Pressure in Fluids and Atmospheric Pressure — Question 10

As we know,

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

and by the principle of hydraulic machine

Pressure on narrow piston = pressure on broader piston

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

Given,

Area of narrow piston (A₁) = 5 cm ²

Area of wider piston (A₂) = 625 cm ²

force (F₂) = 1250 N

Substituting the values in the formula above we get,

$\dfrac{\text F_{1}}{5} = \dfrac{1250}{625} \\[0.5em] \text F_{1} = \dfrac{1250 \times 5 }{625} \\[0.5em] \Rightarrow \text F_{1} = 10\ \text N \\[0.5em]$

Hence, force acting on the smaller piston = 10 N

Assumption — There is no friction and no leakage of liquid.