A load of 500 kg was lifted through a distance of 13 cm. by an effort of 25 kg which moved through a distance of 650 cm. The efficiency of the lifting machine is
A. 50 %
B. 40 %
C. 55 %
D. 30 %
Answer: Option B
Solution (By Examveda Team)
Mechanical advantage (MA) = Load (W)/Effort (P).MA = W/P = 500/25 = 20.
Velocity ratio (VR) = (Distance moved by the effort (Y))/(Distance moved by the load (X)).
VR = Y/X = 650/13 = 50.
Efficiency = MA/VR = 20/50 = 0.4 i.e. 40%.

50 % is the wrong answer
Correct ans is 40% follow see
Mechanical advantage (MA) = Load (W)/Effort (P).
MA = W/P = 500/25 = 20.
Velocity ratio (VR) = (Distance moved by the effort (Y))/(Distance moved by the load (X)).
VR = Y/X = 650/13 = 50.
Efficiency = MA/VR = 20/50 = 0.4 i.e. 40%.
So option B is right
Yes right answer is 40%
This is wrong answer