JAMB Physics 2010 Paper

Concept:In a hydraulic press, pressure applied on a confined liquid is transmitted equally to all parts of the liquid.Therefore, the same pressure acts on the second piston.Explanation:Pressure from the pump piston is $P = 100\, \text{Pa}$.Cross-sectional area of the second piston is $A = 3\, \text{m}^2$.The force produced on the second piston is given by:$$F = P \times A$$Substitute the given values:$$F = 100\, \text{Pa} \times 3\, \text{m}^2$$Since $1\, \text{Pa} = 1\, \text{N/m}^2$, the unit simplifies to newtons:$$F = 300\, \text{N}$$Answer:The force exerted on the second piston is $300\, \text{N}$.This corresponds to Option A.