The correct answer is (B) : 50
Total gravitational PE of water per second
\(=\frac{mgh}{T}\)
\(=\frac{9×10^4×10×40}{3600}=10^4\) J/sec
50% of this energy can be converted into electrical energy so total electrical energy
\(=\frac{10^4}{2}=5000 W\)
So total bulbs lit can be
\(=\frac{5000 W}{100 W}\)
= 50 bulbs
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
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