The surface tension of a soap solution is 3.5 | vedclass

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Question

(D) The work done $W$ in increasing the surface area of a soap bubble is given by $W = T 	imes \Delta A$. Since a soap bubble has two surfaces (inner

Vedclass · Accepted Answer

$264$

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(D) The work done $W$ in increasing the surface area of a soap bubble is given by $W = T 	imes \Delta A$. Since a soap bubble has two surfaces (inner and outer), the change in surface area is $\Delta A = 2 	imes 4\pi (r_2^2 - r_1^2) = 8\pi (r_2^2 - r_1^2)$. Given $T = 3.5 	imes 10^{-2} 	ext{ N/m}$, $r_1 = 1 	ext{ cm} = 0.01 	ext{ m}$, and $r_2 = 2 	ext{ cm} = 0.02 	ext{ m}$. Substituting the values: $W = 8 	imes (22/7) 	imes 3.5 	imes 10^{-2} 	imes ((0.02)^2 - (0.01)^2)$. $W = 8 	imes (22/7) 	imes 3.5 	imes 10^{-2} 	imes (4 	imes 10^{-4} - 1 	imes 10^{-4})$. $W = 8 	imes 22 	imes 0.5 	imes 10^{-2} 	imes 3 	imes 10^{-4}$. $W = 88 	imes 3 	imes 10^{-6} = 264 	imes 10^{-6} 	ext{ J}$. Thus, $\alpha = 264$.

The surface tension of a soap solution is $3.5 \times 10^{-2} \text{ N/m}$. The work required to increase the radius of a soap bubble from $1 \text{ cm}$ to $2 \text{ cm}$ is $\alpha \times 10^{-6} \text{ J}$. The value of $\alpha$ is . . . . . . . $(\pi = 22/7)$

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