A thermodynamics cycle takes in heat energy at a high temperature and rejects energy at a lower temperature. If the amount of energy rejected at the low temperature is 3 times the amount of work done by the cycle, the efficiency of the cycle is:

0.25

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0.33

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0.67

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0.9

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Solution

The correct option is C $0.25$
Let heat taken be $Q_{1}$ , heat rejected be $Q_{2}$ , and work done be W
Then $Q_{2} = 3 W$
Also, $Q_{1} = Q_{2} + W$
$\Rightarrow Q_{1} = 4 W$
Efficiency of cycle=work done/heat taken= $\frac{W}{4 W} = \frac{1}{4} = 0.25$

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