The excess pressure in a soap bubble is doubl | vedclass

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Question

(b)

Excess pressure in soap bubble $=\frac{4 S}{R}$

$P=\frac{4 S}{R}$   $\left\{\begin{array}{l}S 	ext { - Surface tension } \ R 	e

Vedclass · Accepted Answer

$1: 8$

Vedclass · Answer

(b)

Excess pressure in soap bubble $=\frac{4 S}{R}$

$P=\frac{4 S}{R}$   $\left\{\begin{array}{l}S 	ext { - Surface tension } \ R 	ext { - Radius }\end{array}ight.$

For second bubble,

$2 P=\frac{4 S}{x}$   $\left\{\begin{array}{l}S 	ext { - Surface tension } \ R 	ext { - Radius }\end{array}ight.$

Substitute value of $P$

$2 	imes \frac{4 S}{R}=\frac{4 S}{x}$

$\Rightarrow x=\frac{R}{2}$

Ratio of Radii $=\frac{R / 2}{R}=\frac{1}{2}$

So, Ratio of volume $=\left(ight.$ Ratio of Radii )$^3$

$=\left(\frac{1}{2}ight)^3=\frac{1}{8}$

The excess pressure in a soap bubble is double that in other one. The ratio of their volume is .............

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