Question

A scuba diver measures an increase in pressure of around 10^5 Pa upon descending by 10 m, what is the change in force per square centimetre on the diver’s body?

Answer 1

ΔP = 10 N / cm^2

Step-by-step explanation:

Solution:-

- The pressure on an object placed inside any fluid changes linearly with the depth of fluid above the object.

- A fluid has its density. In our case we will take the density of water:

ρw = 1000 kg/m^3

- The pressure ( P ) exerted on an object at the depth of ( h ) is given below:

P = ρ*g*h

Where,

g: The gravitational acceleration constant = 9.81 m/s^2

- The differential expression relating the change in pressure and change in depth of the object under the fluid.

ΔP = ρ*g*Δh

- We are given that the scuba diver experiences an increase in pressure of:

ΔP = 10^5 N / m^2

- We are to determine the change in pressure - ( change in force per square centimetre )

- The conversion of units would be as follows:

ΔP = 10^5 [ N / m^2 ] * [ m^2 / (100cm)^2 ]

ΔP = 100000/10000 [ N / m^2 ] * [ m^2 / cm^2 ]

ΔP = 10 N / cm^2

Answer 2

Step-by-step explanation:

Given that,

The pressure around the diver is

P = 10^5 Pa,

At a height of 10m

h = 10m

We want to find the force per square centimeter of the diver body

Pressure = force / Area

Force / Area = pressure,

And since the pressure is given

Then,

Force / Area = 10^5 Pa

Since 1 pascal = 1 N/m²

Force / Area = 10^5 N/m²

Since, we know that, 100cm = 1m

(100cm)² = (1m)²

10000cm² = 1m²

10⁴cm² = 1m²

Then,

Force / Area = 10^5 N/m² × 1m²/10⁴cm²

Force / Area = 10 N/cm²

The required force per unit area is 10 N /cm²