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Question Number 155561 by peter frank last updated on 02/Oct/21

Answered by mr W last updated on 02/Oct/21

$$p_1=pressureinsidebubbleatradiusr$$$$p_2=pressureinsidebubbleatradius2r$$$$p_1-p_0=\frac{4\gamma }{r}\Rightarrow p_1=p_0+\frac{4\gamma }{r}$$$$p_2-p=\frac{4\gamma }{2r}\Rightarrow p_2=p+\frac{2\gamma }{r}$$$$\frac{p_2}{p_1}=\frac{V_1}{V_2}={\left(\frac{r}{2r}\right)}^3=\frac{1}{8}$$$$p+\frac{2\gamma }{r}=\frac{1}{8}(p_0+\frac{4\gamma }{r})$$$$p=\frac{1}{8}(p_0+\frac{4\gamma }{r})-\frac{2\gamma }{r}$$$$\Rightarrow p=\frac{1}{8}(p_0-\frac{12\gamma }{r})$$

Commented by mr W last updated on 02/Oct/21

Commented by peter frank last updated on 02/Oct/21

$$\mathrm{great}\mathrm{sir};\mathrm{thanks}$$

Commented by Tawa11 last updated on 02/Oct/21

$$\mathrm{great}\mathrm{sir}$$

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