A drop of some liquid of volume 0.04 cm^{3} | vedclass

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(C) Let the thickness of the liquid layer be $x$.The volume $V$ of the liquid is given by $V = A 	imes x$, where $A$ is the area of the layer.Thus, $

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(C) Let the thickness of the liquid layer be $x$.The volume $V$ of the liquid is given by $V = A 	imes x$, where $A$ is the area of the layer.Thus, $x = V / A$.The liquid forms a thin film between two glass plates, creating a concave meniscus with radius of curvature $r$. For a thin film, the thickness $x$ is equal to the diameter of the meniscus, so $x = 2r$, which implies $r = x / 2 = V / (2A)$.The pressure difference $\Delta P$ across the curved surface is given by $\Delta P = T / r$, where $T$ is the surface tension.The force $F$ required to separate the plates is $F = \Delta P 	imes A$.Substituting $\Delta P = T / r$ and $r = V / (2A)$, we get:$F = (T / (V / (2A))) 	imes A = (2AT / V) 	imes A = (2A^{2}T) / V$.Rearranging for $T$: $T = (F 	imes V) / (2A^{2})$.Given: $F = 16 	imes 10^{5} \ dyne$, $V = 0.04 \ cm^{3}$, $A = 20 \ cm^{2}$.$T = (16 	imes 10^{5} 	imes 0.04) / (2 	imes 20^{2}) = (16 	imes 10^{5} 	imes 0.04) / (2 	imes 400) = (0.64 	imes 10^{5}) / 800 = 64000 / 800 = 80 \ dyne \ cm^{-1}$.

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