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Question

(C) soap bubble has two surfaces (inner and outer). Therefore,the work done $W$ in increasing the radius from $r_1$ to $r_2$ is given by the formula:$

Vedclass · Accepted Answer

$720$

Vedclass · Answer

(C) soap bubble has two surfaces (inner and outer). Therefore,the work done $W$ in increasing the radius from $r_1$ to $r_2$ is given by the formula:$W = 2 	imes [T 	imes (A_2 - A_1)] = 2 	imes T 	imes 4\pi(r_2^2 - r_1^2) = 8\pi T(r_2^2 - r_1^2)$Given:$T = 30 	ext{ dynes/cm}$$r_1 = \frac{1}{\sqrt{\pi}} 	ext{ cm}$$r_2 = \frac{2}{\sqrt{\pi}} 	ext{ cm}$Substituting the values:$W = 8 	imes \pi 	imes 30 	imes \left[ \left( \frac{2}{\sqrt{\pi}} ight)^2 - \left( \frac{1}{\sqrt{\pi}} ight)^2 ight]$$W = 240\pi 	imes \left[ \frac{4}{\pi} - \frac{1}{\pi} ight]$$W = 240\pi 	imes \frac{3}{\pi}$$W = 240 	imes 3 = 720 	ext{ ergs}$.

The radius of a soap bubble is increased from $\frac{1}{\sqrt{\pi}} \text{ cm}$ to $\frac{2}{\sqrt{\pi}} \text{ cm}$. If the surface tension of water is $30 \text{ dynes/cm}$,then the work done will be ....... $ergs$.

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