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

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(D) soap bubble has two surfaces (inner and outer),so the change in surface area is $2 	imes 4\pi (R_2^2 - R_1^2)$.Given the initial diameter is $d$,

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$2\pi (D^2 - d^2)T$

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(D) soap bubble has two surfaces (inner and outer),so the change in surface area is $2 	imes 4\pi (R_2^2 - R_1^2)$.Given the initial diameter is $d$,the initial radius is $r_1 = d/2$. The final diameter is $D$,so the final radius is $r_2 = D/2$.The change in surface area $\Delta A = 2 	imes 4\pi (r_2^2 - r_1^2) = 8\pi \left( \frac{D^2}{4} - \frac{d^2}{4} ight)$.Simplifying this,we get $\Delta A = 2\pi (D^2 - d^2)$.The work done $W$ is given by $W = T 	imes \Delta A$.Therefore,$W = T 	imes 2\pi (D^2 - d^2) = 2\pi (D^2 - d^2)T$.

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

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