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
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Correct Answer: Option D
Explanation:
\(\frac{load}{area}\) = \(\frac{effort}{area}\)
\(\frac{load}{5}\) = \(\frac{100}{0.5}\)
load = 1000N
\(\frac{load}{area}\) = \(\frac{effort}{area}\)
\(\frac{load}{5}\) = \(\frac{100}{0.5}\)
load = 1000N