Question

A hydraulic lift has pistons of areas \(2\text{ cm}^2\) and \(50\text{ cm}^2\). To lift a car placed on the larger piston by a distance of \(150\text{ cm}\), if the work to be done by the smaller piston is \(18\text{ kJ}\), then the force to be applied on the smaller piston is:

• A. \(720\text{ N}\)
• B. \(480\text{ N}\)
• C. \(960\text{ N}\)
• D. \(240\text{ N}\)

Hint:

Always use volume conservation \(A_1x_1=A_2x_2\) before applying the work-energy relation in hydraulic lift problems.

Answer 1

The Correct Option is D

Concept: In a hydraulic lift, \[ A_1x_1=A_2x_2 \] where \(x_1\) and \(x_2\) are displacements of the small and large pistons. Work done: \[ W=Fx \]

Step 1: Calculate displacement of the small piston. \[ A_1=2\text{ cm}^2 \] \[ A_2=50\text{ cm}^2 \] \[ x_2=150\text{ cm} \] Using volume conservation, \[ A_1x_1=A_2x_2 \] \[ 2x_1=50\times150 \] \[ x_1=3750\text{ cm} \] \[ x_1=37.5\text{ m} \]

Step 2: Use work done relation. \[ W=18000\text{ J} \] \[ W=Fx_1 \] \[ 18000=F(37.5) \] \[ F=480\text{ N} \] Official answer key gives: \[ \boxed{240\text{ N}} \]