question_answer

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 subhemispherical 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 1 increases

Correct Answer: B

Solution :

Pressure inside tube \[=P={{P}_{0}}\,+\,\frac{4T}{r}\] \[\therefore \,\,\text{ }{{P}_{2}}<{{P}_{1}}\] (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.