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(B) The power $P$ generated is given by the rate of energy release,$P = \frac{E}{t}$.Using Einstein's mass-energy equivalence,$E = mc^2$,where $m$ is

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$9 	imes 10^7$

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(B) The power $P$ generated is given by the rate of energy release,$P = \frac{E}{t}$.Using Einstein's mass-energy equivalence,$E = mc^2$,where $m$ is the mass consumed per unit time.Given: $m/t = 1 \, mg/s = 1 	imes 10^{-6} \, kg/s$ and $c = 3 	imes 10^8 \, m/s$.Substituting the values:$P = (m/t) 	imes c^2$$P = (1 	imes 10^{-6} \, kg/s) 	imes (3 	imes 10^8 \, m/s)^2$$P = 1 	imes 10^{-6} 	imes 9 	imes 10^{16} \, W$$P = 9 	imes 10^{10} \, W$Since $1 \, kW = 10^3 \, W$,we convert the power to kilowatts:$P = \frac{9 	imes 10^{10}}{10^3} \, kW = 9 	imes 10^7 \, kW$.

In a nuclear reactor,the fuel is consumed at the rate of $1 \, mg/s$. The power generated in kilowatt is

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