In a hydraulic machine a force of 3N is applied on the Piston of area of cross section 12 cm2. What force is obtained on its piston of area of cross section 100 cm2.
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Explanation:
pressure = F1/A1 = F2/A2 (Pascal's law)
3 / 12 = F2/100
F2 = 100/4 = 25 N
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