Pressure inside two soap bubbles are 1.02 ,at | vedclass

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(B) The excess pressure inside a soap bubble is given by $\Delta P = P_{in} - P_{out} = \frac{4T}{R}$,where $T$ is the surface tension and $R$ is the

Vedclass · Accepted Answer

$\frac{25}{4}$

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(B) The excess pressure inside a soap bubble is given by $\Delta P = P_{in} - P_{out} = \frac{4T}{R}$,where $T$ is the surface tension and $R$ is the radius of the bubble.Assuming the atmospheric pressure $P_{out} = 1.00 \,atm$,the excess pressures are:$\Delta P_1 = 1.02 - 1.00 = 0.02 \,atm$$\Delta P_2 = 1.05 - 1.00 = 0.05 \,atm$Since $\Delta P \propto \frac{1}{R}$,we have $\frac{\Delta P_1}{\Delta P_2} = \frac{R_2}{R_1} = \frac{0.02}{0.05} = \frac{2}{5}$.The surface area $S$ of a sphere is $4\pi R^2$,so the ratio of surface areas is $\frac{S_1}{S_2} = \left(\frac{R_1}{R_2}\right)^2$.Since $\frac{R_1}{R_2} = \frac{5}{2}$,the ratio of surface areas is $\frac{S_1}{S_2} = \left(\frac{5}{2}\right)^2 = \frac{25}{4}$.

Pressure inside two soap bubbles are $1.02 \,atm$ and $1.05 \,atm$ respectively. The ratio of their surface area is .........

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