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(A) The height of the water column in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$.Since $T$,$ho$,$g$,and the contact angle $\

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The radius of the meniscus increases and therefore water does not overflow.

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(A) The height of the water column in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$.Since $T$,$ho$,$g$,and the contact angle $	heta$ are constant for a given liquid and tube,we have $h \cdot r = 	ext{constant}$.Initially,$h_1 = 8 \ cm$. When the tube is pushed down such that the height above the water level is $h_2 = 5 \ cm$,the water cannot overflow because the radius of curvature $R$ of the meniscus will adjust to satisfy the equilibrium condition.Specifically,$h_1 r_1 = h_2 R_2$,where $R_2$ is the new radius of the meniscus.As $h$ decreases from $8 \ cm$ to $5 \ cm$,the radius of the meniscus $R$ increases to maintain the pressure balance.Therefore,the water does not overflow; it simply adjusts its meniscus shape.

When a capillary tube is dipped in water,it rises up to $8 \ cm$ in the tube. What happens when the tube is pushed down such that its end is only $5 \ cm$ above the outside water level?

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