A12,000-n car is raised using a hydraulic lift, which consists of a u-tube with arms of unequal areas, filled with incompressible oil with a density of 800 kg/m3 and capped at both ends with tight-fitting pistons. the wider arm of the u-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. the car rests on the piston on the wider arm of the u-tube. the pistons are initially at the same level. what is the force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20 m? (for purposes of this problem, you can neglect the weight of the pistons.)

This is a hydraulic system, and these systems rely on the incompressible nature of fluids to transmit pressure through the fluid equally. Therefore, the pressure on the wide side of the U-tube (the side with the car) must be equal to the pressure on the narrow side of the U-tube. We begin by calculating the pressure on the wider side:
Pressure on wide side = pressure due to car + pressure due to difference in height of oil

The pressure by the car is calculated using:
Pressure = force / area
P = 12,000 / (π * 0.18²)
P = 118 kPa

While the pressure due to the oil is given by:
Pressure = density * gravitational field strength * height
P = 800 * 9.81 * 1.2
P = 9.4 kPa

Pressure on wide side:
118 + 9.4 = 127.4 kPa

Pressure on the narrow side will be given by:
Pressure = force / area
Force = area * pressure
Force = (π * 0.05²)(127,000)
Force = 997 Newtons

The force required is about 1 kN

Ithink the answer is a