The areas of the effort and load pistons of a hydraulic press are 0.5m2 and 5m2 respectively. If a force F1 of 100N is applied on the effort piston, the force F2 on the load is
A
10N
B
100N
C
500N
D
1000N
E
5000N
Verified Correct Answer: D.
Physical Principles & Numerical Solution
\(\frac{load}{area}\) = \(\frac{effort}{area}\)
\(\frac{load}{5}\) = \(\frac{100}{0.5}\)
load = 1000N
Source: Myschool Past Questions Archive (Record #28572)