A soap bubble of radius 1.0 text{ cm} is form | vedclass

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(A) The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = \frac{4T}{R}$, where $T$ is the surface tension.Let $R_1 = 2.0 	ex

Vedclass · Accepted Answer

$6.67 	imes 10^{-3}$

Vedclass · Answer

(A) The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = \frac{4T}{R}$, where $T$ is the surface tension.Let $R_1 = 2.0 	ext{ cm}$ and $R_2 = 1.0 	ext{ cm}$.The pressure inside the smaller bubble is $P_{in} = P_0 + \frac{4T}{R_2}$.The pressure between the two bubbles is $P_{mid} = P_0 + \frac{4T}{R_1}$.The pressure difference between the inside of the smaller bubble and the outside of the larger bubble is $\Delta P_{total} = (P_0 + \frac{4T}{R_2}) - P_0 = \frac{4T}{R_2}$.Wait, the question asks for the pressure difference between the inside of the smaller bubble and the outside of the larger bubble. This is $\Delta P = \frac{4T}{R_2} + \frac{4T}{R_1} = 4T(\frac{1}{R_1} + \frac{1}{R_2})$.Let the radius of the equivalent bubble be $R$. Then $\frac{4T}{R} = 4T(\frac{1}{R_1} + \frac{1}{R_2})$.$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{R_1 + R_2}{R_1 R_2}$.$R = \frac{R_1 R_2}{R_1 + R_2} = \frac{2.0 	imes 1.0}{2.0 + 1.0} 	ext{ cm} = \frac{2}{3} 	ext{ cm} = 0.667 	ext{ cm}$.Converting to metres: $R = 0.667 	imes 10^{-2} 	ext{ m} = 6.67 	imes 10^{-3} 	ext{ m}$.

$A$ soap bubble of radius $1.0 \text{ cm}$ is formed inside another soap bubble of radius $2.0 \text{ cm}$. The radius of another soap bubble which has the same pressure difference as that between the inside of the smaller and outside of the large soap bubble, in metres is:

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