Question

4:​A submarine descends into the sea to a depth of 200 m below the surface of water.

​The density of the sea water is 1.03 g/cm3. Atmospheric pressure is 1.0 x105 Pa.​[10 marks]

Calculate:

(a) The increase in pressure as it descends from the surface to a depth of 70 m.​

(b) On another dive, the submarine experiences a total pressure of 6.5 x105 Pa.​

A hatch cover on the submarine has an area of 3.5 m2. Calculate the force on the outside of

the cover.​​​​​​​​​​

(c) The submarine undergoes tests in fresh water of density 1000 kg/m3. Explain why the pressure on the submarine is less at the same depth.​​​​​

Answer 1

a. Increase in the pressure = density*depth*acceleration due to gravity = 1.03*1000*9.8*70 = 706580 Pascal

multiplied by 1000 to convert g/cm^3 to kg/m^3

b. Force=Pressure*Area= 6.5*10^5 *3.5 = 2275000 N

c. Using the equation used in the first question, The pressure depends on the density of the liquid if the depth is same. And as mentioned in the question the density of fresh water is 1000 kg/m^3, while the density of sea water is 1030 kg/m^3, So since the density of fresh water is lesser , So the pressure is lesser compared to the pressure in sea water.