If the excess pressure inside a soap bubble i | vedclass

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Question

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

Vedclass · Accepted Answer

$3.9 	imes 10^{-2}\; N/m$

Vedclass · Answer

(B) The excess pressure inside a soap bubble is given by $\Delta P = \frac{4T}{R}$.This pressure is balanced by the pressure exerted by the oil column,which is $\Delta P = h ho g$.Equating the two,we get $\frac{4T}{R} = h ho g$.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\; g/cm^3 = 800\; kg/m^3$,and $g \approx 9.8\; m/s^2$.Substituting these values: $T = \frac{10^{-2} 	imes 2 	imes 10^{-3} 	imes 800 	imes 9.8}{4}$.$T = \frac{1.568 	imes 10^{-2}}{4} = 0.392 	imes 10^{-1} = 3.92 	imes 10^{-2}\; N/m$.Rounding to the nearest option,$T \approx 3.9 	imes 10^{-2}\; N/m$.

If the excess pressure inside a soap bubble is balanced by an oil column of height $2\; mm$,then the surface tension of the soap solution will be ($r = 1\; cm$ and density $d = 0.8\; g/cm^3$).

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