In a hydraulic press , a force of 40N is applied on the effort piston of area 0.4m². If the force exerted on the load piston is 400N, the area of the large piston is

4m²

8m²

2m²

1m²

Correct answer is A

Pressure developed in the effort piston = ^F/_a
∴^F/_a = ⁴⁰/_0.4 = ⁴⁰⁰/₄ = 100Nm²
this pressure is transmitted to the larger piston of area A; => P = ^F/_A
∴ 100 = ⁴⁰⁰/_A
=> A = ⁴⁰⁰/₁₀₀ = 4m²

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