A hydraulic lift is used to raise a car of mass 2000 kg. The area of cross-section of piston on which the car is placed is 375 cm². What pressure the smaller piston has to bear if bigger piston is 2m higher than the smaller piston? The density of oil used in the lift is 800 kg/m³ and g = 9.8 m/s².

Concept:

According to Blaise Pascal, a French scientist observed that the pressure in a fluid at rest is the same at all points if they are at the same height and it is termed Pascal’s Law.

i.e., the pressure exerted by the fluid on an object at a certain height will be the same in all direction

and hence it can be expressed as

\({P_a} = {P_b} = {p_c} \Rightarrow \frac{{{F_a}}}{{{A_a}}} = \frac{{{F_b}}}{{{A_b}}} = \frac{{{F_c}}}{{{A_c}}}\)

From the above fig. we can see that the force against area within a fluid at rest will always experience pressure perpendicular to their surface area, and the object will experience equal pressure throughout the surface.

And there are a number of devices, such as hydraulic lift and hydraulic brakes, are based on Pascal’s law. these devices used fluids for transmitting pressure.

In a hydraulic lift, as shown in the figure below, two pistons are separated by the space filled with a liquid

Now as Pascal suggested both the platform will have some pressure although the platform has a different surface area hence the force exerted on surface 2 will be much greater than on force one

\(i.e.,\;{P_1} = {P_2} \Rightarrow \frac{{{F_1}}}{{{A_1}}} = \frac{{{F_2}}}{{{A_2}}}\)

Here,

F1, F2 = force exerted on areas A1 and A2 respectively

P1, P2 = Pressure exerted on areas A1 and A2 respectively

Calculation:

Given:

Mass of car = 2000 kg

Area of cross-section of the piston = 375 cm², g = 9.8 m/s²

Force = Weight of car 

⇒ Force on bigger piston = mg

= 2000 × 9.8 = 19600 N

∴ Pressure = 19600 /(375 × 10^-4)

⇒ Pressure, P0= 5.22 × 10⁵ Pa

Pressure 2 m below = P₀ + ρgh

= 5.22 × 105 + 800 × 9.8 × 2

⇒ Pressure 2 m below = 5.37 × 10⁵ Pa