Q.
A glass tube of uniform internal radius ( $r$ ) 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 $r$. End $2$ has sub-hemispherical soap bubble as shown in figure. Just after opening the valve
Mechanical Properties of Fluids
Solution:
Pressure inside tube $=P=P_0+\frac{4 T}{r}$
$\therefore P_2 < P_1\left(\right.$ since $r _2> r _1$ )
Hence pressure on side $1$ will be greater than side $2$ .
So air from end $1$ flows towards end $2$
