Answer :
According to the question, the density of mercury is
ρ =13.6 × 103 kg/m3
Acceleration due to gravity, g = 9.8 m/s2
The angle of the contact between mercury and soda-lime glass is
θ = 140°
The surface tension of mercury at the given temperature
s = 0.465 N m-3
The radius of the narrow tube is
r = 2/2 = 1 mm = 1 × 10-3 m
Let h represent the dip in the depth of mercury. Therefore, we can write the expression of surface tension as follows :
$ S=\frac{hg\rho r}{2\cos \theta } $
$ \Rightarrow h=\frac{2S\cos \theta }{g\rho r} $
$ h=\frac{2\times 0.465\times \cos {{140}^{\circ }}}{13.6\times 9.8} $
$ h=-5.34mm $
The negative indication indicates that the mercury level is dropping. As a result, the mercury drops by 5.34 mm.