Question

The lower end of a capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, up to what height will the water rise in the capillary?

Answer 1

Let T be the surface tension, r be the inner radius of the capillary tube and ρ bethe density of the liquid.

For cos θ = 1, height (h) of the liquid level is given as:

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

Now, for mercury:

\[\text{h}_{\text{Hg}} = \frac{2\text{T}_{\text{Hg}}cosθ}{\text{r}\rho_{\text{Hg}} \text{ g}}\]   .........(i)

For water:

\[\text{h}_\text{w} = \frac{2 \text{T}_\text{w}cosθ}{\text{r} \rho_\text{w g}}\]    ...(ii)

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

\[\frac{h_w}{h_{Hg}} = \frac{T_w}{T_{Hg}} \times \frac{\rho_{Hg}}{\rho_w}\] ` xx (cos0°)/(cos 140°)`

`h_w/(- 0.02) = (7.5 xx 10^-2 xx 1 xx 13.6 xx 10^3)/(10^3 xx cos 140 xx 0.465)`

h_w = 0.0576 m
h_w = 5.73 cm
Hence, the required rise in the water level in the capillary tube is 5.73 cm.