JEE Main 2023 · 8 April · evening · Single correct · +4 / −1
A hydraulic automobile lift is designed to lift vehicles of mass 5000 kg. The area of cross section of the cylinder carrying the load is 250 cm². The maximum pressure the smaller piston would have to bear is [ Assume g=10 m / s²]
- (a) 20 × 10⁺⁶ Pa
- (b) 200 × 10⁺⁶ Pa
- (c) 2 × 10⁺⁵ Pa
- (d) 2 × 10⁺⁶ Pa
Answer and solution
Answer (d) 2 × 10⁺⁶ Pa
Solution
Physics sir · 3 steps
Pascal's law says the pressure is the same on both pistons, so find it at the big one.
- Same pressure everywhere. By Pascal's law, pressure applied to an enclosed liquid reaches every part of it unchanged. So the small piston bears the same pressure as the large piston carrying the car. Only the forces differ.
- Load and area. The largest load is the weight of the heaviest vehicle: F = mg = 5000 × 10 = 5 × 10⁴ N. The area is 250 cm² = 250 × 10⁻⁴ m² = 2.5 × 10⁻² m².
P = F/A = (5 × 10⁴ N)/(2.5 × 10⁻² m²) = 2 × 10⁶ Pa
- So. The maximum pressure on the smaller piston is 2 × 10⁶ Pa, which is option (d).
The idea to keep. In a hydraulic lift the force changes from piston to piston, but the pressure does not. Compute F/A wherever you know both, and convert cm² to m² with 10⁻⁴.
Answer from the NTA final answer key. Solution by PrepWiser.