A hydraulic jack is used in a car workshop to lift a car. The jack consists of two connected pistons A(area = 5 cm²) and B(area = 250 cm²), filled completely with an incompressible hydraulic fluid (such as hydraulic oil). The fluid transmits pressure uniformly throughout the system. A force of 120 N is applied vertically downward on piston A.

Assuming the system to be ideal, answer the following questions:

(a) Calculate the pressure produced in the liquid by piston A.

(b) Determine the force exerted on piston B. Name the law used.

(c) Does this hydraulic jack provide a gain in force or a gain in distance? Justify your answer.

(d) If piston B rises by a height of 2 cm, calculate the distance moved by piston A.

(e) What would happen if a small air bubble enters the liquid of the hydraulic jack?

Given,

• Area of piston A ($\text A_\text A$) = 5 cm²
• Area of piston B ($\text A_\text B$) = 250 cm²
• Force applied on piston A ($\text F_\text A$) = 120 N

(a) Converting cm² into m²

100 cm = 1 m

And, 100 cm x 100 cm = 1 m²

Therefore, 1 cm² = $\dfrac{1}{10000}$ m²

Hence,

• Area of piston A = 5 x 10^-4 m²
• Area of piston B = 250 x 10^-4 m²

As pressure on a surface is given by,

$\text {Pressure} = \dfrac{\text {Force}}{\text {Area}}$

Then,

$\text {Pressure on piston A} = \dfrac{\text {Force applied on piston A}}{\text {Area of piston A}}\\[1em] = \dfrac{120}{5 \times 10^{-4}} \\[1em] = \dfrac{120 \times 10^4}{5} \\[1em] = 24 \times 10^4 \text { Pa} \\[1em] = 2.4 \times 10^5 \text { Pa}$

Hence, the pressure produced in the fluid by piston A is 2.4 x 10⁵ Pa.

(b) As Pascal's law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.

Thus,

Pressure on piston A = Pressure on piston B

Now,

$\text {Pressure on piston B} = \dfrac{\text {Force applied on piston B}}{\text {Area of piston B}} \\[1em] \Rightarrow \text {Force applied on piston B} = \text {Pressure on piston B} \times \text {Area of piston B} \\[1em] = \text {Pressure on piston A} \times \text {Area of piston B} \\[1em] = 2.4 \times 10^5 \times 250 \times 10^{-4} \\[1em] = 24 \times 10^4 \times 250 \times 10^{-4} \\[1em] = 24 \times 250 \\[1em] = 6000\ \text N \\[1em]$

Hence, the force exerted on piston B is 6000 N and the law is Pascal's law.

(c) The hydraulic jack provides gain in force. This is because a small force of 120 N applied on the smaller piston produces a much larger force of 6000 N on the larger piston. Since the output force is greater than the input force, the machine acts as a force multiplier. It does not give a gain in distance.

(d) Given,

• Distance moved by piston B = 2 cm
• Area of piston A = 5 cm²
• Area of piston B = 250 cm²

From conservation of volume.

Volume of fluid displaced in piston A = Volume of fluid displaced in piston B

⇒ Area of piston A × Distance moved by piston A = Area of piston B × Distance moved by piston B

⇒ Distance moved by piston A = $\dfrac{\text {Piston B Area} \times \text {Dist. moved Piston B}}{\text {Piston A Area}}$

$= \dfrac{250 \times 2}{5} \\[1em] = \dfrac{250 \times 2}{5} \\[1em] = 50 \times 2 \\[1em] = 100 \text { cm}$

Hence, the distance moved by piston A is 100 cm.

(e) If a small air bubble enters the liquid, the hydraulic jack will not work efficiently.
This is because air is compressible, whereas hydraulic fluid is incompressible. When force is applied, some part of it will be used to compress the air bubble instead of being fully transmitted through the liquid.
As a result, the pressure transmitted to piston B will decrease, and the jack may lift the car less effectively or work sluggishly.