The areas of the effort and load pistons of a hydraulic press are 0.5m² and 5m² respectively. If a force F₁ of 100N is applied on the effort piston, the force F₂ on the load is

10N

100N

500N

1000N

5000N

Correct answer is D

\(\frac{load}{area}\) = \(\frac{effort}{area}\)

\(\frac{load}{5}\) = \(\frac{100}{0.5}\)

load = 1000N

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The areas of the effort and load pistons of a hydraulic press are 0.5m2 and 5m2 respectively. If a force F1 of 100N is applied on the effort piston, the force F2 on the load is