A circular film of a liquid has an area of 10 | vedclass

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(C) Let the initial radius be $r$. The initial area is $A_1 = \pi r^2 = 10 \,cm^2$.When the radius becomes $2r$, the new area is $A_2 = \pi (2r)^2 = 4

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$3$

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(C) Let the initial radius be $r$. The initial area is $A_1 = \pi r^2 = 10 \,cm^2$.When the radius becomes $2r$, the new area is $A_2 = \pi (2r)^2 = 4\pi r^2 = 4A_1$.The change in area is $\Delta A = A_2 - A_1 = 4A_1 - A_1 = 3A_1$.Given $A_1 = 10 \,cm^2 = 10 	imes 10^{-4} \,m^2 = 10^{-3} \,m^2$.So, $\Delta A = 3 	imes 10^{-3} \,m^2$.$A$ liquid film has two surfaces, so the work done $W$ is given by $W = 2 	imes T 	imes \Delta A$, where $T$ is the surface tension.Given $W = 8 	imes 10^{-3} \,J$.Substituting the values: $8 	imes 10^{-3} = 2 	imes T 	imes (3 	imes 10^{-3})$.$8 = 6T \implies T = \frac{8}{6} = \frac{4}{3} \,N/m$.We are given $T = \left(1 + \frac{1}{\alpha}ight) \,N/m$.So, $1 + \frac{1}{\alpha} = \frac{4}{3} = 1 + \frac{1}{3}$.Therefore, $\alpha = 3$.

$A$ circular film of a liquid has an area of $10 \,cm^2$. If the work done in increasing its radius to two times the initial radius is $8 \times 10^{-3} \,J$, the surface tension of the liquid is $\left(1+\frac{1}{\alpha}\right) \,N/m$. The value of $\alpha$ is

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