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Q.
Pressure inside two soap bubbles are $1.01$ and $1.02$ atmospheres. Ratio between their volumes is
Punjab PMETPunjab PMET 2001Mechanical Properties of Fluids
Solution:
From the question we have
Excess pressure inside Ist bubble is
$=(1.01-1)$ atmosphere
$=0.01$ atmosphere
Excess pressure inside IInd bubble is
$=(1.02-1)=0.02$ atmosphere
While we know excess pressure are given by
$\frac{4 T}{r_{1}}$ and $\frac{4 T}{r_{2}}$ respectively
So, $\frac{\frac{4 T}{r_{1}}}{\frac{4 T}{r_{2}}}=\frac{0.01}{0.02}$
or $\frac{r_{2}}{r_{1}}=\frac{1}{2}$
Let volumes of first and second bubbles are $V_{1}$ and $V_{2}$ respectively
So, $\frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{4}{3} \pi r_{2}^{3}}$
$=\left(\frac{r_{1}}{r_{2}}\right)^{2}=\left(\frac{2}{1}\right)^{3}=\frac{8}{1}$
Hence, $V_{1}: V_{2}=8: 1$