Question

In a hydraulic lift, the surface area of the input piston is 6 cm² and that of the output piston is 1500 cm². If 100 N force is applied to the input piston to raise the output piston by 20 cm, then the work done is kJ.

Answer 1

Correct Answer: 5

To determine the work done, we start with Pascal's principle, which states that the pressure applied to a confined fluid is transmitted undiminished throughout the fluid.

Given:

• Surface area of input piston, A₁ = 6 cm²
• Surface area of output piston, A₂ = 1500 cm²
• Force on input piston, F₁ = 100 N
• Output piston raised by h = 20 cm = 0.2 m

The pressure applied at the input piston is:

P = F₁ / A₁ = 100 N / 6 cm²

Since 1 cm² = 0.0001 m²:

A₁ = 6 × 0.0001 = 0.0006 m²

P = 100 N / 0.0006 m² = 166666.67 N/m²

This pressure is transmitted to the output piston:

F₂ = P × A₂ = 166666.67 N/m² × (1500 cm² × 0.0001 m²/cm²) = 25000 N

The work done, W, is given by:

W = F₂ × h = 25000 N × 0.2 m = 5000 J

Converting 5000 J to kJ:

W = 5 kJ

The work done is 5 kJ.