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IB MYP 4-5 Physics- Pressure in liquids – Study Notes

IB MYP 4-5 Physics- Pressure in liquids - Study Notes - New Syllabus

IB MYP 4-5 Physics-Pressure in liquids – Study Notes

Key Concepts

  • Pressure in liquids 

IB MYP 4-5 Physics Study Notes – All topics

Pressure in Liquids

Pressure in Liquids

Liquids exert pressure due to the weight of the liquid above a certain point. This pressure increases with depth and acts equally in all directions at a given depth.

The formula for pressure in liquids is:

P = \(\rho g h\)

  • \(\rho\) = density of the liquid (kg/m³)
  • g = acceleration due to gravity (m/s²)
  • h = depth below the surface (m)
  • P = pressure (Pa)

Key Points:

 

  • Pressure in liquids increases linearly with depth.
  • Pressure depends on the density of the liquid and gravity, but not on the shape or volume of the container.
  • At the same depth, pressure is the same in all directions (Pascal’s principle).
  • Liquids are incompressible, so density is constant for most practical purposes.

Applications:

  • Hydraulic presses and brakes (Pascal’s principle).
  • Design of dams and submarines (pressure increases with depth).
  • Measurement using manometers and barometers.

Example:

A diver is 20 m below the surface of seawater (\(\rho = 1025 \, \text{kg/m}^3\)). Calculate the pressure due to the water at this depth.

▶️ Answer/Explanation

Using \(P = \rho g h\)

\(P = 1025 \times 9.8 \times 20\)

\(P = 200,900 \, \text{Pa}\)

\(\boxed{P = 2.01 \times 10^5 \, \text{Pa}}\)

Example:

If a submarine descends from 50 m to 150 m in the ocean, calculate the increase in pressure on its hull. Assume seawater density \(\rho = 1025 \, \text{kg/m}^3\).

▶️ Answer/Explanation

Change in depth: \(h = 150 – 50 = 100 \, \text{m}\)

Using \(P = \rho g h\):

\(P = 1025 \times 9.8 \times 100\)

\(P = 1,004,500 \, \text{Pa}\)

\(\boxed{\Delta P = 1.00 \, \text{MPa}}\)

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