Question

$A$ glass tube of uniform internal radius $\left(r\right)$ 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 $r_1$. End $2$ has sub- hemispherical soap bubble of radius $r_2$ as shown in figure. Just after opening the valve,

• A. air from end $1$ flows towards end $2$. No change in the volume of the soap bubbles.
• B. air from end $1$ flows towards end $2$. Volume of the soap bubble at end $1$ decreases.
• C. no change occurs.
• D. air from end $2$ flows towards end $1$. Volume of the soap bubble at end $1$ increases.

Answer 1

The correct option is B air from end $1$ flows towards end $2$. Volume of the soap bubble at end $1$ decreases.

Pressure inside tube, $P=P_0+\frac{4T}{r}$

$\therefore P_2<P_1$
(since $r_2>r_1$, as sub-hemispherical bubble will have greater radius than that of hemisphrical bubble)

Hence, pressure on side $1$ will be greater than that on side $2$



So, air from end $1$ flows towards end $2$, so that volume of the soap bubble at end $1$ decreases.

Hence, $\left(B\right)$ is the correct answer.