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Q.
The excess pressure in a soap bubble is double that in other one. The ratio of their volume is
Mechanical Properties of Fluids
Solution:
Excess pressure in soap bubble $=\frac{4 S}{R}$
Let for first bubble,
$P=\frac{4 S}{R}$
$S -$ Surface tension
$R$ - Radius
For second bubble,
$2 P=\frac{4 S}{x}$
$S-$ Surface tension
$x-$ Radius
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 $=$ (Ratio of Radii) $^{3}$.
$=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$