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Question
In a single movable pulley system, a load of 125 kgf is lifted by an effort of 75 kgf. Find the percentage efficiency of system.
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Solution
V.R. of single movable pulley = No. of supporting segments of string = 2
M.A. of single movable pulley =`"L"/"E"=125/75=5/3`
% efficiency =`"M.A."/"V.R."xx100`
= `(5xx100)/(3xx2)=83%.`
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