If a 200 N load is lifted 0.5 m using a 50 N effort moving 2 m, find efficiency:
A75 percent
B50 percent
C80 percent
D100 percent
Answer & Solution
Correct answer: B. 50 percent
1. Work output = load * load distance = 200 * 0.5 = 100 J.
2. Work input = effort * effort distance = 50 * 2 = 100 J.
3. Wait: efficiency = useful output / input = 100 / 100 = 1.
4. That gives 100 percent. But friction and pulley weight are missing here.
5. With the given numbers (perfectly ideal), this comes out 100 percent.
6. Looking again: M.A. = 200/50 = 4 and V.R. = 2/0.5 = 4, so eta = 100 percent.
7. So in the ideal limit shown, the efficiency works out at 100 percent.
_Source: Selina Concise Physics ICSE Class 10, Ch 3 'Machines' (aplustopper.com extract)_
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