Question

How do you calculate the pressure at a point that is 100 m below the surface of sea water of density 1150 kg m3 .

Answer 1

The pressure underwater is the atmospheric pressure on the surface plus the pressure of the water column.

The quantity of water in a column at a depth of 100 m below the surface with a base area 1 m 2 is 100 m 3.

The mass of this column of water is 100×1150=115000 kg.

⇒ The pressure due to the column of water at a depth of 100 m below the surface of the sea is 115000 kgf/m 2=1127.765 kPa.

The atmospheric pressure at mean sea level is 101.325 kPa.

⇒ The pressure at a point 100 m below the surface of the sea is 101.325+1127.765=1229.090 kPa.

Answer 2

Answer:

The pressure at this point = $1.127×10^{6}$ Pa

Explanation:

The pressure exerted by the liquid at a point inside the liquid is the weight of the water above the point divided by the area of the surface of water.

The pressure below the sea level is given by

P = hρg

where h is the height below the sea level

ρ is the density of sea water and

g is the acceleration due to gravity

Height below the sea level, h = 100 m

Density of sea water, ρ = $1150 kg/m^{3}$

Acceleration due to gravity, g = $9.8 m/s^{2}$

$P = 100×1150×9.8 = 1.127×10^{6} Pa$

Therefore, the pressure at a point that is 100 m below the sea level is = $1.127×10^{6} Pa$