A capillary tube of radius r is dipped in a l | vedclass

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(B) The pressure difference across a curved liquid surface is given by the formula for excess pressure.For a capillary tube of radius $r$,the radius o

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$\frac{2S}{r}\cos 	heta $

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(B) The pressure difference across a curved liquid surface is given by the formula for excess pressure.For a capillary tube of radius $r$,the radius of the meniscus $R$ is related to the tube radius by $R = \frac{r}{\cos 	heta}$.The excess pressure $\Delta P$ inside the capillary is given by $\Delta P = \frac{2S}{R}$.Substituting the value of $R$,we get $\Delta P = \frac{2S}{r/\cos 	heta} = \frac{2S}{r} \cos 	heta$.Thus,the pressure difference between the liquid surface in the beaker (atmospheric pressure) and the liquid surface inside the capillary is $\frac{2S}{r} \cos 	heta$.

$A$ capillary tube of radius $r$ is dipped in a liquid of density $\rho$ and surface tension $S$. If the angle of contact is $\theta$,what is the pressure difference between the two surfaces in the beaker and the capillary?

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