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(D) The excess pressure inside a soap bubble of radius $R$ is given by $P = 4T/R$,where $T$ is the surface tension.Let $R_1$ and $R_2$ be the radii of

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$27/8$

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(D) The excess pressure inside a soap bubble of radius $R$ is given by $P = 4T/R$,where $T$ is the surface tension.Let $R_1$ and $R_2$ be the radii of the first and second soap bubbles,respectively.Given that the excess pressure in the first bubble is $1.5$ times the excess pressure in the second bubble:$P_1 = 1.5 P_2$$\frac{4T}{R_1} = 1.5 	imes \frac{4T}{R_2}$$\frac{1}{R_1} = \frac{3}{2R_2} \implies R_2 = 1.5 R_1 = \frac{3}{2} R_1$The volume of a soap bubble is $V = \frac{4}{3} \pi R^3$.$V_1 = \frac{4}{3} \pi R_1^3$ and $V_2 = \frac{4}{3} \pi R_2^3$.Given $V_2 = x V_1$,we have:$x = \frac{V_2}{V_1} = \left( \frac{R_2}{R_1} ight)^3 = \left( \frac{3}{2} ight)^3 = \frac{27}{8}$.

The excess pressure inside a soap bubble is $1.5$ times the excess pressure inside a second soap bubble. The volume of the second bubble is '$x$' times the volume of the first bubble. The value of '$x$' is

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