A soap bubble of radius frac{1}{sqrt{pi}} tex | vedclass

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(C) The work done $(W)$ in expanding a soap bubble is given by the formula: $W = T 	imes \Delta A$, where $T$ is the surface tension and $\Delta A$ i

Vedclass · Accepted Answer

$1600$

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(C) The work done $(W)$ in expanding a soap bubble is given by the formula: $W = T 	imes \Delta A$, where $T$ is the surface tension and $\Delta A$ is the change in surface area.Since a soap bubble has two surfaces (inner and outer), the surface area $A = 2 	imes (4\pi r^2) = 8\pi r^2$.Initial radius $r_1 = \frac{1}{\sqrt{\pi}} 	ext{ cm}$, so initial area $A_1 = 8\pi \left(\frac{1}{\sqrt{\pi}}ight)^2 = 8 	ext{ cm}^2$.Final radius $r_2 = \frac{3}{\sqrt{\pi}} 	ext{ cm}$, so final area $A_2 = 8\pi \left(\frac{3}{\sqrt{\pi}}ight)^2 = 8\pi 	imes \frac{9}{\pi} = 72 	ext{ cm}^2$.Change in area $\Delta A = A_2 - A_1 = 72 - 8 = 64 	ext{ cm}^2$.Work done $W = T 	imes \Delta A = 25 	imes 64 = 1600 	ext{ erg}$.

$A$ soap bubble of radius $\frac{1}{\sqrt{\pi}} \text{ cm}$ is expanded to radius $\frac{3}{\sqrt{\pi}} \text{ cm}$. The surface tension of the soap solution is $25 \text{ dyne/cm}$. The work done during expansion in $\text{erg}$ is:

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