Surface tension of a water bubble

When we heat water in a vessel, air bubbles are formed at the bottom. These bubbles detach from the base and rise.

Take the bubbles to be spheres of radius R R and making a circular contact of radius r r with the bottom of the vessel. If r < < R r < < R , and the surface tension of water is T T .

If the density of water is ρ \rho , find the value of r r just before the bubbles detach.

R 2 3 ρ g 2 T R^2 \sqrt{ \frac{3\rho \cdot g } { 2T} } R 2 ρ g T R^2 \sqrt{ \frac{\rho \cdot g } { T} } R 2 2 ρ g T R^2 \sqrt{ \frac{2\rho \cdot g } { T} } R 2 3 ρ g T R^2 \sqrt{ \frac{3\rho \cdot g } { T} } None of the above

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2 solutions

Aniket Sanghi
Dec 29, 2016

If I go on with actual solution , then

Surface tension force

Now, Buoyancy Force : F = V i ρ g F = V_i \rho g

Volume can be calculated by using volume of dome's formula V = π h 2 ( 3 R h ) 3 V =\frac { \pi h^2 ( 3R - h) }{3}

But in this case due to the approximation , everything goes neglected and you end up with volume of sphere's formula .

These two forces balance each other !

Can you please check my solution and tell what's incorrect?

A Former Brilliant Member - 4 years, 5 months ago

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Nothing is incorrect , actually first we should calculate the volume of dome and at end when we will apply condition that r,<<R we will land on volume of sphere !

Aniket Sanghi - 4 years, 5 months ago

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Thanks for your help!!!!

A Former Brilliant Member - 4 years, 5 months ago

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