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The pressure at the bottom of a lake due to water is `4.9xx 10^(6)N//m^(2)`. What is the depth of the lake? |
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Answer» Pressure `P=hrhog=4.9xx10^(6)N//m^(2)` `rho` density of water `=1000km//m^(3), g=9.8m//s^(2)` Hence, `h=(P)/(rhog)=(4.9xx10^(6))/(1000xx9.8)=500m` |
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