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IIT JEE Advanced 2008 Paper 2
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© examsnet.com
Question : 5
Total: 66
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,
air from end 1 flows towards end 2. No change in the volume of the soap bubbles.
air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases.
no change occurs.
air from end 2 flows towards end 1. Volume of the soap bubble at end 1 increases.
Validate
Solution:
The air flows from high-pressure region to low-pressure region.
Therefore,
P =
P
0
+
4
T
R
where
P
0
is the atmospheric pressure, T the surface tension and R the radius of curvature of the bubble. Thus,
P
2
=
P
0
+
4
T
R
2
However,
R
1
<
R
2
; thus, the air flows from 1 to 2.
© examsnet.com
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