Pressure in Fluids and Atmospheric Pressure — Question 9

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,

$\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \\[0.5em] \dfrac{0.5}{\text A_{1}} = \dfrac{15}{\text A_{2}} \\[0.5em] \dfrac{\text A_{1}}{\text A_{2}} = \dfrac{0.5}{15} \\[0.5em] \Rightarrow \dfrac{\text A_{1}}{\text A_{2}} = \dfrac{5}{150} \\[0.5em] \Rightarrow \dfrac{\text A_{1}}{\text A_{2}} = \dfrac{1}{30} \\[0.5em]$

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