IIT JEE , AIEEE Forum - Physics , Mechanics

audittn8

Capillary tube sealed at the top has an internal radius of 0.05cm.The tube is placed verically in water,open end first.What should be the length of such a tube for the water column to rise in it to a length of 1 cm? Surface tension of water = 70 X ${10}^{-3}$ N/m.

Froggy

A straight forward sum that makes use of the formula:

$h = \frac{2S cos \theta}{r \rho g}$

S = Surface Tension of the liquid, r = inner radius of the capillary tube, etc. So go ahead!

gauravtargetsiit

Since the tube is sealed at the top, so the pressure above water level in the tube will be zero .
hence we can't use the formula : h=2Scos(theta)/rpg .

So, Let W be the weight , S be the surface tension and P be the pressure at the bottom of the capillary tube in level with the water surface, then
W = (pie)*r^2 h*p*g

At equilibrium,
W = P*(pie)* r^2 + 2*(pie)* r*S
put the values and get the answer.....

Froggy

Oh yes, good work. I made a grave mistake there.

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