IIT-JEE 2008 · Paper 2 · Physics Q28

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PhysicsSingle correct+3 / -1Question 28 of 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,

An inverted-U glass tube with a valve at the top; both ends point down. End 1 (right) carries a hemispherical soap bubble of radius r; end 2 (left) carries a shallower, sub-hemispherical bubble.

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

Answer and worked solution

Official answer: B

Excess pressure in a soap bubble is $\frac{4T}{R}$ : smaller radius, higher pressure.

Radii. At end 1 the bubble is a hemisphere of radius $r$ , the smallest possible radius for a cap on a tube of radius $r$ . The sub-hemispherical bubble at end 2 is flatter, so its radius is larger and its pressure lower.

On opening the valve air flows from end 1 to end 2. Bubble 1 shrinks (becoming flatter, with a larger radius) and bubble 2 grows.

(B)

Topics:Surface tension

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IIT-JEE 2008 · Paper 2, Physics question 28