Question

In a glass capillary tube, water rises upto a height of 1-.0cm while mercury fall down by 5.0 cm in the same capillary . If the angles contact for mercury glass is 60 degree and water glass is 0 degree, then find the ratio of surface tension of mercury and water?

Answer 1

Hi

Here is your answer,

For water,   h₁ - 10.0 cm = 0.1 m 

                   ρ₁ = 10³ kg/m³ , θ = 0°

For mercury, h₂ = 5.0 cm = 0.05 m 

                    ρ₂ = 13.6 × 10³ kg/m³ , θ = 60° 

Suppose S₁ and S₂ are the surfaces tensions for water and mercury, respectively, then 

                                S₁ = h₁ Rρ₁ g/ 2 cos θ₁ and S₂ = h₂Rρg/ 2 cos θ₂


The ratio of surface tension of mercury and water,


                 →    S₂/S₁ = h₂ Rρ₂ g / 2 cos θ₂ × 2 cos θ₁ / h₁ Rρ₁ g

                 →     S₂/S₁ = h₂ ρ₂ cos θ₁ / h₁ ρ₁ cos θ₂ 


                             = 0.05 × 13.6 × 10³ × cos 0°/ 0.1 × 1000 × cos 60°


                      = 13.6 : 1 





Hope it helps you !