question_answer

At a given place a barometer records a pressure of 70 cm of mercury. If the mercury in the barometer is replaced by water, what would be the height of the water column? Take density of mercury \[=13600\text{ }kg/{{m}^{3}}\], density of water \[=1000kg\text{ }/{{m}^{3}}\]

Answer:

\[{{h}_{1}}=70\text{ }cm=0.7\text{ }m\] \[{{d}_{1}}=13600\text{ }kg/{{m}^{3}}\] \[{{h}_{2}}=?\] \[{{d}_{2}}=1000\text{ }kg/{{m}^{3}}\] We know, \[P=hdg\] As the pressure is the same, \[{{P}_{1}}={{P}_{2}}\] i.e., \[{{h}_{1}}{{d}_{1}}g={{h}_{2}}{{d}_{2}}g\] \[\Rightarrow \]\[{{h}_{2}}=\frac{{{h}_{1}}{{d}_{1}}}{{{d}_{2}}}=\frac{0.7\times 13600}{1000}=9.52\] Therefore, the height of the water column is 9.52 m.