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 s⁻²]:
Options

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 s⁻²]:

Options

G1 Physics · Accepted Answer

By Pascal's law, the same pressure is transmitted through the hydraulic fluid. Required pressure: P = F/A = Mg/A Given: M = 5000 kg g = 10 m s⁻² A = 250 cm² = 250×10⁻⁴ m² = 2.5×10⁻² m² Therefore, P = (5000×10)/(2.5×10⁻²) = 5.0×10⁴ / 2.5×10⁻² = 2×10⁶ Pa Hence option D follows.