Question

In a hydraulic machine, the two pistons are of the area of the cross section in the
ratio 1:4. What force is needed in the narrow piston to overcome the force of
200 N?​

Answer 1

We know that, Pressure=

Area

Force

For Larger Piston, P

l

=

625

1250

=2Ncm

−2

So, Pressure on the smaller piston should be equal to 2Ncm

−2

So, Force on smaller piston X=Pressure×Area=2×5=10N

∴

2

X

=

2÷10 =5N