# Consider fresh water lake. Density of water is 10^3 kg/m^3 and gravitational force=10m/s^2, and it is at sea level. Then what is the pressure at 100m in the lake?

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Given the density (`rho` ) of a fluid, its depth (h) and the acceleration due to gravity (g), we can calculate the pressure exerted by this volume of fluid as:

P = `rho gh`

That is, the denser the fluid, more pressure it will exert. Similarly, the greater the depth of fluid, more will be the pressure exerted by it. This can be explained by the bigger volume of fluid, as the depth increases and hence the higher pressure.

In the given question, density = 1000 kg/m^3 and the depth of water is 100 m, thus the static pressure due to this column of water is:

Ps = 1000 kg/m^3 x 10 m/s^2 x 100 m = 10^6 kg/m^3/s^2 = 10^6 N/m^2.

The total pressure is the sum total of the static pressure and the atmospheric pressure (which is equal to 1.01 x 10^5 N/m^2),

Ptotal = Ps + Pa = 10^6 N/m^2 + 1.01 x 10^5 N/m^2 = 1.101 x 10^6 N/m^2

= **1101 kPa.**

Hope this helps.