What should be the ratio of area of cross section of the master cylinder and wheel cylinder of a hydraulic brake so that a force of 15 N can be obtained at each of it's brake shoe by exerting a force of 0.5 N on the pedal ?

As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

Let,

area of cross section of the master cylinder = A₁

area of cross section of the wheel cylinder = A₂

force applied on pedal be F₁ = 0.5 N

force applied on brake shoe F₂ = 15 N

By the principle of hydraulic brakes which works on Pascal's law

Pressure on narrow piston = pressure on broader piston

Hence,

$\dfrac{F${1}}{A{1}} = \dfrac{F{2}}{A{2}} \\[0.5em] \dfrac{0.5}{A{1}} = \dfrac{15}{A{2}} \\[0.5em] \dfrac{A{1}}{A{2}} = \dfrac{0.5}{15} \\[0.5em] \Rightarrow \dfrac{A{1}}{A{2}} = \dfrac{5}{150} \\[0.5em] \Rightarrow \dfrac{A{1}}{A{2}} = \dfrac{1}{30} \\[0.5em]

Hence, the ratio of area of cross section = 1 : 30