Pressure in Fluids and Atmospheric Pressure — Question 14

As we know, by the principle of hydraulic machine

Pressure on piston A = pressure on piston B

Hence,

$\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \\[0.5em]$

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,

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

Hence, force applied = 25 N