Question

2 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 1.68 m, one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluidlevelin the left container when equilibrium is re-established. Hm Valve Water Mercury Type your answer Previous Next

Answer 1

EHE 168 m, Given initial height of me mercury = initial height of water Sung 2 13500 kg/m3 density of water Sw= ggg leg/m3 density of merg cury Let mere by for equilibrium container from right mereusy lowers distance x & come to side Left side upto height and for equilibrium pressure due to left container at is op to pressure due to right container at bottom PR Se botom equal so Ē Pw+ Pu water # PE Pieg = Hewig+ & my g mercury 1 mercury PL PR 4 Pe = PHAR c(H-22.3 Segi g & PL=PR 2 H Sw g + & Sing g = (1-x) Ing. g of Swt & Sug HI Hg & Sug 3 22 Seg = H (Png - Sw) HESHg3w) 2 Sug

NAMA - x H(SHg - Sw) 23 Hg 1.68 (13500-997) a 2x 13500 a=0.78 m so Height of liquid in left column is h= Hta 2 h = 1.68 +0.78 m ha 2.46 m from the bottom.