Multiple Choice Question. A hydraulic lift is designed to lift heavy objects of maximum mass 2000 kg. The area of cross-section of piston carrying the load is 2.25 $\times$ 10-2 m2. What is the maximum pressure the smaller piston would have to bear?

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0.8711 $\times$ 106 N/m2 Explanation: Apply Newton’s second law of motion to find the force: F = mg Substitute the value mass as m = 2000 kg and acceleration due to gravity as g = 9.81 m/s2 to find the force exerted by the piston. F = mg F = 2000 kg $\times$ 9.81 m/s2 F = 19620 N Use the law of pressure p = F/A Substitute the value of force as F = 19620 N, and area as A = 2.25 $\times$ 10–2m2 to calculate the maximum pressure the hydraulic lift can bear. Pmax = (19620N)/(2.25 xx 10^-2m^2) Pmax = 0.8711 $\times$ 106Pa

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