A thermodynamic 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
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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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