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# A household refrigerator with a coefficient of performance $1.2$ removes heat from the refrigerated space at the rate of $60kJ/\min$. What would be cost of running this fridge for one month ($30days$)(assuming each day it is used for$4$ hours and cost of one electrical unit is$6Rs.$ )A. $180Rs.$B. $300Rs.$C. $480Rs.$D. $600Rs.$

Last updated date: 13th Jun 2024
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Hint: As in the given question we know we are given with energy per minute and coefficient of performance and we know the formula $\dfrac{Q}{W} = 1.2$ to calculate the power and then we will calculate for given days and then total cost.

Complete step by step answer:
As in the question we are given with
coefficient of performance as, $1.2$
And energy per minute as, $60kJ/\min$
And total no. of days as, $30days$
And no. of hours it is used in each day as, $4$
And we are also given cost of each unit as, $6Rs.$
So now substituting values in formula we get,
$\dfrac{Q}{W} = 1.2$, where $Q$is energy per unit minute
$\dfrac{{60}}{W} = 1.2$
$W = 50kJ/\min$
$W = 50000J/\min$
$W = \dfrac{{50000}}{{60}}J/\sec$
$W = \dfrac{5}{6}kW$
So now to calculate total power
${W_t} = \dfrac{5}{6} \times 4 \times 30$
${W_t} = 100kWh$, as we have calculated total so now to calculate total cost
$C = 100 \times 6$
$C = 600Rs$
So the correct option is D.

Note: Remember the concept of power and energy and energy per unit time and for coefficient of performance and basic units of electricity that are $kWh$ which we have calculated above in the form of $W$.