Mercury in a capillary tube suffers a depression of 13.2 mm. Find the diameter of the tube. If angle of contact of mercury is `140^@` and density `13.6 xx 10^3 kg m^(-3)` Surface tension of mercury is `540 xx 10^(-3)Nm^(-1)`

Answer

`9.406 xx 10^(-4)m.`

Explanation

`h = 13.2 mm = - 13.2 xx 10^(-3) m , theta = 140^@` and `cos 140^@ = cos (180^@ - 40^@)`
`=- cos 40^@ = 0.7660`
`h =2 S cos theta//r rho g`
` or 2 r = 4S cos theta //h rho g`