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The pressure inside two soap bubbles is 1.01 atmosphere and 1.02 atmosphere. The ratio of their respective volume is

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Detailed Solution

$\frac{Δ p_{1}}{Δ p_{2}} = \frac{0.01}{0.02} = \frac{1}{2}$

or $\frac{4 T / r_{1}}{4 T / r_{2}} = \frac{1}{2}$ or $\frac{r_{1}}{r_{2}} = 2$

$∴ \frac{v_{1}}{v_{2}} = {(\frac{r_{1}}{r_{2}})}^{3} = 8$

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