1
\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}
2
\mathrm{P}=\mathrm{h}\mathrm{\rho}\mathrm{g}
3
\mathrm{P}=\mathrm{P}_0+\mathrm{h}\mathrm{\rho}\mathrm{g}
4
\frac{\mathrm{F}_1}{\mathrm{A}_1}=\frac{\mathrm{F}_2}{\mathrm{A}_2}
FORMULA CHAPTER
Chapter 04 – Pressure in Fluids and Atmospheric Pressure
ICSEClass 09Physics4 formulas
Revise ICSE Class 9 Chapter 04 formulas for thrust, pressure, liquid pressure, density, depth, Pascal’s law and hydraulic machines.
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Pressure
\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}
Worked example
Given: Thrust = 200 N; area = 0.5 m² Calculation: P = 200/0.5 Answer: P = 400 Pa
Pressure Due to a Liquid Column
\mathrm{P}=\mathrm{h}\mathrm{\rho}\mathrm{g}
Worked example
Given: h = 2 m; ρ = 1000 kg/m³; g = 9.8 m/s² Calculation: P = 2 × 1000 × 9.8 Answer: P = 19600 Pa
Total Pressure in a Liquid
\mathrm{P}=\mathrm{P}_0+\mathrm{h}\mathrm{\rho}\mathrm{g}
Worked example
Given: P₀ = 101325 Pa; hρg = 19600 Pa Calculation: P = 101325 + 19600 Answer: P = 120925 Pa
Pascal’s Law for a Hydraulic Machine
\frac{\mathrm{F}_1}{\mathrm{A}_1}=\frac{\mathrm{F}_2}{\mathrm{A}_2}
Worked example
Given: F₁ = 50 N; A₁ = 0.01 m²; A₂ = 0.20 m² Calculation: F₂ = 50 × 0.20/0.01 Answer: F₂ = 1000 N