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.
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 increases
Explanation
(b) $\mathrm{\Delta}p_{1} = \frac{4T}{r_{1}}\ \ \ \ \ \ and\ \mathrm{\Delta}p_{2} = \frac{4T}{r_{2}}$
$r_{1} < r_{2}$
$\therefore\ \mathrm{\Delta}p_{1} > \mathrm{\Delta}p_{2}$
Air will flow from 1 to 2 and volume of bubble at end-1 will decreases.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.