Question

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. 2 × 10⁺⁵ Pa
• C. 200 × 10⁺⁶ Pa
• D. 2 × 10⁺⁶ Pa

Answer 1

The Correct Option is D

To determine the maximum pressure that the smaller piston would have to bear in a hydraulic lift designed to lift vehicles of mass 5000 kg with a given cross-sectional area, we will use the relationship between force, pressure, and area. Here are the steps to solve the problem:

1. First, calculate the weight of the vehicle, which acts as the force required to lift it. The weight \( W \) can be calculated using the equation: W = m \times g, where:
   • m = 5000 \, \text{kg} (mass of the vehicle)
   • g = 10 \, \text{m/s}^2 (acceleration due to gravity)
2. Substitute the values: W = 5000 \times 10 = 50000 \, \text{N}
3. The maximum pressure \( P \) exerted can be calculated using the formula: P = \frac{F}{A}, where:
   • F = 50000 \, \text{N} (force or weight)
   • A = 250 \, \text{cm}^2 = 0.025 \, \text{m}^2 (area in square meters)
4. Substitute the values into the equation: P = \frac{50000}{0.025} = 2000000 \, \text{Pa}
5. The maximum pressure that the smaller piston would have to bear is \( 2 \times 10^6 \, \text{Pa} \).

Therefore, the correct option is:

• 2 × 10⁺⁶ Pa

Explanation of Options:

• 2 × 10⁺⁵ Pa: This would be an incorrect option as it underestimates the actual pressure.
• 20 × 10⁺⁶ Pa: This would overestimate the required pressure.
• 200 × 10⁺⁶ Pa: This is an unrealistically high value based on the given information.