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(B) The work done $(W)$ in increasing the area of a liquid film is given by $W = T 	imes \Delta A 	imes 2$,where $T$ is the surface tension and the

Vedclass · Accepted Answer

$0.04 	ext{ N/m}$

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(B) The work done $(W)$ in increasing the area of a liquid film is given by $W = T 	imes \Delta A 	imes 2$,where $T$ is the surface tension and the factor of $2$ accounts for the two surfaces of the film.Initial area $A_1 = 5 	ext{ cm} 	imes 4 	ext{ cm} = 20 	ext{ cm}^2 = 20 	imes 10^{-4} 	ext{ m}^2$.Final area $A_2 = 7 	ext{ cm} 	imes 8 	ext{ cm} = 56 	ext{ cm}^2 = 56 	imes 10^{-4} 	ext{ m}^2$.Change in area $\Delta A = A_2 - A_1 = (56 - 20) 	imes 10^{-4} 	ext{ m}^2 = 36 	imes 10^{-4} 	ext{ m}^2$.Given $W = 3 	imes 10^{-4} 	ext{ J}$.Substituting the values: $3 	imes 10^{-4} = T 	imes (36 	imes 10^{-4}) 	imes 2$.$3 = T 	imes 72$.$T = 3 / 72 = 1 / 24 \approx 0.0416 	ext{ N/m}$.Rounding to the nearest option,$T \approx 0.04 	ext{ N/m}$.

$A$ rectangular film of liquid is expanded from $(5 \text{ cm} \times 4 \text{ cm})$ to $(7 \text{ cm} \times 8 \text{ cm})$. If the work done is $3 \times 10^{-4} \text{ J}$,the surface tension of the liquid is (nearly):

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