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Applied Mechanics and Graphic Statics
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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

Answer & Solution
Correct Answer: Option B
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%.
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3 Comments
Siddhi Ghuge
Siddhi Ghuge 5 years ago
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
Virender Sharma
Virender Sharma 6 years ago
Yes right answer is 40%
Ashwani Kumar
Ashwani Kumar 6 years ago
This is wrong answer