COMEDK · Physics · 29. Nuclear Physics
To get 300 MW electric power for half an hour, how much mass is to be completely converted into energy?
- A \(3 \times 10^{-6} \mathrm{~kg}\)
- B \(6 \times 10^{-6} \mathrm{~kg}\)
- C \(6 \times 10^{-2} \mathrm{~kg}\)
- D \(6 \times 10^{-3} \mathrm{~kg}\)
Answer & Solution
Correct Answer
(B) \(6 \times 10^{-6} \mathrm{~kg}\)
Step-by-step Solution
Detailed explanation
The total energy \(E\) required is given by the product of power \(P\) and time \(t\). \(P = 300 \text{ MW} = 300 \times 10^{6} \text{ W}\) \(t = 30 \text{ minutes} = 30 \times 60 \text{ seconds} = 1800 \text{ s}\)…
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