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The height of a mercury barometer is 75 cm at sea level and 50 cm at the top of a hill. Ration of density of mercury to that of air is `10^(4)`. The height of the hill isA. `250 m`B. `2.5 km`C. `1.25 km`D. `750 m` |
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Answer» Correct Answer - B Difference of pressure between sea level and the top of hill `DeltaP =(h_(1)-h_(2))xxrho_(Hg)xxg=(75-50)xx10^(-2)xxrho_(Hg)xxg" "……(i)` and pressure difference due to h meter of air `DeltaP =hxxrho_("air")xxg" "….(ii)` By equating (i) and (ii) we get `hxx rho_("air")xx g =(75-50) xx 10^(-2) xx rho_(Hg) xx g` `therefore h=25xx10^(-2)((rho_(Hg))/(rho_(air))) = 25xx10^(-2)xx10^(4) = 2500 m therefore " Height of the hill" = 2.5 km`. |
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