A reservoir is filled with liquid of density 2000 kgm−1 − 1 . Calculate the depth at which the pressure in the liquid will be equal to 9100 Nm−2 − 2 (g = 10...

A reservoir is filled with liquid of density 2000 kgm−1 ${}^{-1}$. Calculate the depth at which the pressure in the liquid will be equal to 9100 Nm−2 ${}^{-2}$ (g = 10 ms−2 ${}^{-2}$)

Answer Details

The pressure at a depth "h" in a liquid of density "ρ" can be calculated using the formula: P = ρ * g * h where "g" is the acceleration due to gravity (9.8 m/s^2). So, we can rearrange the formula to find the depth "h" at which the pressure will be 9100 N/m^2: h = P / (ρ * g) Plugging in the given values, we get: h = 9100 N/m^2 / (2000 kg/m^3 * 9.8 m/s^2) = 0.455 m So, the pressure in the liquid will be equal to 9100 N/m^2 at a depth of 0.455 m.