If work done in blowing a bubble of volume V | vedclass

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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 $T$ is the surface tension and $\Delta A$ is th

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$2^{2/3}W$

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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 $T$ is the surface tension and $\Delta A$ is the change in surface area. Since a soap bubble has two surfaces,$\Delta A = 2 	imes (4\pi R^2) = 8\pi R^2$.Thus,$W = 8\pi R^2 T$.The volume of the bubble is $V = \frac{4}{3}\pi R^3$,which implies $R = (\frac{3V}{4\pi})^{1/3}$.Substituting $R$ in the expression for $W$,we get $W \propto R^2 \propto V^{2/3}$.If the new volume is $V' = 2V$,the new work done $W'$ is given by $W' = W 	imes (\frac{V'}{V})^{2/3}$.$W' = W 	imes (\frac{2V}{V})^{2/3} = 2^{2/3}W$.

If work done in blowing a bubble of volume $V$ is $W$,then the work done in blowing another bubble of volume $2V$ will be

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