The surface tension of a soap bubble is 0.03 | vedclass

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Question

(D) The work done $W$ in increasing the surface area of a soap bubble is given by $W = T 	imes \Delta A 	imes 2$, where $T$ is the surface tension a

Vedclass · Accepted Answer

$7.68$

Vedclass · Answer

(D) The work done $W$ in increasing the surface area of a soap bubble is given by $W = T 	imes \Delta A 	imes 2$, where $T$ is the surface tension and $2$ accounts for the two surfaces of the soap bubble.$\Delta A = 4\pi(r_2^2 - r_1^2)$.Given: $T = 0.03 	ext{ N/m}$, $r_1 = 1 	ext{ cm} = 0.01 	ext{ m}$, $r_2 = 3 	ext{ cm} = 0.03 	ext{ m}$.$\Delta A = 4 	imes 3.14 	imes ((0.03)^2 - (0.01)^2) = 4 	imes 3.14 	imes (0.0009 - 0.0001) = 12.56 	imes 0.0008 = 0.010048 	ext{ m}^2$.$W = 0.03 	imes 0.010048 	imes 2 = 0.06 	imes 0.010048 = 0.00060288 	ext{ J} = 6.0288 	imes 10^{-4} 	ext{ J}$.Note: Based on the provided options, there appears to be a discrepancy in the calculation or the options provided. If we assume the question implies a single surface or a different radius change, $7.68$ is often the intended answer in similar textbook problems where $W = T 	imes 8\pi(r_2^2 - r_1^2)$ is used with different constants. Given the standard formula, $\alpha \approx 6.03$.

The surface tension of a soap bubble is $0.03 \text{ N/m}$. The work done in increasing the diameter of the bubble from $2 \text{ cm}$ to $6 \text{ cm}$ is $\alpha \times 10^{-4} \text{ J}$. The value of $\alpha$ is . . . . . . (Take $\pi = 3.14$).

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