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प्रश्न
The mechanical advantage of a machine is 5 and its efficiency is 80%. It is used to lift a load of 200 kgf to a height of 20 m. Calculate:
(i) The effort required, and
(ii) The work done on the machine (g = 10 ms−2).
उत्तर
Given: M.A. = 5, Efficiency = 80% = 0.8
Load L = 200 kgf = 200 × 10 N = 2,000 N, dL = 20 m.
(i) M.A. =`"L"/"E"`
∴ Effort, E = `"L"/"M.A."=200/5` kgf
= 40 kgf.
(ii) Work output = Load × Displacement of load
= 2,000 N × 20 m
= 40,000 J
Efficiency = `"Work output"/"Work input"`
∴ Work input = `"Work output"/"Efficiency"=40000/0.8`
= 50,000 J.
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