In a hydraulic press, the small piston has area 5 cm² and the large piston 400 cm². A force of 200 N is applied to the small piston. (a) What force does the large piston exert? (b) If the small piston moves down 40 cm, how far does the large piston rise? Step 1: (a) F₂ = F₁(A₂/A₁) = 200 × (400/5) = 200 × 80 = 16,000 N. Step 2: (b) Volume displaced is the same: A₁d₁ = A₂d₂ → d₂ = d₁(A₁/A₂) = 40 × (5/400) = 0.5 cm. Step 3: Check work: in = 200 × 0.4 = 80 J; out = 16,000 × 0.005 = 80 J. Equal, as expected. Answer: (a) 16,000 N (80× multiplication), (b) 0.5 cm. Huge force, tiny movement — the signature of hydraulics.