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(C) The excess pressure inside a soap bubble of radius $r$ is given by $\Delta P = \frac{4S}{r}$,where $S$ is the surface tension.For a soap bubble of

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$2$

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(C) The excess pressure inside a soap bubble of radius $r$ is given by $\Delta P = \frac{4S}{r}$,where $S$ is the surface tension.For a soap bubble of radius $r_2 = 3 \, cm$ formed inside a soap bubble of radius $r_1 = 6 \, cm$,the pressure inside the smaller bubble relative to the atmospheric pressure is the sum of the excess pressure due to the smaller bubble and the excess pressure due to the larger bubble.$\Delta P_{total} = \frac{4S}{r_2} + \frac{4S}{r_1}$We want to find the radius $R_{eq}$ of an equivalent single soap bubble such that its excess pressure equals $\Delta P_{total}$:$\frac{4S}{R_{eq}} = \frac{4S}{r_2} + \frac{4S}{r_1}$Dividing both sides by $4S$:$\frac{1}{R_{eq}} = \frac{1}{r_2} + \frac{1}{r_1}$Substituting the given values $r_1 = 6 \, cm$ and $r_2 = 3 \, cm$:$\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$Therefore,$R_{eq} = 2 \, cm$.

$A$ soap bubble of radius $3 \, cm$ is formed inside another soap bubble of radius $6 \, cm$. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is .......... $cm$.

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