JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Fluid Mechanics

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✓ Correct
(C)no change occurs
(D)air from end 2 flows towards end 1. Volume of the soap bubble at end 1 increases

Answer: (B)

Solution

The excess pressure inside a soap bubble is

Δ P = \frac{4 T}{R}

, where T is surface tension and R is radius.

For the hemispherical bubble at end 1 (radius

r

P_{1} = P_{0} + \frac{4 T}{r}

For the sub-hemispherical bubble at end 2 (radius

R > r

P_{2} = P_{0} + \frac{4 T}{R}

Since

R > r

, we have

P_{1} > P_{2}

. Therefore, air flows from end 1 (higher pressure) to end 2 (lower pressure), and the volume of the bubble at end 1 decreases.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option