The correct option is C No, Loss in mechanical energy =2πT2ρg
For the vertical equilibrium of liquid column in capillary tube,
2πrTcosθ=πr2hρg ...(i)
⇒h=2Tcosθρgr
For water( perfectly wetting nature), θ=0∘
∴h=2Tρgr
Since force due to surface tension acting in vertically upward direction is,
F=2πrTcosθ=2πrT
where 2πr= perimeter of meniscus;
Work done by surface tension when liquid rises through height h,
W=(2πrT)×h=4πT2ρg ...(ii)
The potential energy of water in the tube,
U=mgh′
U=(πr2hρ)×g×h2 ...(iii)
It is multiplied by h′=h2 because the whole mass of liquid in the tube is concentrated at the centre of mass at a height of h2.
Substituting in Eq.(iii) in terms of T,
U=2πT2ρg
Consider U=0 at the initial level of water, the increase in PE of water in the tube is ΔU=U−0=U
Since all the work done by the force due to surface tension has not been converted into gain in PE, mechanical energy is not conserved.
W≠ΔU
Hence, loss in mechanical energy :
ΔE=W−ΔU
⇒ΔE=4πT2ρg−2πT2ρg=2πT2ρg
The loss in mechanical energy is dissipated as heat.