Question

Two soap bubbles are blown such that the excess of pressure in the first bubble is 3 times that in the second. The ratio of surface area of the two bubbles is

Answer 1

Hlw mate


            Excess pressure in a soap bubble is given by \[P=\frac{4T}{R}=\Rightarrow P\propto \frac{1}{R}\]                 Hence,                  \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}\]                 Given:                   \[{{P}_{1}}=4{{P}_{2}}\]                 \[\frac{4{{P}_{2}}}{{{P}_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}\Rightarrow \frac{{{R}_{2}}}{{{R}_{1}}}=4\Rightarrow \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{4}=1:4\

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