Water rises to a height of 10.0 cm in a capillary tube and mercury gal

1.	Water rises to a height of 10.0 cm in a capillary tube and mercury galls to a depth of 3.42 cm in the same capillary tube. If the density of mercury is 13.6 g cm-3 and the angle of contact is 135°, find the ratio of surface tension for water and mercury.
Answer» Here, For water, h₁ = 10.0 cm, θ₁ = 0°, P₁=1 g cm^-3; r₁ = r. For mercury, h₂ = −3.42 cm; θ₂ = 135° P₂ = 13.6 g cm^-3; r₂ = r = r₁ S₁ = \(\frac{r_1h_1ρ_1g}{2\,cos\, θ_1}\) S₂ = \(\frac{r_2h_2ρ_2g}{2\,cos\, θ_2}\) \(\frac{S_1}{S_2}\) = \(\frac{h_1ρ_1\,cos\,θ_2}{h_2ρ_2\,cos\,θ_1}\) = \(\frac{10×1×(−\frac{1}{\sqrt{2}})}{ −3.421×13.6×1}\) = 0.15

Water rises to a height of 10.0 cm in a capillary tube and mercury galls to a depth of 3.42 cm in the same capillary tube. If the density of mercury is 13.6 g cm-3 and the angle of contact is 135°, find the ratio of surface tension for water and mercury.

Answer»

Here,

For water, h₁ = 10.0 cm, θ₁ = 0°, P₁=1 g cm^-3; r₁ = r.

For mercury, h₂ = −3.42 cm; θ₂ = 135°

P₂ = 13.6 g cm^-3; r₂ = r = r₁

S₁ = \(\frac{r_1h_1ρ_1g}{2\,cos\, θ_1}\)

S₂ = \(\frac{r_2h_2ρ_2g}{2\,cos\, θ_2}\)

\(\frac{S_1}{S_2}\) = \(\frac{h_1ρ_1\,cos\,θ_2}{h_2ρ_2\,cos\,θ_1}\)

= \(\frac{10×1×(−\frac{1}{\sqrt{2}})}{ −3.421×13.6×1}\)

= 0.15