A soap bubble has radius R and thickness d ( < < R) as shown. It … - Vidyadip

MCQ

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

A
${\left( {\frac{R}{{3d}}} \right)^{\frac{1}{3}}}$
B
${\left( {\frac{R}{{6d}}} \right)^{\frac{1}{3}}}$
✓
${\left( {\frac{R}{{24d}}} \right)^{\frac{1}{3}}}$
D
None

✓

Answer

Correct option: C.

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

c
Let $r$ be the radius of the water drop formed.

Since the volume of the water forming bubble and drop is same,

$\frac{4}{3} \pi\left(R^{3}-(R-d)^{3}\right)=\frac{4}{3} \pi r^{3}$

$\Longrightarrow r^{3} \approx 3 R^{2} d\left(\text { neglecting } d^{2} \text { and } d^{3}\right)$

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 $=\frac{1}{2}\left(\frac{R}{r}\right)$

Substituting the value of $\mathrm{r}$ gives $\left(\frac{R}{24 d}\right)^{1 / 3}$

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