NB :

If the load on the large piston is not sufficient to keep equilibrium with the small piston , the small piston will move a distance (S1) down & the large piston will move a small distance (S2) upward till equilibrium between the two pistons occurs.

          Px           =             Py

           F/a           =        F/A + rgh

·        since  The liquid is incompressible

     then   :

(I) The volume of the liquid displaced = The volume of the liquid displaced by the small piston by the large piston .

                    aS1              =               AS2

(ii) Work done by small piston = work out by the large piston

                     fS1              =                FS2   

·        The mechanical advantage  (e)

e = A/a = r^2/R^2= F/f = S1/S2

                           work out by large piston  

·        Efficiency =  work in on the small one