Question

Mercury has an angle of contact equal to `140^(@)` with soda lime galss. A narrow tube of radius `1.00mm` made of this glass is dipped in a through containing mercury. By what amount does the mercury dip down in the tube relative to the mercury surface outside? Surface tension of mercury at the temperature of the experiment is `0.465 Nm^(-1)`. Density of mercury = `13.6xx10^(3) kg m^(-3)`.

Answer 1

Here, `theta = 140^(@), r =10^(-3)m, S=0465 Nm^(-1), rho = 13.6 xx10^(3) kg, h=?`
`cos 140^(@) = -cos 40^(@) = -0.7660`
Now, `h=(2S cos theta)/(r rho g) = (2xx0.465xxcos 140^(@))/(10^(-3)xx13.6xx10^(3)xx9.8) = 2xx0.465xx(-0.7660)0/(10^(-3)xx13.6 xx 10^(3)xx9.8) = -5.34 xx 10^(-3)m = -5.34mm`.
Negative value of h shows that the mercury level is depressed in the tube.