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

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$6$

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(D) soap bubble has two surfaces (inner and outer). The work done $W$ in changing the surface area 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 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 $W = T 	imes 6 \pi D^2 = 6 \pi TD^2$.

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

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