Question

A force of 50kgf id applied to the dmaller plston of a hydraulic machine neglecting friction.Find the force exerted on the large piston,the diameters of the piston being 5cm and 25cm redpectively

Answer 1

for hydraulic lift
force is directly proportional to area
in the pic there is the relation between force and area .
and then area is converted in m^2

Answer 2

Ratio of diameter of smaller piston to bigger piston = 5:25

.•. Ratio of area of smaller piston to bigger piston = 25:625

Force applied on smaller piston, F1 = 50kgf

Let F2 be the force on the bigger piston.

By the principle of hydraulic machine,

Pressure on narrow piston = Pressure on wider piston

$\frac{F1}{A1} = \frac{F2}{A2} \\ \\ \implies \: \frac{F1}{F2} = \frac{A1}{A2} \\ \\ \implies \: \frac{50}{F2} = \frac{25}{625} \\ \\ \implies \: F2 \: = \: \cancel{50} \: \: ²\times \frac{625}{ \cancel{25} } \\ \\ \ = 1250 \: kgf \: \: \: \: \: (ans)$