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Q. A nuclear reactor delivers a power of $10^9$ W. What is the amount of fuel consumed by the reactor in one hour ?

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Solution:

P = $\frac{E}{t} = \frac{mc^2}{t}$ $\therefore $ $m = \frac{Pt}{c^2} = \frac{10^9 \times 60 \times 60}{(3 \times 10^8)^2} = \frac{36 \times 10^{11}}{9 \times 10^{16}} = 4 \times 10^{-5} $ kg = 0.04 g