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

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Question

(D) soap bubble has two surfaces (inner and outer). Therefore,the total change in surface area is given by:$\Delta A = 2 	imes [4 \pi (R_2^2 - R_1^2)

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) soap bubble has two surfaces (inner and outer). Therefore,the total change in surface area is given by:$\Delta A = 2 	imes [4 \pi (R_2^2 - R_1^2)]$Given the diameters are $D$ and $d$,the radii are $R_2 = D/2$ and $R_1 = d/2$.$\Delta A = 2 	imes 4 \pi [(\frac{D}{2})^2 - (\frac{d}{2})^2]$$\Delta A = 8 \pi [\frac{D^2 - d^2}{4}] = 2 \pi (D^2 - d^2)$Work done $W = T 	imes \Delta A$$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 solution).

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