Q. In nuclear fusion, One gram hydrogen is converted into 0.993gm. If the efficiency of the generator be 5%, then the energy obtained in KWH is
(a) 8.75 ×10³
(b) 4.75 × 10³
(c) 5.75 × 10³
(d) 3.73 × 10³
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Ans: (a)
Sol: Δm = 1−0.993 = 0.007g =7×10-6 kg
E = Δmc2
= (7×10-6)(3×108)2
= 63 × 10-10 J
Energy required = (5/100)×63 × 10-10
= 63 × 5× 10-8
Since , 1KWH = 3.6× 106 J
Hence, Energy obtained = (63 × 5× 10-8)/(3.6× 106 )
= 8.75 ×10³ KWH