Question

How much $U^{235}$ is consumed in a day in an atomic power house operating at $400 \, MW, \, $ provided the whole of the mass of $U^{235}$ is converted into energy?

• A. 0.384g
• B. 0.386g
• C. 0.382g
• D. 0.381g

Answer 1

Answer: (A)

Here, rate of production of energy at the atomic power house,
$P=400 \, MW=400 \, \times 10^{6}J \, s^{- 1}$
Therefore, total energy produced in a day i.e., $24 \, \times 60\times 60s$ ,
$E=P \, \times 24 \, \times 60 \, \times 60=400 \, \times 10^{6}\times 24 \, \times 60\times 60$
$=3.456 \, \times 10^{13}J$
If mass of $U^{235}$ consumed per day in m (in kg) so as to produce the required amount of energy, then
$E=mc^{2}$
Or $3.456 \, \times 10^{13}=mc^{2}$
Or $m=\frac{3.456 \times 10^{13}}{c^2}=\frac{3.456 \times 10^{13}}{\left(3 \times 10^8\right)^2}$
$=0.384 \, \times 10^{- 3}kg$
$=0.384 \, g$