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A glass tube of uniform internal radius ...

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. Vole of the soap bubble at end 1 decreases

C

No change occurs

D

Air from end 2 flows towards end 1 decreases the soap bubble at end 1 increases

Text Solution

Verified by Experts

The correct Answer is:
B

End 1 has hemispherical soap bubble, hence its radius is r which is same as radius of tube. Pressure inside bubble 1 can be written as follows: `P_(1)=P_(0) +(4T)/(r) ...........(i)`
Let radius of bubble at end 2 is R then pressure inside bubble 2 can be written as follows:
`P_(2)=P_(0)=(4T)/(4) ......(ii)`
Sincer `lt R," hence "P_1 gt P_2,` hence air from end 1 flows towards end 2 and volume of bubble 1 decreases.
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