A load of 300N is to be lifted by a machine with a velocity ratio of 2 and an efficiency of 60%. What effort will be applied to lift the load?
A
100N
B
150N
C
250N
D
400N
Verified Correct Answer: C.
Step-by-Step Mathematical Solution
Efficiency = \(\frac{\text{ M.A}}{\text{V.R}}\) x 100
M.A = \(\frac{\text{LOAD}}{\text{EFFORT}}\)
V.R = 2, Eff. = 60% = 0.6, Load = 300N
0.6 = \(\frac{\text{M.A}}{2}\)
M.A = 1.2
1.2 = \(\frac{300}{\text{E}}\)
Effort = \(\frac{300}{1.2}\) = 250N
Source: WAEC / JAMB / NECO Classroom Archive (Record #70423)