When two soap bubbles coalesce relation between angle?
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Excess pressure in a soap bubble ( in concave part ) is 4S/r .
Pressure inside a soap bubble will be = P + 4S/r.
S is the surface tension of the soap liquid and r is the radius of the bubble and P is the atmospheric pressure.
When two soap bubble will come in contact then contact area will be bulge towards the bigger sphere. air pressure inside the smaller soap bubble will be greater as compared to the bigger sphere.
If pressure inside bigger bubble is P1 and inside smaller bubble is P2
P1= P +4S/r1,
P2= P+ 4S/r2
P2>P1 as r1> r2
Pressure inside a soap bubble will be = P + 4S/r.
S is the surface tension of the soap liquid and r is the radius of the bubble and P is the atmospheric pressure.
When two soap bubble will come in contact then contact area will be bulge towards the bigger sphere. air pressure inside the smaller soap bubble will be greater as compared to the bigger sphere.
If pressure inside bigger bubble is P1 and inside smaller bubble is P2
P1= P +4S/r1,
P2= P+ 4S/r2
P2>P1 as r1> r2
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