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]:
- 20 × 10+6 Pa
- 2 × 10+5 Pa
- 200 × 10+6 Pa
- 2 × 10+6 Pa
The Correct Option is D
Solution and Explanation
Step 1: Write the formula for pressure. The pressure \( \Delta P \) exerted by the smaller piston is given by: \[ \Delta P = \frac{F}{A}, \] where: - \( F \) is the force exerted due to the weight of the vehicle, - \( A \) is the area of cross-section of the cylinder.
Step 2: Calculate the force \( F \). The force \( F \) is the weight of the vehicle: \[ F = m \cdot g, \] where: - \( m = 5000 \, \text{kg} \), - \( g = 10 \, \text{m/s}^2 \). \[ F = 5000 \cdot 10 = 50000 \, \text{N}. \]
Step 3: Convert the area to SI units. The given area is: \[ A = 250 \, \text{cm}^2. \] Convert it to square meters: \[ A = 250 \times 10^{-4} \, \text{m}^2 = 0.025 \, \text{m}^2. \]
Step 4: Calculate the pressure. Substitute \( F = 50000 \, \text{N} \) and \( A = 0.025 \, \text{m}^2 \) into the formula for pressure: \[ \Delta P = \frac{F}{A} = \frac{50000}{0.025}. \] Simplify: \[ \Delta P = 2 \times 10^6 \, \text{Pa}. \]
Final Answer: The maximum pressure the smaller piston would have to bear is: \[ \boxed{2 \times 10^6 \, \text{Pa}}. \]
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