Question

Two vessels containing mercury are joined together. The diameter of one vessel is four times the diameter of the other (Fig. 79) A column of water of height h0 is poured into the vessel on the left. By what distance will the level of the mercury in the vessel on the right rise if h0 is equal to 70cm? How far will the level of mercury in the vessel on the left drop? By what height will the level of the mercury column in the narrow vessel rise if a column of water of the same height is poured in the wide one?

Answer 1

Answer: h2 = 0.3 cm; h1 = 4.8cm.

Explanation:

If the changes in the level of the mercury in the vessels on the left and right are denoted by h1 and h2, respectively (h1 + h2 = x), and if the pressure is measured in centimetres, then the condition of equilibrium of the liquids takes the form

h1 + h2 = h0d0/d.

where do is the density of water and d is the density of mercury. The condition of incompressibility of liquids gives

S1h1 = S2h2,

where S1 and S2 are areas of the cross sections of the vessels, which in this case are connected by the relation S2 = 16 S1. The first equation determines the condition of equilibrium in the tube and the second expresses the fact that the volume of the mercury passing from the left-hand limb to the right-hand limb *** constant. From these equations we obtain

h1 = 16 h0d0/17d and h2 = h0d0/17d.