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 25 cm respectively.
Ans. 1250 kgf

please explain the question nicely​

Answer 1

$\text{We know that pressure on both the}$

$\text{pistol should be same}$

$\therefore Pressure\:on\:smaller\:piston = Pressure\:on\:larger\:piston$

$\dfrac{force\:on\:smaller\:piston}{Area\:of\: smaller\:piston} = \dfrac{force\:on\: larger\:piston}{Area\:of\:larger\:piston}$

$\therefore force\:on\: larger\:piston =$

$\dfrac{force\:on\:smaller\:piston}{Area\:of\: smaller\:piston} × Area\:of\:larger\:piston$

$\Rightarrow \dfrac{50}{2.5 × 2.5 × 3.14} × 12.5 × 12.5 × 3.14 = 1250kgf$

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

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