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A glass tube of radius 0.80 mm is dipped in mercury…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A glass tube of radius $0.80\ \text{mm}$ is dipped in mercury. The angle of contact is $140^\circ$ , the surface tension $0.465\ \text{N/m}$ and the density $13\,600\ \text{kg/m}^3$ . How far does the mercury in the tube stand below the level outside? ( $g = 9.8\ \text{m/s}^2$ .)

Need a reminder? The key ideaShow

Capillary rise: $h = \dfrac{2S\cos\theta}{\rho gr}$ , where $\theta$ is the angle of contact. A liquid that wets the tube ( $\theta < 90^\circ$ ) rises; one that does not ( $\theta > 90^\circ$ ) is pushed down.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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