In a house there are $2$ bulbs each of $50 W$ and $2$ fans each of $50 W$ . If they are used at the rate of $10$ hours a day, the electrical energy consumed in a month of $30$ days is

64 KWH

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54 KWH

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72 KWH

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42 KWH

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Solution

The correct option is D 42 KWH
$\begin{matrix} 2 b u l b s o f 50 W = 2 \times 50 = 100 w e n e r g y c o n s u m e d f o r 100 w f o r 4 f a n s = 100 \times 4 \times 3600 = 140000 j 2 f a n o f 50 w = 2 \times 50 = 100 w e n e r g y c o n s u m e d f o r 100 w f o r 10 h = 100 \times 10 \times 3600 = 3600000 j T o t a l e n e r g y c o n s u m e d i n o n e d a y = 1440000 + 3600000 = 50400000 = \frac{5040000}{3.6 \times 10^{6}} = 1.4 K W h o r u n i t s s o, u n i t s i n 360 d a y s = 1.4 \times 30 = 42 u n i t s h e n c e, o p t i o n D i s t h e c o r r e c t a n s w e r . \end{matrix}$

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