The energy in MeV is released due to transformation of 1 kg mass completely into energy (c = 3 × 108 m/s)
Answer & explanation
Correct answer: option 4
$E=\Delta m c^2=1 \times\left(3 \times 10^8\right)^2=9 \times 10^{10} \mathrm{~J}$
$\Rightarrow E=\frac{9 \times 10^{16}}{1.6 \times 10^{-19}}=5.625 \times 10^{35} \mathrm{eV}=5.625 \times 10^{29} \mathrm{MeV}$