Question

On dipping a capillary of radius r in water water rises upto height h amd potential energy of water is u1 if a capillary of radius 2r

Answer 1

Potential energy's ratio of two surfaces would be 1:1

i.e. u₁ : u₂ = 1 : 1

Solution:

In case of interfaces (liquid - solid), as per Jurin's law:

hr = constant value

h₁r₁ = h₂r₂

h₂ = hr/2r = h/2

h of H20 in  2nd capillary will be 1/2 of the h of the 1st capillary.

P.E. of liquid will be:

U ∝ h²r²

u₁/u₂ = h₁²r₁²/h₂²r₂²

= h²r²/(h/2)²(2r)²

= h²r²/h²r²

u₁ : u₂ = 1 : 1