The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is

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KollegeApply · Accepted Answer

Let the excess pressure inside the second bubble be P. ​$\therefore$ Excess pressure inside the first bubble = 2P Let the radius of the second bubble be R. Let the radius of the first bubble be x. \text{ Excess pressure inside the 2 nd soap bubble :} P = 4S/R . . . (1) \text{ Excess pressure inside the 1st soap bubble: } 2P = 4S/x \text{ From (1), we get :} 2\left( 4S/R \right) = 4S/x \Rightarrow x = R/2 \text{ Volume of the first bubble }= $\frac{4}{3}$ $\pi$ x^3 \text{ Volume of the second bubble } = $\frac{4}{3}$ $\pi$ R^3 \Rightarrow $\frac{4}{3}$ $\pi$ x^3 = n4/3 $\pi$ R^3 \Rightarrow x^3 = {nR}^3 \Rightarrow \left( R/2 \right)^3 = {nR}^3 \Rightarrow n = $\frac{1}{8}$ = 0 . 125

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