Mercury is poured into a U-tube as in Figure (a). The left armof the tube has a
ID: 1734359 • Letter: M
Question
Mercury is poured into a U-tube as in Figure (a). The left armof the tube has a cross-sectional area A1 of92.0 cm2, and the right arm hasa cross-sectional area A2 of 6.00 cm2. Fourhundred grams of water are then poured into the right arm, as inFigure (b). (a) Determine the length of the water column inthe right arm of the U-tube.cm
(b) Given that the density of mercury is 13.6 g/cm3,what distance h does the mercury rise in the left arm?
cm (a) Determine the length of the water column inthe right arm of the U-tube.
cm
(b) Given that the density of mercury is 13.6 g/cm3,what distance h does the mercury rise in the left arm?
cm Mercury is poured into a U-tube as in Figure (a). The left armof the tube has a cross-sectional area A1 of92.0 cm2, and the right arm hasa cross-sectional area A2 of 6.00 cm2. Fourhundred grams of water are then poured into the right arm, as inFigure (b). (a) Determine the length of the water column inthe right arm of the U-tube. (b) Given that the density of mercury is 13.6 g/cm3,what distance h does the mercury rise in the left arm?
Explanation / Answer
We know, mass = volume * density = area*height*density Here m = 400 g Area = 6 cm2 density = = 1g/ cm 3 so height = 400 / 6 = 66.67 cm (This is the height of water column ) ii)Let height of mercury be h , Then volume of mercury moving from right to left column = 92*hcm3 so height of mercury lost in right column = 92h / 6 = 15.33 h (Here ,we are calculating the level of water below the initiallevel relative to volume increased in left column ) Also height of water =66.67 c m so additional height form initial position of equilibrium = 66.67 -15.33 h (This is the height of water above the initial level ) weight due to it is balanced by h cm of mercury . so (66.67-15.33h)*1 = 13.6 * h On solving we get , h=2.3 cm