Two pistons of a hydraulic machine have diameters of 30.0 cm and 2.5 cm. What is the force exerted on the larger piston when 40 kg wt. is placed on the smaller piston? If the smaller piston moves in through 6 cm, how much does the other piston move out ?

Explanation

For smaller piston, area `A_(1)=pixx(1 cm)^(2)`
For larger piston, area `A_(2)=pixx(10 cm)^(2)`
`therefore` Force exerted on the larger piston,
`F_(2)=(A_(2))/(A_(1))xxF_(1)=(pi(10 cm)^(2))/(pi(1 cm)^(2))xx50xx9.8`
`=100xx50xx9.8=5xx10^(4) N`
This is the force exerted on the large piston. The liquids are considered incompressible. Therefore, volume covered by movement of smaller piston inwards equal to the outward movement of larger piston.
` therefore L_(1)A_(1)=L_(2)A_(2) implies L_(2)=((A_(1))/(A_(2)))L_(1)=(pi(1 cm)^(2))/(pi (10 cm)^(2))xx5cm`
`=(1)/(100)xx5 cm =0.05 cm`
So, the distance moved out by the larger piston is 0.05 cm.