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Question
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
Options
4
2
1
0.125
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Solution
Let the excess pressure inside the second bubble be P.
∴ 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.
\[\text{ Excess pressure inside the 2 nd soap bubble :} \]
\[P = \frac{4S}{R} . . . (1)\]
\[\text{ Excess pressure inside the 1st soap bubble: }\]
\[2P = \frac{4S}{x}\]
\[\text{ From (1), we get :} \]
\[2\left( \frac{4S}{R} \right) = \frac{4S}{x}\]
\[ \Rightarrow x = \frac{R}{2}\]
\[\text{ Volume of the first bubble }= \frac{4}{3} \pi x^3 \]
\[\text{ Volume of the second bubble } = \frac{4}{3} \pi R^3 \]
\[ \Rightarrow \frac{4}{3} \pi x^3 = n\frac{4}{3} \pi R^3 \]
\[ \Rightarrow x^3 = {nR}^3 \]
\[ \Rightarrow \left( \frac{R}{2} \right)^3 = {nR}^3 \]
\[ \Rightarrow n = \frac{1}{8} = 0 . 125\]
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