Mercury has an angle of contact of `120^(@)` with glass. A narrow tube of radius `1.0mm` made of this glass is dipped in a through containig mercury. By what amount dies the mercury dip down in the tube relative to the liquid surface outside. Surface tension of mercury at the temperature of the experiment os `0.5 N//m` and density of mercury is `13.6xx10^(3) kg//m^(3)`.
(Take `g = 9.8 m//s^(2))`.

Answer

C

Explanation

`h=(2T costheta)/(r rho g)`
Substituting the values, we get `h=(2xx0.5xxcos 120^(@))/(10^(-3)xx13.6xx10^(3)xx9.8)`
`=-3.75xx10^(-3)m`
`h=-3.75 mm`.