The amount of work done in blowing a soap bub | vedclass

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Question

(C) soap bubble has two surfaces (inner and outer). Therefore,the total surface area of a soap bubble of radius $r$ is $A = 2 	imes (4 \pi r^2) = 8 \

Vedclass · Accepted Answer

$2 \pi (d_2^2 - d_1^2) T$

Vedclass · Answer

(C) soap bubble has two surfaces (inner and outer). Therefore,the total surface area of a soap bubble of radius $r$ is $A = 2 	imes (4 \pi r^2) = 8 \pi r^2$.Given the diameter changes from $d_1$ to $d_2$,the radii change from $r_1 = d_1/2$ to $r_2 = d_2/2$.The initial surface area is $A_1 = 8 \pi (d_1/2)^2 = 2 \pi d_1^2$.The final surface area is $A_2 = 8 \pi (d_2/2)^2 = 2 \pi d_2^2$.The change in surface area is $\Delta A = A_2 - A_1 = 2 \pi (d_2^2 - d_1^2)$.The work done $W$ is equal to the surface tension $T$ multiplied by the change in surface area $\Delta A$.Thus,$W = T 	imes \Delta A = 2 \pi (d_2^2 - d_1^2) T$.

The amount of work done in blowing a soap bubble such that its diameter increases from $d_1$ to $d_2$ is ($T=$ surface tension of soap solution).

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