1. In a hydraulic lift, the diameters of the effort piston and the load piston are in the ratio 1 : 5. If a force of 120 N is applied downward on the effort piston, causing it to move down by 50 cm, what is the work done on the load assuming no friction or fluid compressibility?
- 300 J
- 60 J
- 1500 J
- 12 J
Show answer and explanation
Answer: (B) 60 J
According to Pascal's principle and conservation of energy for an ideal hydraulic machine, the work done on the input piston equals the work done on the output piston (Work = Force × displacement). Here, Work = 120 N × 0.50 m = 60 J. Although the force is multiplied by a factor of 25, the displacement of the load piston decreases by the same factor of 25, keeping the work done equal to 60 J.