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Question

Pressure inside two soap bubbles are $$1.01$$ and $$1.02$$ atmospheres. Ratio between their volumes is


A
102:101
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B
(102)3:(101)3
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C
8:1
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D
2:1
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Solution

The correct option is C $$8:1$$
Outside pressure = $$1$$ $$atm$$ 
Pressure inside first bubble = $$1.01$$ $$atm$$ 
Pressure inside second bubble = $$1.02$$ $$atm$$ 
Excess pressure $$\Delta P_{1}=1.01-1=0.01$$ $$atm$$ 
Excess pressure $$\Delta P_{2}=1.02-1=0.02$$ $$atm$$

$$\Delta P\propto \dfrac{1}{r}\Rightarrow r\propto \dfrac{1}{\Delta P}\Rightarrow \dfrac{r_{1}}{r_{2}}=\dfrac{\Delta P_{2}}{\Delta P_{1}}=\dfrac{0.02}{0.01}=\dfrac{2}{1}$$

Since, $$V=\dfrac{4}{3}\pi r^{3}\therefore \dfrac{V_{1}}{V_{2}}=\left ( \dfrac{r_{1}}{r_{2}} \right )^{3}=\left ( \dfrac{2}{1} \right )^{3}=\dfrac{8}{1}$$

Physics

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