Question

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 1 increases

Answer 1

Answer: (B)

Pressure inside tube $=P=P_0+\frac{4T}{r}$
$\therefore P_2<P_1($ since $r_2>r_1$ )
Hence pressure on side $1$ will be greater than side $2$ .
So air from end $1$ flows towards end $2$