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Question

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

Vedclass · Accepted Answer

$(4)^{1/3} W$

Vedclass · Answer

(D) The work done in blowing a soap bubble is given by $W = T \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area. Since a soap bubble has two surfaces,the total surface area is $A = 2 	imes (4 \pi r^2) = 8 \pi r^2$.For a spherical bubble,the volume is $V = \frac{4}{3} \pi r^3$,which implies $r = (\frac{3V}{4\pi})^{1/3}$.Substituting $r$ into the area formula: $A = 8 \pi (\frac{3V}{4\pi})^{2/3} \propto V^{2/3}$.Since $W \propto A$,we have $W \propto V^{2/3}$.Let $W'$ be the work done for volume $2V$. Then $\frac{W'}{W} = (\frac{2V}{V})^{2/3} = 2^{2/3} = (2^2)^{1/3} = 4^{1/3}$.Therefore,$W' = 4^{1/3} W$.

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

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