In a single movable pulley system, a load of 125 kgf is lifted by an effort of 75 kgf.

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asked 6 days ago in Class X Science by navnit40 (-1,119 points)

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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answered 6 days ago by priya12 (-10,126 points)

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. x 100

= 5 x 100/3 x 2 = 83%.

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