In air,a charged soap bubble of radius R brea | vedclass

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(D) The volume of the big bubble is equal to the sum of the volumes of $27$ small bubbles: $\frac{4}{3} \pi R^3 = 27 	imes \frac{4}{3} \pi r^3$. Taki

Vedclass · Accepted Answer

$1/9$

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(D) The volume of the big bubble is equal to the sum of the volumes of $27$ small bubbles: $\frac{4}{3} \pi R^3 = 27 	imes \frac{4}{3} \pi r^3$. Taking the cube root on both sides: $R = 3r$. The mechanical force per unit area (electrostatic pressure) on a charged soap bubble is given by $P = \frac{\sigma^2}{2\epsilon_0}$,where $\sigma = \frac{Q}{4\pi R^2}$. Since the total charge $Q$ is conserved,the surface charge density $\sigma \propto \frac{1}{R^2}$. Thus,the pressure $P \propto \frac{1}{R^4}$. However,the question asks for the ratio of mechanical force per unit area,which is the electrostatic pressure $P = \frac{Q^2}{32\pi^2 \epsilon_0 R^4}$. For the big bubble: $P_B = \frac{Q^2}{32\pi^2 \epsilon_0 R^4}$. For a small bubble: $q = Q/27$,so $P_S = \frac{(Q/27)^2}{32\pi^2 \epsilon_0 r^4} = \frac{Q^2}{729 	imes 32\pi^2 \epsilon_0 (R/3)^4} = \frac{Q^2}{729 	imes 32\pi^2 \epsilon_0 (R^4/81)} = \frac{Q^2}{9 	imes 32\pi^2 \epsilon_0 R^4}$. Therefore,the ratio $\frac{P_B}{P_S} = \frac{1/R^4}{1/(9R^4)} = 9/1$.

In air,a charged soap bubble of radius $R$ breaks into $27$ small soap bubbles of equal radius $r$. Then,the ratio of mechanical force acting per unit area of the big soap bubble to that of a small soap bubble is

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