Ncert Exercises & Solutions

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? The surface tension of mercury at the temperature of the experiment is 0.465 N m^–1. Density of mercury = 13.6 × 10³ kg m^–3.

The angle of contact, θ = 140°

Radius of the narrow tube, r = 1 mm = 1 × 10^–3 m

The surface tension of mercury, S = 0.465 N m^–1

Density of mercury, ρ =13.6 × 10³ kg/m³

Surface tension is given by:

$s = \frac{hρgr}{2 cosθ} ∴ h = \frac{2 s cosθ}{rρg} = \frac{2 \times 0.465 \times \cos 140^{0}}{1 \times 10^{- 3} \times 13.6 \times 10^{3} \times 9.8} = - 0.00534 m = - 5.34 mm$

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