Question

09:58:55 Time Remaining < Return Next * 2 ard 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 1.49 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. -es 1 ndar 2 Hm 3 Valve 3 BOX Water Mercury 4 Help Type your answer "Next Previous

Answer 1

Gnitially bolte liquido have same level but as the valve open liquids will adjust their height Sothat prepourure cit same level remaini same on at the bottom common pressure should be same 179 water Tiagra mercury Now at A prcoouse from left arms PAL Z Petmalog (7.49)+ Sigga preppure at A from Right Sider PAz = Patm + Study f (H-x)

(-49-) : Punt Putnt SHg = Now preppure will be same PAI=PA2 3w3 f *149 +3498 (15ma x2 Jw - demoity of water z loookg/m3 mercury z 13600kg/m3 3w x 1.49 + 3Hgx 2 = 349 (1-49-2) 1000X1.497 13600 23 13600 (1-49->) 2 x 13600 x = 13600X1.49 . 1000x1-49 x= 1960 x 1449 =oom 0.69m 2x 13600 In left container water level. = 1.49m Ano and mercury level=0.69m Ano' and miri ghat continer 2 Hg level=-8 1-49m, water - V by reg 0-69m oor