Mercury has an angle of contact equal to `140^(@)` with soda lime glass. 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)`.

Options

A. 5.34 mm
B. 2.35 mm
C. 6.25 mm
D. 1.44 mm

Answer

A

Explanation

Here, `theta=140^(@), r=1xx10^(-3)`m
`S=0.465 N m^(-1), rho=13.6xx10^(3) kg m^(-3)`
As `h=(2Scostheta)/(r rhog)=(2xx0.465xx(-cos40^(@)))/(10^(-3)xx13.6xx10^(3)xx9.8) " "(because cos 140^(@)=-cos 40^(@))`
`=5.34xx10^(-3)m=-5.34` mm
Note: Negative value of h shows that the mercury level is depressed in the tube.