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(D) The work done $W$ in blowing a soap bubble of radius $R$ is given by $W = T 	imes \Delta A$,where $\Delta A$ is the change in surface area. For a

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$\sqrt[3]{4} W$

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(D) The work done $W$ in blowing a soap bubble of radius $R$ is given by $W = T 	imes \Delta A$,where $\Delta A$ is the change in surface area. For a soap bubble,there are two surfaces,so $\Delta A = 2 	imes (4 \pi R^2) = 8 \pi R^2$.Thus,$W = 8 \pi R^2 T$.Since the volume of the bubble is $V = \frac{4}{3} \pi R^3$,we have $R^3 \propto V$,which implies $R \propto V^{1/3}$.Substituting this into the work equation: $W \propto R^2 \propto (V^{1/3})^2 = V^{2/3}$.Therefore,$\frac{W_2}{W_1} = \left( \frac{V_2}{V_1} ight)^{2/3}$.Given $V_1 = V$ and $V_2 = 2V$,we get $\frac{W_2}{W} = \left( \frac{2V}{V} ight)^{2/3} = 2^{2/3} = (2^2)^{1/3} = 4^{1/3} = \sqrt[3]{4}$.Hence,$W_2 = \sqrt[3]{4} W$.

If the work done in blowing a bubble of volume $V$ is $W$,then the work done in blowing the bubble of volume $2V$ from the same soap solution will be

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