Question Number 14139 by tawa tawa last updated on 28/May/17
Commented by ajfour last updated on 28/May/17
Answered by ajfour last updated on 28/May/17
$${first}\:{case}: \\ $$$${P}_{\mathrm{0}} +\rho{gh}={P}_{\mathrm{1}} =\frac{{nRT}}{{V}_{\mathrm{1}} }\:\:\:….\left({i}\right) \\ $$$${second}\:{case}: \\ $$$${P}_{\mathrm{0}} −\rho{gh}={P}_{\mathrm{2}} =\frac{{nRT}}{{V}_{\mathrm{2}} }\:\:\:….\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\frac{{P}_{\mathrm{0}} +\rho{gh}}{{P}_{\mathrm{0}} −\rho{gh}}=\frac{{V}_{\mathrm{2}} }{{V}_{\mathrm{1}} } \\ $$$$\Rightarrow\:{P}_{\mathrm{0}} \left({V}_{\mathrm{2}} −{V}_{\mathrm{1}} \right)=\rho{gh}\left({V}_{\mathrm{1}} +{V}_{\mathrm{2}} \right) \\ $$$$\frac{{P}_{\mathrm{0}} }{\rho{g}}={h}\left(\frac{{V}_{\mathrm{1}} +{V}_{\mathrm{2}} }{{V}_{\mathrm{2}} −{V}_{\mathrm{1}} }\right)\:={h}\left(\frac{{l}_{\mathrm{1}} +{l}_{\mathrm{2}} }{{l}_{\mathrm{2}} −{l}_{\mathrm{1}} }\right) \\ $$$$\:\:\:\:=\left(\mathrm{14}{cm}\right)\left(\frac{\mathrm{15}+\mathrm{22}}{\mathrm{22}−\mathrm{15}}\right) \\ $$$$\:\:\:\:=\left(\mathrm{14}{cm}\right)\left(\frac{\mathrm{37}}{\mathrm{7}}\right)\:=\:\mathrm{74}\boldsymbol{{cm}}\:\boldsymbol{{H}}{g}\:. \\ $$
Commented by tawa tawa last updated on 28/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by tawa tawa last updated on 28/May/17
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}. \\ $$