A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is___________________ cm.

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Vishal Baghel | Contributor-Level 10

2 months ago

$p_{1} - p_{2} = \frac{4 T}{r_{1}} . . . . . . . (1)$

$p_{2} - p_{0} = \frac{4 T}{r_{2}} . . . . . . . (2)$

(1) + (2), p₁ – p₀ = 4T $(\frac{1}{r_{1}} + \frac{1}{r_{2}})$

$\Rightarrow \frac{4 T}{r_{e q}} = 4 T (\frac{1}{r_{1}} + \frac{1}{r_{2}})$

$r_{e g} = \frac{r_{1} r_{2}}{r_{1} + r_{2}} = \frac{3 \times 6}{3 + 6} = 2 c m$

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A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is___________________ cm.

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