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Q.
Pressure inside two soap bubble are 1.01 and 1.03 aim, ratio between their volume is
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Solution:
Excess pressure as compared to atmosphere inside bubble A is $ \Delta {{p}_{1}}=1.01-1 $ $ =0.01\,\text{atm} $ Inside bubble B is $ \Delta {{p}_{2}}=1.03-1 $ $ =0.03\,\text{atm} $ $ \rho =\frac{4T}{r} $ Let $ {{r}_{1}} $ and $ {{r}_{2}} $ be the radii of bubbles A and 8 respectively, then $ \frac{{{p}_{1}}}{{{p}_{2}}}=\frac{4T/{{r}_{1}}}{4T/{{r}_{2}}} $ $ =\frac{0.01}{0.03} $ $ \frac{{{r}_{2}}}{{{r}_{1}}}=\frac{1}{3} $ Since, bubbles are spherical in shape, their volumes are in the ratio $ \frac{{{V}_{1}}}{{{V}_{2}}}=\frac{4/3\pi r_{1}^{3}}{4/3\pi r_{2}^{3}} $ $ \frac{{{V}_{1}}}{{{V}_{2}}}={{\left( \frac{3}{1} \right)}^{3}} $ $ =\frac{27}{1} $ $ {{v}_{1}}:{{v}_{2}}=27:1 $