A hydraulic auto mobile lift is designed to unit vehicles of mass 500 kg. The area of cross section of the cylinder carrying the wad is 250 cm2 . The maximum pressure the smaller piston would have to bear is ( g = 10m/s2 )

The problem is asking for the maximum pressure that the smaller piston would have to bear. This can be found using the principle of hydraulic pressure, which states that pressure is transmitted undiminished in an enclosed static fluid.

The formula for pressure is P = F/A, where P is pressure, F is force, and A is area.

1. First, we need to calculate the force exerted by the vehicle. The force is the weight of the vehicle, which is its mass times the acceleration due to gravity. So, F = m*g = 500 kg * 10 m/s^2 = 5000 N.

2. Next, we need to convert the area from cm^2 to m^2 because the standard unit of area in this context is m^2. 1 m^2 = 10,000 cm^2, so A = 250 cm^2 * (1 m^2 / 10,000 cm^2) = 0.025 m^2.

3. Finally, we can calculate the pressure. P = F/A = 5000 N / 0.025 m^2 = 200,000 Pa or 200 kPa.

So, the maximum pressure the smaller piston would have to bear is 200 kPa.

The problem is asking for the maximum pressure that the smaller piston would have to bear. This can be found using the formula for pressure in a hydraulic system, which is P = F/A, where P is the pressure, F is the force, and A is the area.

1. First, we need to calculate the force exerted by the vehicle. The force is the weight of the vehicle, which is its mass times the acceleration due to gravity. So, F = m*g = 500 kg * 10 m/s^2 = 5000 N.

2. Next, we need to convert the area from cm^2 to m^2 because the standard unit of measurement for area in this context is m^2. 1 m^2 = 10,000 cm^2, so A = 250 cm^2 * (1 m^2 / 10,000 cm^2) = 0.025 m^2.

3. Finally, we can calculate the pressure. P = F/A = 5000 N / 0.025 m^2 = 200,000 Pa.

So, the maximum pressure the smaller piston would have to bear is 200,000 Pa.