A vertical glass capillary tube of radius r o | vedclass

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(A) The water column in the capillary tube forms two concave menisci at both ends.Let $P_0$ be the atmospheric pressure.The pressure just below the up

Vedclass · Accepted Answer

$L=\frac{4 T}{r \rho g}$

Vedclass · Answer

(A) The water column in the capillary tube forms two concave menisci at both ends.Let $P_0$ be the atmospheric pressure.The pressure just below the upper meniscus is $P_{upper} = P_0 - \frac{2T}{r}$.The pressure just below the lower meniscus is $P_{lower} = P_0 - \frac{2T}{r}$.Moving from the top meniscus to the bottom meniscus through the water column of length $L$,the pressure changes as:$P_{lower} = P_{upper} + \rho g L$Substituting the values:$P_0 - \frac{2T}{r} = (P_0 - \frac{2T}{r}) + \rho g L$Wait,this implies the column is in equilibrium under the pressure difference. The pressure at the top is $P_0 - \frac{2T}{r}$ and at the bottom is $P_0 - \frac{2T}{r}$.Actually,for a water column of length $L$ to be in equilibrium,the pressure difference between the two ends must balance the hydrostatic pressure $\rho g L$.Since both ends have concave menisci,the pressure at the top is $P_0 - \frac{2T}{r}$ and the pressure at the bottom is $P_0 + \frac{2T}{r}$ (if the meniscus is convex relative to the water). However,in a capillary tube,both ends form concave menisci. The pressure at the top is $P_0 - \frac{2T}{r}$ and the pressure at the bottom is $P_0 - \frac{2T}{r}$. This would mean the column cannot be in equilibrium unless the pressure difference is balanced by gravity.Correct approach: The pressure at the top is $P_0 - \frac{2T}{r}$. The pressure at the bottom is $P_0 - \frac{2T}{r}$. The pressure difference across the column is $\rho g L$. Thus,$\rho g L = \frac{2T}{r} - (-\frac{2T}{r}) = \frac{4T}{r}$.Therefore,$L = \frac{4T}{r \rho g}$.

$A$ vertical glass capillary tube of radius $r$ open at both ends contains some water (surface tension $T$ and density $\rho$). If $L$ be the length of the water column,then:

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