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

4

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2

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8

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0.125

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Solution

The correct option is D $0.125$
$Δ P_{1} = 2 Δ P_{2} \Rightarrow \frac{4 S}{r_{1}} = 2 (\frac{4 S}{r_{2}}) \Rightarrow \frac{1}{r_{1}} = \frac{2}{r_{2}} \Rightarrow {r_{1}}^{3} = \frac{{r_{2}}^{3}}{8} \Rightarrow \frac{4}{3} π {r_{1}}^{3} = \frac{1}{8} \times \frac{4}{3} π {r_{2}}^{3} \Rightarrow V_{1} = 0.125 V_{2} = n V_{2} \Rightarrow n = 0.125$

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

The correct option is D 0.125ΔP1=2ΔP2⇒4Sr1=2(4Sr2)⇒1r1=2r2⇒r13=r238⇒43πr13=18×43πr23⇒V1=0.125V2=nV2⇒n=0.125

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