question_answer

Calculate the total pressure at the bottom of a lake of depth 5.1 m (atm. Pressure \[={{10}^{5}}\text{ }P\]), density of water \[={{10}^{3}}kg/{{m}^{g}}=10m/{{s}^{2}}\].

Answer:

Atmospheric pressure \[({{P}_{0}})={{10}^{5}}\text{ }P\] Depth \[(h)=5.1\text{ }m\] Density \[(d)=1\text{ }g/cc=1000\text{ }kg/{{m}^{3}}\] \[g=10\text{ }m/{{s}^{2}}\] \[{{P}_{depth}}={{P}_{atmosphere}}+{{P}_{water\text{ }column}}={{P}_{0}}+hdg\]_(1) Substituting the above values in (1), we get \[P={{10}^{5}}+5.1\times {{10}^{3}}\times 10\text{ }Pa\] \[=15.1\times {{10}^{4}}P\] Therefore, the total pressure at the bottom of the lake is \[15.1\times {{10}^{4}}P\]