9.2 , surface tension — Physics STD 11 Science — Question

Excess pressure in soap bubble $=\frac{4 S}{R}$

$P=\frac{4 S}{R}$   $\left\{\begin{array}{l}S \text { - Surface tension } \\ R \text { - Radius }\end{array}\right.$

For second bubble,

$2 P=\frac{4 S}{x}$   $\left\{\begin{array}{l}S \text { - Surface tension } \\ R \text { - Radius }\end{array}\right.$

Substitute value of $P$

$2 \times \frac{4 S}{R}=\frac{4 S}{x}$

$\Rightarrow x=\frac{R}{2}$

Ratio of Radii $=\frac{R / 2}{R}=\frac{1}{2}$

So, Ratio of volume $=\left(\right.$ Ratio of Radii )$^3$

$=\left(\frac{1}{2}\right)^3=\frac{1}{8}$