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(D) Given: Coefficient of performance $(	ext{COP}) = 1.2$,Rate of heat removal $(Q_2) = 60 \ kJ/min = 1000 \ J/s = 1 \ kW$.The relationship between w

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$600$

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(D) Given: Coefficient of performance $(	ext{COP}) = 1.2$,Rate of heat removal $(Q_2) = 60 \ kJ/min = 1000 \ J/s = 1 \ kW$.The relationship between work done $(W)$,heat removed $(Q_2)$,and $COP$ is given by $	ext{COP} = \frac{Q_2}{W}$.Therefore,the power input $(W)$ is $W = \frac{Q_2}{	ext{COP}} = \frac{1 \ kW}{1.2} = \frac{1}{1.2} \ kW$.Total time of operation for one month = $30 	ext{ days} 	imes 4 	ext{ hours/day} = 120 	ext{ hours}$.Total energy consumed in $kWh$ = $	ext{Power} 	imes 	ext{Time} = \left(\frac{1}{1.2} \ kWight) 	imes 120 	ext{ hours} = 100 	ext{ kWh}$ (or $100 	ext{ units}$).Cost of energy = $	ext{Total units} 	imes 	ext{Cost per unit} = 100 	imes 6 = 600 \ Rs$.Thus,the correct option is $D$.

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