A force of 50 kgf is applied to smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, the diameters of the pistons being 5 cm and 25 cm respectively.

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Solution

Let, F ₁ be the force applied to the smaller piston of cross-sectional area A ₁.
F ₁ = 50 kgf
A ₁ = Ï€d ²/4 = (3.14)(0.05 ²)/4 = 0.0019625 m ²
Let, F ₂ be the force exerted on the large piston of cross-sectional area A ₂.
A ₂ = Ï€d ²/4 = (3.14)(0.25 ²)/4 = 0.0490625 m ²
Now,
F ₁/A ₁ = F ₂/A ₂
=> F ₂ = (F ₁/A ₁)A ₂ = 1250 kgf

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