Question

A submarine with a total outer area of 2,100 m² is at a depth of 64.0 m below ocean surface. The weight density of ocean water is 10,100 N/m³. The pressure due to water at that depth is

a. 610 kPa

b. 566 kPa

c. 10,100 Pa

d. 646 kPa

e. 86.0 kPa

A submarine with a total outer area of 2,100 m² is at a depth of 64.0 m below ocean surface. The weight density of ocean water is 10,100 N/m³. The total (absolute) pressure (i.e., including pressure due to the air column above the surface) at that depth is

a. 187 kPa

b. 667 kPa

c. 747 kPa

d. 711 kPa

e. 172 kPa

Answer 1

Part A

Pressure due to water at depth is given by:

P1 = rho*g*h

h = 64.0 m

rho*g = weight density = 10100 N/m^3

So,

P1 = 10100*64 = 646400 N/m^2

1 Pa = 1 N/m^2

P1 = 646.4 kPa

Correct option is D.

Part 2

Absolute pressure = Patm + P1

Patm = 1.013*10^5 Pa

Absolute pressure = 1.013*10^5 + 646400 = 747700 Pa

Absolute pressure = 747.7 kPa

Correct option is C.