question_answer

Calculate the equivalent height of water barometer if the pressure recorded by mercury barometer is 76 cm. Density of mercury is \[13600\text{ }kg\text{ }{{m}^{-3}}\]and density of water is\[1000\text{ }kg\text{ }n{{r}^{3}}\].

Answer:

               

Mercury Barometer	Water Barometer
\[{{h}_{1}}=76\,cm=\frac{76}{100}m\]	\[{{h}_{2}}=?\]
\[{{d}_{1}}=13,600\,kg\,{{m}^{-3}}\]	\[{{d}_{2}}=1000\,kg\,{{m}^{-3}}\]

According to the problem, the pressure exerted by the water barometer is equal to mercury barometer i.e., \[{{P}_{mercury}}={{P}_{water}}\] We know, pressure \[(P)=hdg\] \[\Rightarrow \]\[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{d}_{1}}}{{{d}_{2}}}\] \[\Rightarrow \]\[\frac{{{P}_{2}}}{{{P}_{1}}}\times {{d}_{1}}=\frac{10}{1}\times 1.293\]\[{{d}_{2}}=12.93\,kg\,{{m}^{-3}}\] By applying this to both the cases, we get \[{{h}_{1}}{{d}_{1}}{{g}_{1}}={{h}_{2}}{{d}_{2}}{{g}_{2}}\] Acceleration due to gravity is same in both the cases i.e., \[(\because {{g}_{1}}={{g}_{2}})\] \[{{h}_{1}}{{d}_{1}}={{h}_{2}}{{d}_{2}}\] \[{{h}_{2}}=\frac{{{h}_{1}}{{d}_{1}}}{{{d}_{2}}}=\frac{\frac{76}{100}\times 13,600}{1000}\]\[\Rightarrow \]\[h=10.34\,m\] Therefore, equivalent height of water barometer is 10.34 m.