Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. the ratio of volumes will be:
 
Option: 1 4:1
Option: 2 0.8:1
Option: 3 8:1  
Option: 4 2:1

Answers (1)

\\ \text{Outside pressure} = 1 atm\\ \text{Pressure inside first bubble} = 1.01 atm\\ \text{ Pressure inside second bubble} =1.02 atm \\ \text{Excess pressure in 1st bubble} \Delta P_{1}=1.01-1=0.01 \mathrm{atm} \\ \text{Excess pressure in 2nd bubble} \Delta P_{2}=1.02-1=0.02 \mathrm{atm} \\ \Delta P \propto \frac{1}{r} \Rightarrow r \propto \frac{1}{\Delta P} \\ \Rightarrow \frac{r_{1}}{r_{2}}=\frac{\Delta P_{2}}{\Delta P_{1}}=\frac{0.02}{0.01}=\frac{2}{1} \\ since \ V=\frac{4}{3} \pi r^{3} \\ \therefore \quad \frac{V_{1}}{V_{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{3}=\left(\frac{2}{1}\right)^{3}=\frac{8}{1}

So, option 3 is correct

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