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, if the diameters of the pistons are 5 cm and 25cm respectively.

Answer 1

Answer:

We know that Pressure on both the Piston should be same,

∴Pressure on smaller Piston=Pressure on larger Piston

Area of smaller

Force on smaller

=

Area of larger

Force on larger

∴ Force on Larger=

Area of smaller

Force on smaller

×Area of larger

⇒

2.5×2.5×3.14

50

×12.5×12.5×3.14=1250kgf

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

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