Question

A long glasss capillary tube is dipped in water . it is known that water wets glass . the water level rises by h in tube . the tube is now pushed down that only a length h/2 is outside the water surface . the angle of contact of the water surface is
ans: 60 deg

Answer 1

The angle of contact at the water surface at the upper end of the tube will be θ′  =60°.

Explanation:

Given data:

Rises the level of water in the tube = h

Again, length outside the water = h/2

Here, h= 2s cos(0°)/rpg

= 2s / rpg

According to question,

h / 2 =  2scosθ' / rpg  

Dividing equation (ii) by (i) we get,.

1/2 =cosθ  

θ′  =60°

Thus The angle of contact at the water surface at the upper end of the tube will be θ′  =60°.

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