Question

In a typical hydraulic press, a force of 20 N is exerted on small piston of area 0.050 m2. Force exerted by piston on load if it has an area of 0.50 m2 will be
a. 200N
b. 100N
c. 50N
d. 10N​

Answer 1

$\huge{\underline{\tt{\blue{Answer-}}}}$

a. 200N

Explanation:

        F1 / A1 = F2 / A2

        20 / 0.05 = F2 / 0.5

     

          F2 = 200 N

Answer 2

Answer:

F1=20N

A1=0.050m2

A2=0.50m2

F2=?

According to the principle of hydraulic machine,

F1/A1=F2/A2

20/0.050=F2N/0.50

F2=20×0.50/0.050

=20×50/5

=20×10

=200N

HOPE YOU HAVE GOT SOME HELP FROM MY SOLUTION