Question

The lower end of a capillary tube of radius 1 mm is dipped vertically into mercury. (a) Find the depression of mercury column in the capillary. (b) If the length dipped inside is half the answer of part (a), find the angle made by the mercury surface at the end of the capillary with the vertical. Surface tension of mercury = 0.465 N m⁻¹ and the contact angle of mercury with glass −135 °.

Answer 1

Given:
Radius of tube r = 1 mm = 10⁻³ m
Contact angle of mercury with glass θ = 135°
Surface tension of mercury T = 0.465 N/m
Let ρ be the density of mercury.

(a) Depression (h) of mercury level is expressed as follows: 

\[\text{ h } = \frac{2T\cos\theta}{r\rho g}\]     ....(i)

\[\Rightarrow \text{ h } = \frac{2 \times 0 . 465 \times \cos 135^\circ }{{10}^{- 3} \times 13600 \times \left( 9 . 8 \right)}\]

\[ = 0 . 0053 \text{ m = 5 . 3 mm }\]

(b) If the length dipped inside is half the result obtained above:
New depression h'= \[\frac{h}{2}\]

Let the new contact angle of mercury with glass be θ'.

∴ \[h' = \frac{2T\cos\theta'}{r\rho g}\]    ....(ii) 

Dividing equation (ii) by (i), we get:

\[\frac{h'}{h} = \frac{\cos\theta'}{\cos\theta}\]

\[ \Rightarrow \cos\theta' = \frac{\cos\theta}{2}\]

\[\Rightarrow \theta = 112^\circ\]