The excess pressure inside a spherical soap b | vedclass

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Question

(B) The excess pressure inside a soap bubble is given by $p = \frac{4T}{R}$.This pressure is balanced by the pressure exerted by the oil column,which

Vedclass · Accepted Answer

$0.0392$

Vedclass · Answer

(B) The excess pressure inside a soap bubble is given by $p = \frac{4T}{R}$.This pressure is balanced by the pressure exerted by the oil column,which is $p = h ho g$.Equating the two,we get $h ho g = \frac{4T}{R}$.Rearranging for surface tension $T$,we have $T = \frac{R h ho g}{4}$.Given values: $R = 1 ~cm = 10^{-2} ~m$,$h = 2 ~mm = 2 	imes 10^{-3} ~m$,$ho = 0.8 	imes 10^3 ~kg/m^3$,and $g = 9.8 ~m/s^2$.Substituting these values into the formula:$T = \frac{10^{-2} 	imes 2 	imes 10^{-3} 	imes 0.8 	imes 10^3 	imes 9.8}{4}$$T = \frac{2 	imes 0.8 	imes 9.8 	imes 10^{-2}}{4}$$T = 0.4 	imes 0.8 	imes 9.8 	imes 10^{-2} = 0.392 	imes 10^{-1} = 0.0392 ~N/m$.

The excess pressure inside a spherical soap bubble of radius $1 ~cm$ is balanced by a column of oil (specific gravity $= 0.8$),$2 ~mm$ high. The surface tension of the bubble is: (in $~N/m$)

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