A) \[2\times {{10}^{-6}}g/hr\]
B) \[9\times {{10}^{12}}g/hr\]
C) \[8\times {{10}^{-9}}g/hr\]
D) \[2\times {{10}^{-9}}g/hr\]
Correct Answer: D
Solution :
Having efficiency of 20% by the reactor, the fuel consumed by the reactor \[E=\frac{100\times 10}{20}=50\,J/s\] Again by mass energy equivalence relation, \[E=m{{c}^{2}}\] or \[m=\frac{E}{{{c}^{2}}}\] \[=\frac{50}{{{(3\times {{10}^{8}})}^{2}}}kg/s=\frac{50\times 3600\times {{10}^{3}}}{{{(3\times {{10}^{8}})}^{2}}}g/hr\] \[=2\times {{10}^{-9}}\,g/hr\]
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