The correct option is B 2πσ2ρg
The mass of the liquid in the capillary tube can be considered to be concentrated at the centre of mass of the liquid column of height h i.e COM is at a height h2.
The potential energy of the liquid in the capillary tube is,
U=mg×h2
U=(πr2hρ)×g×(h2)
U=π(rhg)2ρ2g ....(1)
The equation for capillary rise,
h=2σcosθρrg
For perfectly wetting liquid, cosθ=1
⇒σ=hρrg2
⇒rhg=2σρ ....(2)
Substituting Eq.(2) in Eq.(1), the PE of a liquid in the capillary tube can be written as,
U=π2g(2σρ)2ρ
∴U=2πσ2ρg
Hence, the correct choice is (b)