Q. When _{92}U^{235} undergoes fission. About 0.1% of the original mass is converted into energy. Then the amount of _{92}U^{235} should undergo fission per day in a nuclear reactor so that it provides energy of 200 mega water electric power is

(a) 9.6 × 10^{-2} kg

(b) 4.8 × 10^{-2} kg

(c) 19.2 × 10^{-2} kg

(d) 1.2 × 10^{-2} kg

Ans: (c)

Sol: E = mc^{2}

= 0.001 × (3×10^{8})^{2}

E = 10^{-3}×9×10^{16} = 9×10^{13} J

Total mass of U required/sec

m= M×0.1% = M/1000

Power P = 200×10^{6} J/s

200×10^{6} = (M/1000)c^{2}

M = 200×10^{6} ×1000/c^{2}

M = 0.22×10^{-5}

Mass required per day = 0.22×10^{-5} ×24×60×60 = 19.2 ×10^{-2} kg