JEE Mains · Physics · STD 12 - 13. Nuclei
The mass defect in a particular reaction is \(0.4 \mathrm{~g}\). The amount of energy liberated is \(\mathrm{n} \times 10^7 \mathrm{kWh}\), where\(\mathrm{n}=\)________. (speed of light \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\) )
- A \(10\)
- B \(1\)
- C \(5\)
- D \(11\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\Delta \mathrm{mc}^2\) \(=0.4 \times 10^{-3} \times\left(3 \times 10^8\right)^2\) \(=3600 \times 10^7 \mathrm{kWs}\) \(=\frac{3600 \times 10^7}{3600} \mathrm{kWh}=1 \times 10^7 \mathrm{kWh}\)
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