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Two soap bubbles of radii a and b coales...

Two soap bubbles of radii `a` and `b` coalesce to form a single bubble of radius `c`. If the external pressure is `P`, find the surface tension of the soap solution.

A

`sqrt(x^2 + y^2)`

B

`sqrt(x + y)`

C

`x + y`

D

`(x + y)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Volume of soap bubble of radius x is `V_x = 4/3 pi x^3`
Volume of soap bubble of radius y is `V_y = 4/3 pi y^3`
Let T be the surface tension of soap bubble. Let `P_x` and` P_y` are the excess of pressure inside these two soap bubbles, then
` P_x = (4T)/(x)` and `P_y = (4T)/(y)`
These two soap bubbles coalesce to form a new soap bubble of radius z under isothermal conditions. Let `V_z` and `P_z` be the volume and excess of pressure inside this new soap bubble, then
`V_z = 4/3 pi z^3 , P_z = (4T)/(z)`
Assuming the external pressure is negligible as compared to internal pressure and the new bubble is formed under isothermal conditions, so Boyle.s law holds good.
` therefore P_xV_x + P_y V_y = P_z V_z`
`(4T)/(x) xx 4/3 pi x^3 + (4T)/(y) xx 4/3 pi y^3 = (4T)/(z) xx 4/3 pi z^3`
`x^2 + y^2 = z^2 " or " z = sqrt(x^2 + y^2)`
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