Question

I 06:26:29 Time Remaining < Return Next sunt > 2 board 3 points Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is closed. Both containers are filled initially to the same height of 1.61 m, one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is re-established. rses 1 ndar 2 Hm Valve оқ 3 Water Mercury Type your answer... Previous "Next

Answer 1

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l.flm Ler, height of the mere mercury at to the left contained after the h که valve is removed, is PL (1.bl-hf = Hydrostatic pressure at the boltom of the container with water = pressure at the bottom of the container with meru cury (right) P = P2 In ght 1.61 x Iw x y = 11:61-h) x lm xg Im h + 1.61 Pw = 1.61 km - I'm h → h 2 sm h = 1.61 lm - to 1.612 1.611 lm - Pw) 2 Pm sh 1061 (13600-1000) 2 X 13600 m 0.79 The fluid level in the left container in equilibrium is (1.61+0.71) =2.35 m