In a hydraulic press , a force of 40N is applied on the effort piston of area 0.4m2. If the force exerted on the load piston is 400N, the area of the large piston is
-
A.
4m2 -
B.
8m2 -
C.
2m2 -
D.
1m2
Correct Answer: Option A
Explanation
Pressure developed in the effort piston = F/a
∴F/a = 40/0.4 = 400/4 = 100Nm2
this pressure is transmitted to the larger piston of area A; => P = F/A
∴ 100 = 400/A
=> A = 400/100 = 4m2