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Question
An electric iron of 1100 W is operated for 2 hrs daily. What will be the electrical consumption expenses for that in the month of April? (The electric company charges Rs 5 per unit of energy).
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Solution
Electric power required for working of iron, P = 1100 W
Duration for which the iron is operated daily = 2 h = \[2 \times 60 \times 60 = 7200 s \]
Electric energy consumed by iron in 7200 s is
or, \[1 J = \frac{1}{3 . 6 \times {10}^6} \] unit
Thus,
\[237600000 J = \frac{237600000}{3 . 6 \times 1000000} = 66\] units
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