10.29 Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 N m^–1. Density of mercury = 13.6 × 10³kg m^–3.

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Vishal Baghel | Contributor-Level 10

5 months ago

Radius of the narrow tube, r = 1 mm= 1 $\times 10^{- 3}$ m

Surface tension of mercury at the given temperature, s= 0.465 N/m

Density of mercury $ρ = 13.6 \times 10^{3}$ kg/ $m^{3}$

Dipping depth = h

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

Surface tension related with angle of contact and dipping depth is given by:

s = $\frac{h ρ g r}{2 \cos ? θ}$ or h = $\frac{2 scos ? θ}{ρ g r}$ = $\frac{2 \times 0.465 \times \cos ? 140 °}{13.6 \times 10^{3} \times 9.8 \times 1 \times 10^{- 3}}$ = -5.345 $\times 10^{- 3}$ m= -5.345 mm

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