A liquid of density ` rho` and surface tension ` sigma ` rises in a capillary tube of diameter `d`. Angle of contact between the tube and liquid is zero.
In the above problem, the increase in gravitational potential energy of the liquid due to rise in the capillary is :

Options

A. ` sigma g rho d `
B. ` (2pi sigma ^2 )/( rho g )`
C. ` 2pi sigma ^2 rho g `
D. ` (2pi sigma ^2 )/( rho g d ) `

Answer

B

Explanation

P.E. ` = mg (h)/(2) = (pi sigma d h ) /(2) = (pi sigma d )/(2) ((4 sigma)/(rho d g )) = (2 sigma ^ 2 pi )/(rho g )`