AIIMS 2017 Physics Nuclear Fission MCQ Question

To determine the amount of U²³⁵ fuel needed to produce 1 MW of power for one year, we first calculate the total energy required: $1 \, \text{MW} = 10^6 \, \text{W}$ means $10^6 \, \text{J/s}$. Over one year (approximately $3.15 \times 10^7 \, \text{s}$), the total energy is $3.15 \times 10^{13} \, \text{J}$. Given that each fission of U²³⁵ releases about $200 \, \text{MeV} = 3.2 \times 10^{-11} \, \text{J}$, the number of fissions required is $3.15 \times 10^{13} \, \text{J} / 3.2 \times 10^{-11} \, \text{J/fission} \approx 9.84 \times 10^{23} \, \text{fissions}$. Since one mole of U²³⁵ (approximately $235 \, \text{g}$) contains Avogadro's number of atoms ($6.022 \times 10^{23}$), the mass of U²³⁵ needed is about $0.01 \, \text{kg}$, making option B correct. Other options are incorrect because they either