrm{~atm} \\ & \Delta \mathrm{P}_2=1.03 \mathrm{~atm}-1 \mathrm{~atm}=0.03 \mathrm{~atm}\end{aligned}\)
\(\begin{aligned} & \therefore \frac{\Delta \mathrm{P}_1}{\Delta \mathrm{P}_2}=\frac{\mathrm{r}_2}{\mathrm{r}_1} \\ & \frac{0.03}{0.01}=3=\frac{\mathrm{r}_2}{\mathrm{r}_1} \\ & \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^3=(3)^3=27\end{aligned}\)
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