What force is applied on a piston of area of cross section 2 cm ² to obtain a force 150 N on the piston of area of cross section 12 cm ² in a hydraulic machine ?

As we know, by the principle of hydraulic machine

Pressure on piston A = pressure on piston B

Hence,

$\dfrac{F${1}}{A{1}} = \dfrac{F{2}}{A{2}} \\[0.5em]

Given,

A₁ = 2 cm ²

Converting cm ² into m ²

100 cm = 1 m

So, 100 cm x 100 cm = 1 m ²

Hence, 2 cm ² = $\dfrac{1 \times 2}{10000}$

Therefore, A₁ = 2 x 10^-4 m ²

A₂ = 12 cm ²

Hence, 12 cm ² = $\dfrac{1 \times 12}{10000}$

Therefore, A₂ = 12 x 10^-4 m ²

F₂ = 150 N

Substituting the values in the formula above we get,

$\dfrac{F$1}{2 \times 10^{-4}} = \dfrac{150}{12 \times 10^{-4}} \\[0.5em] \Rightarrow F1 = \dfrac{150 \times 2 \times 10^{-4}}{12 \times 10^{-4}} \\[0.5em] \Rightarrow F_1 = 25 N \\[0.5em]

Hence, force applied = 25 N