Question

A force of 50 kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, the diameters of the piston being 5 cm and 25 cm respectively. 

(1) 1250 kgf 

(2) 250 kgf 

(3) 1000 kgf 

(4) 1500 kgf

Answer 1

Correct option is (1) 1250 kgf 

Ratio of diameter of smaller piston to bigger piston \(=5: 25\)

\(\therefore\) Ratio of area of smaller piston to bigger piston \(=25: 625\)

Force applied on smaller piston, \(\mathrm{F}_1=50 \mathrm{kgf}\)

Let \(F_2\) be the force on the bigger piston.

By the principle of hydraulic machine,

Pressure on narrow piston = pressure on wider piston

or, \(\frac{F_1}{A_1}=\frac{F_2}{A_2}\)

or, \(\frac{F_1}{F_2}=\frac{A_1}{A_2}\)

or, \(\frac{50}{F_2}=\frac{25}{625}\)

or, \(F_2=50 \times \frac{625}{25}=1250 \mathrm\,{kgf}\).