If T is the surface tension of a soap solutio | vedclass

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(C) soap bubble has two surfaces (inner and outer). The work done $W$ in changing the surface area $A$ is given by $W = T 	imes \Delta A 	imes 2$.In

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$6$

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(C) soap bubble has two surfaces (inner and outer). The work done $W$ in changing the surface area $A$ is given by $W = T 	imes \Delta A 	imes 2$.Initial diameter $d_1 = D$,so initial radius $r_1 = D/2$.Initial surface area $A_1 = 4\pi r_1^2 = 4\pi (D/2)^2 = \pi D^2$.Final diameter $d_2 = 2D$,so final radius $r_2 = D$.Final surface area $A_2 = 4\pi r_2^2 = 4\pi D^2$.Change in surface area $\Delta A = A_2 - A_1 = 4\pi D^2 - \pi D^2 = 3\pi D^2$.Since the bubble has two surfaces,the total change in area is $2 	imes \Delta A = 2 	imes 3\pi D^2 = 6\pi D^2$.Therefore,the work done $W = T 	imes (6\pi D^2) = 6\pi D^2T$.

If $T$ is the surface tension of a soap solution,the amount of work done in blowing a soap bubble from a diameter $D$ to $2D$ is (in $\pi D^2T$)

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