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The mass defect in a particular nuclear reaction is $0.3 \mathrm{~g}$. The amount of energy liberated in kilowatt hour is :
(Velocity of light $=3 \times 10^8$ )
Options:
(Velocity of light $=3 \times 10^8$ )
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Verified Answer
The correct answer is:
$7.5 \times 10^6$
$\Delta m=0.3 \mathrm{~g}$
$=0.3 \times 10^{-3} \mathrm{~kg}=3 \times 10^{-4} \mathrm{~kg}$
Energy liberated, $E=\Delta m c^2$
$=3 \times 10^{-4} \times\left(3 \times 10^8\right)^2$
$=3 \times 10^{-4} \times 9 \times 10^{16}$
$=27 \times 10^{12} \mathrm{~J}$
$=\frac{27 \times 10^{12}}{3.6 \times 10^6} \mathrm{kWh}$
$=7.5 \times 10^6 \mathrm{kWh}$
$=0.3 \times 10^{-3} \mathrm{~kg}=3 \times 10^{-4} \mathrm{~kg}$
Energy liberated, $E=\Delta m c^2$
$=3 \times 10^{-4} \times\left(3 \times 10^8\right)^2$
$=3 \times 10^{-4} \times 9 \times 10^{16}$
$=27 \times 10^{12} \mathrm{~J}$
$=\frac{27 \times 10^{12}}{3.6 \times 10^6} \mathrm{kWh}$
$=7.5 \times 10^6 \mathrm{kWh}$
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