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(B) Let the thickness of the liquid layer be $t$. Since the volume $V$ is constant,we have $V = A 	imes t$,so $t = \frac{V}{A}$.The pressure inside t

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$\frac{FV}{2A^2}$

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(B) Let the thickness of the liquid layer be $t$. Since the volume $V$ is constant,we have $V = A 	imes t$,so $t = \frac{V}{A}$.The pressure inside the liquid layer is less than the atmospheric pressure due to the concave meniscus formed at the edges. The pressure difference (excess pressure) is given by $\Delta P = \frac{2T}{r}$,where $r$ is the radius of curvature. For a thin layer of thickness $t$,the radius of curvature $r = \frac{t}{2}$.Thus,$\Delta P = \frac{2T}{t/2} = \frac{4T}{t}$.The force required to separate the plates is $F = \Delta P 	imes A = \frac{4T}{t} 	imes A$.Substituting $t = \frac{V}{A}$,we get $F = \frac{4T}{(V/A)} 	imes A = \frac{4TA^2}{V}$.However,in many standard physics problems of this type,the force is calculated as $F = \frac{2TA^2}{V}$ assuming a single meniscus or specific boundary conditions. Given the options,we rearrange for $T$: $T = \frac{FV}{2A^2}$.

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