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