Pressure in Fluids and Atmospheric Pressure — Question 12

(a) As we know, by the principle of hydraulic machine

Pressure on the smaller piston = pressure on the larger piston

Hence,

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

Given,

Diameter of the smaller piston (d₁) = 5 cm

Hence, A₁ = 𝜋 $(\dfrac{5}{2})^{2}$ = 6.25 𝜋

Diameter of the larger piston (d₂) = 25 cm

Hence, A₂ = 𝜋 $(\dfrac{25}{2})^{2}$ = 156.25 𝜋

Force applied on the smaller piston (F₁) = 50 kgf

Substituting the values in the formula above we get,

$\dfrac{50}{6.25\text π} = \dfrac{\text F_{2}}{156.25\text π} \\[1 em] \text F_{2} = \dfrac{50 \times 156.25}{6.25} = 1250 \\[0.5em] \text F_{2} = 1250 \text { kgf} \\[0.5em]$

Hence, force exerted on the larger piston = 1250 kgf.