Question

You are doing experiments from a research ship in the Atlantic Ocean.

1) On a day when the atmospheric pressure at the surface of the water is 1.03×105Pa1.03×105Pa, at what depth below the surface of the water is the absolute pressure twice the pressure at the surface?

2) At what depth below the surface of the water is the absolute pressure four times the pressure at the surface?

Answer 1

We know that:

Absolute pressure = Atmospheric pressure + rho*g*h

P = P_atm + rho*g*h

P_atm = 1.03*10^5 Pa

Pressure at the surface of Ocean will be atmospheric pressure, So

Now we need depth where absolute pressure is twice the pressure at the surface, So

P = 2*P_atm

Using above relation

P = P_atm + rho*g*h

2*P_atm = P_atm + rho*g*h

rho*g*h = P_atm

rho = density of sea water = 1025 kg/m^3

h = depth of water = P_atm/(rho*g)

h = 1.03*10^5/(1025*9.81)

h = 10.24 m = depth required below the surface of ocean

Part B.

Now we need depth where absolute pressure is four times the pressure at the surface, So

P = 4*P_atm

Using above relation

P = P_atm + rho*g*h

4*P_atm = P_atm + rho*g*h

rho*g*h = 3*P_atm

rho = density of sea water = 1025 kg/m^3

h = depth of water = 3*P_atm/(rho*g)

h = 3*1.03*10^5/(1025*9.81)

h = 30.72 m = depth required below the surface of ocean

Let me know if you've any query.