A glass tube of uniform internal radius ( $\mathrm{r}$ ) has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End $1$ has a hemispherical soap bubble of radius $\mathrm{r}$. End $2$ has sub-hemispherical soap bubble as shown in figure. Just after opening the valve,
Figure: $Image$
IIT 2008, Advanced
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$P_1=$ pressure just inside the bubble at the end $2=P_0+\frac{4 T}{R}$
$\mathrm{P}_2=$ pressure just inside the bubble at the end $1=\mathrm{P}_0+\frac{4 \mathrm{~T}}{\mathrm{r}}$
$\mathrm{R}>\mathrm{I} \Rightarrow \mathrm{P}_2<\mathrm{P}_1 \Rightarrow$ Air will flow from end $1$ to end $2$
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