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
Verified Correct Answer: A.
Physical Principles & Numerical Solution
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
Source: Myschool Past Questions Archive (Record #12893)