418 Fluid MechanicsP14.18(a)Using the definition of density, we havehmAw===waterwater23g5.00 cmg cmcm210010020 0ρ..ej(b)Sketch (b) at the right represents the situation afterthe water is added. A volume Ah22bgof mercuryhas been displaced by water in the right tube. Theadditional volume of mercury now in the left tubeis 1. Since the total volume of mercury has notchanged,FIG. P14.181=orhAAh212=(1)At the level of the mercury–water interface in the right tube, we may write the absolutepressure as:PPghw=+0waterThe pressure at this same level in the left tube is given byghh Pghw+ =+020ρρHgwaterwhich, using equation (1) above, reduces toHgwaterhAAhw112+LNMOQP=or hhwAA=+waterHg112.Thus, the level of mercury has risen a distance ofh=+=20013 61049010 0500......gcmcmcm33afchabove the original level.P14.19∆∆Pgh03266 10−×. Pa:PP P=+ =−×=×00551 013 0 026 6
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .