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

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Question

Let $r$ be the radius of the water drop formed.

Since the volume of the water forming bubble and drop is same,

$\frac{4}{3} \pi\left(R^{3}-(R-d)

Vedclass · Accepted Answer

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

Vedclass · Answer

Let $r$ be the radius of the water drop formed.

Since the volume of the water forming bubble and drop is same,

$\frac{4}{3} \pi\left(R^{3}-(R-d)^{3}ight)=\frac{4}{3} \pi r^{3}$

$\Longrightarrow r^{3} \approx 3 R^{2} d\left(	ext { neglecting } d^{2} 	ext { and } d^{3}ight)$

Ratio of excess pressure in the drop to the excess pressure inside the bubble is given by,

Ratio $=\frac{2 \sigma / r}{4 \sigma / R}$

Ratio $=\frac{1}{2}\left(\frac{R}{r}ight)$

Substituting the value of $\mathrm{r}$ gives $\left(\frac{R}{24 d}ight)^{1 / 3}$

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

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