JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2024
- Hard
- Numerical
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 \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\) )
Pick an option to check your answer.
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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