Question

With the help of a neat diagram, derive the capillary rise in a vertical capillary tube of circular

cross section (radius a) inserted into an open vessel of water.​

Answer 1

Explanation:

sorry don't know the answer

Answer 2

Answer:

Consider a capillary tube of radius r partially immersed into a wetting liquid of density ρ. Let the capillary rise be h and be the angle of contact at the edge of contact of the concave meniscus and glass figure. If R is the radius of curvature of the meniscus then from the figure, r=R cosθ. 

Analysing capillary capillary action using Laplace's law for a spherical membrane Surface tension T is the tangential force per unit length acting along the contact line. It is directed into the liquid making an angle with the capillary wall. We ignore the small volume of the liquid in the meniscus. The gauge pressure within the liquid at a depth h, i.e., at the level of the free liquid surface open to the atmosphere, is 

p−P0=ρ gh            ...(1) 

By Laplace's law for a spherical membrane, this gauge pressure is 

P−P0=R2T      ...(2) 

∴  hρ g=R2T=r2T cosθ

∴  h=rρ g2T cosθ     ...(3) 

Thus, narrower the capillary tube, the greater is the capillary rise. 

From Eq. (3), 

= 2T cosθhρ rg     ...(4) 

Equations (3) and (4) are also valid for capillary depression h of a non-wetting liquid. In this case, the meniscus is convex and is obtuse. Then,  cosθ is negative but so is h, indicating a fall or depression of the liquid in the capillary. T is positive in both cases.