JEE Advanced · Physics · 11. Properties of Fluids
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 & Solution
Correct Answer
(B) air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases
Step-by-step Solution
Detailed explanation
\(\Delta p_1=\frac{4 T}{r_1}\) 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).
\(
\begin{array}{cc}
& r_1 < r_2 \\
\therefore & \Delta p_1>\Delta p_2
\end{array}
\)
Air will flow from 1 to 2 and volume of bubble at end- 1 will decrease.
Therefore, correct option is (b).
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