Imagine that a reactor converts all given mass into energy and that it operates at a power level of \(10^9\) watt. The mass of the fuel consumed per hour in the reactor will be : (velocity of light, c is \(3\times10^8\) m/s)
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- 0.96 gm
- 0.8 gm
- \(4\times10^{-2}\) gm
- \(6.6\times10^{-5}\) gm
The Correct Option is C
Solution and Explanation
All mass is converted: \(E = mc^2\), so \(m = \frac{3.6\times10^{12}}{(3\times10^8)^2} = \frac{3.6\times10^{12}}{9\times10^{16}} = 4\times10^{-5}\) kg.
\(m = 4\times10^{-5}\) kg \(= 4\times10^{-2}\) g.
Answer: (3) \(4\times10^{-2}\) gm
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