A soap bubble has radius R and thickness d (d | vedclass

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(C) Let $r$ be the radius of the water drop formed.Since the volume of the water forming the bubble and the drop is the same,the volume of the soap fi

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$(\frac{R}{24d})^{1/3}$

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(C) Let $r$ be the radius of the water drop formed.Since the volume of the water forming the bubble and the drop is the same,the volume of the soap film is $V = 4\pi R^2 d$.Equating this to the volume of the spherical drop: $4\pi R^2 d = \frac{4}{3} \pi r^3$.Thus,$r^3 = 3R^2d$,which implies $r = (3R^2d)^{1/3}$.The excess pressure in a spherical drop is $\Delta P_{	ext{drop}} = \frac{2\sigma}{r}$.The excess pressure inside a soap bubble is $\Delta P_{	ext{bubble}} = \frac{4\sigma}{R}$.The ratio is $\frac{\Delta P_{	ext{drop}}}{\Delta P_{	ext{bubble}}} = \frac{2\sigma/r}{4\sigma/R} = \frac{1}{2} \cdot \frac{R}{r}$.Substituting $r = (3R^2d)^{1/3}$,we get the ratio $= \frac{1}{2} \cdot \frac{R}{(3R^2d)^{1/3}} = \frac{1}{2} \cdot (\frac{R^3}{3R^2d})^{1/3} = \frac{1}{2} \cdot (\frac{R}{3d})^{1/3} = (\frac{1}{8} \cdot \frac{R}{3d})^{1/3} = (\frac{R}{24d})^{1/3}$.

$A$ soap bubble has radius $R$ and thickness $d$ $(d \ll R)$ as shown. It collapses into a spherical drop. The ratio of excess pressure in the drop to the excess pressure inside the bubble is

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