418 Fluid MechanicsP14.18(a)Using the definition of density, we havehmAw===waterwater23g5.00 cmg cmcm21001 0020 0ρ..ej(b)Sketch (b) at the right represents the situation afterthe water is added. A volume A h22bgof mercuryhas been displaced by water in the right tube. Theadditional volume of mercury now in the left tubeis A h1. Since the total volume of mercury has notchanged,FIG. P14.18A hA h221=orhAAh212=(1)At the level of the mercury–water interface in the right tube, we may write the absolutepressure as:PPghw=+0ρwaterThe pressure at this same level in the left tube is given byPPg hhPghw=++=+020ρρHgwaterbgwhich, using equation (1) above, reduces toρρHgwaterhAAhw112+LNMOQP=or hhwAA=+ρρwaterHg112ej.Thus, the level of mercury has risen a distance ofh=+=1 0020 013 610 49010 05 00......gcmcmg cmcm33ejafejchabove the original level.P14.19∆∆Pg h032 6610== −×ρ.Pa :PPP=+=
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