Pressure inside two soap bubbles are 1.01 and | vedclass

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(A) The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = P_{in} - P_{out} = \frac{4T}{R}$.Assuming the atmospheric pressure

Vedclass · Accepted Answer

$8$

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(A) The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = P_{in} - P_{out} = \frac{4T}{R}$.Assuming the atmospheric pressure $P_{atm} = 1 	ext{ atm}$, the excess pressures are:$\Delta P_1 = 1.01 - 1 = 0.01 	ext{ atm} = \frac{4T}{R_1} \quad \dots(1)$$\Delta P_2 = 1.02 - 1 = 0.02 	ext{ atm} = \frac{4T}{R_2} \quad \dots(2)$Dividing equation $(1)$ by $(2)$:$\frac{0.01}{0.02} = \frac{R_2}{R_1} \implies \frac{1}{2} = \frac{R_2}{R_1} \implies R_1 = 2R_2$.The ratio of the volumes $V_1$ and $V_2$ is:$\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi R_1^3}{\frac{4}{3}\pi R_2^3} = \left(\frac{R_1}{R_2}ight)^3 = (2)^3 = 8$.Thus, the ratio is $8 : 1$.

Pressure inside two soap bubbles are $1.01$ and $1.02$ atmosphere, respectively. The ratio of their volumes is (in $ : 1$)

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