Hydraulic Pressure And Pascal's Law

Question

In a hydraulic lift the small piston has an area of 0.01 m² and the large piston has an area of 0.5 m²; what force on the small piston is needed to support a car weighing 20000 N?

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NCERT Class 11, Chapter 10 (Mechanical Properties of Fluids) presents Pascal's law, which states that pressure applied to an enclosed incompressible fluid is transmitted undiminished to every point. In a hydraulic lift the pressure under both pistons is equal, so $\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$. Rearranging for the small-piston force: $F_1 = F_2 	imes \dfrac{A_1}{A_2}$. Substituting $F_2 = 20000\ 	ext{N}$, $A_1 = 0.01\ 	ext{m}^2$ and $A_2 = 0.5\ 	ext{m}^2$ gives $F_1 = 20000 	imes \dfrac{0.01}{0.5} = 20000 	imes 0.02 = 400\ 	ext{N}$. The option 800 N wrongly doubles the area ratio. The option 200 N halves the correct value through an arithmetic slip. The option 1000 N uses an inverted ratio. The essential idea is that pressure, not force, is the same on both sides, so the force scales directly with the piston area; the larger piston experiences a proportionally larger force. As a magnitude check, the small piston force of 400 N is fifty times smaller than the supported weight of 20000 N, exactly matching the fifty-to-one area ratio, which is the force multiplication that makes hydraulic lifts and car brakes so useful. The trade-off, by conservation of energy, is that the small piston must move fifty times farther than the large one.

Hydraulic Pressure And Pascal's Law

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