Water rises to height h in a capillary tube. | vedclass

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(C) The height to which a liquid rises in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$. This height $h$ is the equilibrium heigh

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water rises up to the top of the capillary tube and stays there without overflowing

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(C) The height to which a liquid rises in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$. This height $h$ is the equilibrium height determined by the balance between surface tension forces and the weight of the liquid column. If the physical length of the tube is less than $h$,the liquid will rise to the top of the tube. At the top,the radius of curvature of the meniscus will adjust such that the new height $h'$ equals the length of the tube. The liquid will not overflow because the contact angle $	heta$ will increase to satisfy the equilibrium condition $h' = \frac{2T \cos 	heta'}{r ho g}$,where $	heta' > 	heta$. Thus,the water stays at the top without overflowing.

Water rises to height $h$ in a capillary tube. If the length of the capillary tube above the surface of water is made less than $h$,then

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The correct relation for the capillary rise $h$ in a tube of radius $r$ is:

$A$ capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of water in the capillary tube is $5 \, g$. Another capillary tube of radius $2r$ is immersed in water. The mass of water that will rise in this tube is ...... $g$

When a capillary tube is dipped vertically in water,the rise of water in the capillary is $h$. The angle of contact is $0^{\circ}$. Now,the tube is depressed so that its length above the water surface is $\frac{h}{3}$. The new apparent angle of contact is $(\cos 0^{\circ} = 1)$.

$A$ capillary tube made of glass is dipped into mercury. Then

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