A glass tube of uniform internal radius $(r)$ has a valve separating the two identical ends. Initially, the…
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,
air from end 1 flows towards end 2. No change in the volume of the soap bubbles
air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases
no change occurs
air from end 2 flows towards end 1. Volume of the soap bubble at end 1 increases
Solution
$\Delta p_1=\frac{4 T}{r_1}$ and $\Delta p_2=\frac{4 T}{r_2}$
$
\begin{array}{cc}
& r_1 < r_2 \\
\therefore & \Delta p_1>\Delta p_2
\end{array}
$
Air will flow from 1 to 2 and volume of bubble at end- 1 will decrease.
Therefore, correct option is (b)
