The areas of pistons in a hydraulic machine are 5 cm 2 and 625 cm 2. What force on the smaller piston will support a load of 1250 N on the larger piston ? State any assumption which you make in your calculation.

Area of narrow piston (A₁) = 5 cm ²

Area of wider piston (A₂) = 625 cm ²

force (F₂) = 1250 N

⇒ {F_1\over 5 } = {1250\over 625}

⇒ {F_1} = {1250\over 625}×5 = 10 N

Hence, force acting on the smaller piston = 10 N

Assumption — There is no friction and no leakage of liquid.