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(D) The coefficient of performance $(COP)$ is defined as $COP = \frac{Q_2}{W}$, where $Q_2$ is the heat removed and $W$ is the work done.Given $Q_2 =

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$600$

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(D) The coefficient of performance $(COP)$ is defined as $COP = \frac{Q_2}{W}$, where $Q_2$ is the heat removed and $W$ is the work done.Given $Q_2 = 60 \ kJ/min = \frac{60 	imes 10^3 \ J}{60 \ s} = 1000 \ J/s = 1000 \ W$.Given $COP = 1.2$, the power input $W$ is $W = \frac{Q_2}{COP} = \frac{1000}{1.2} \ W$.The total energy consumed in one month $(30 \ days)$ with $4 \ hours$ of usage per day is:$E = W 	imes 	ext{time} = \left( \frac{1000}{1.2} ight) 	imes (4 	imes 30 \ hours) = \frac{1000}{1.2} 	imes 120 \ Wh = 100,000 \ Wh = 100 \ kWh$.Since $1 \ kWh = 1 \ 	ext{unit}$, the total units consumed is $100 \ units$.The total cost is $100 \ units 	imes 6 \ Rs./unit = 600 \ Rs.$

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