If the efficiency of a machine is 40%, and it does 120 J of work, the input of the work would be:

300 J

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500 J

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600 J

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800 J

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Solution

The correct option is A 300 J
Given:
Efficiency, E = 40% OR 0.4
Output work = 120 J

We know that,
Efficiency of a machine, E = $\frac{O u t p u t w o r k}{I n p u t w o r k}$
$\Rightarrow 0.4$ = $\frac{120 J}{I n p u t w o r k}$
$\Rightarrow Input work$ = $\frac{120 J}{0.4}$
$\Rightarrow Input work$ = $300 J$

Hence, the input work would be 300 J.

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