KCET · Physics · Nuclear Physics
A nuclear reactor delivers a power of \(10^9 \mathrm{~W}\), the amount of fuel consumed by the reactor in one hour is
- A \(0.08 \mathrm{~g}\)
- B \(0.72 \mathrm{~g}\)
- C \(0.96 \mathrm{~g}\)
- D \(0.04 \mathrm{~g}\)
Answer & Solution
Correct Answer
(D) \(0.04 \mathrm{~g}\)
Step-by-step Solution
Detailed explanation
Given, power delivered by nuclear reactor,
\[
\begin{aligned}
& P=10^9 \mathrm{~W} \\
& c=3 \times 10^8 \mathrm{~m} / \mathrm{s} \\
& t=1 \mathrm{~h}=60 \times 60 \mathrm{~s}
\end{aligned}
\]
Since,
\[
P=\frac{E}{t}=\frac{m c^2}{t} \quad\left[\because E=m c^2\right]
\]
\[
\begin{aligned}
\Rightarrow \quad m & =\frac{P t}{c^2}=\frac{10^9 \times 60 \times 60}{\left(3 \times 10^8\right)^2} \\
& =\frac{10^9 \times 3600}{9 \times 10^{16}}=4 \times 10^{-5} \mathrm{~kg} \\
& =0.04 \times 10^{-3} \mathrm{~kg}=0.04 \mathrm{~g}
\end{aligned}
\]
\[
\begin{aligned}
& P=10^9 \mathrm{~W} \\
& c=3 \times 10^8 \mathrm{~m} / \mathrm{s} \\
& t=1 \mathrm{~h}=60 \times 60 \mathrm{~s}
\end{aligned}
\]
Since,
\[
P=\frac{E}{t}=\frac{m c^2}{t} \quad\left[\because E=m c^2\right]
\]
\[
\begin{aligned}
\Rightarrow \quad m & =\frac{P t}{c^2}=\frac{10^9 \times 60 \times 60}{\left(3 \times 10^8\right)^2} \\
& =\frac{10^9 \times 3600}{9 \times 10^{16}}=4 \times 10^{-5} \mathrm{~kg} \\
& =0.04 \times 10^{-3} \mathrm{~kg}=0.04 \mathrm{~g}
\end{aligned}
\]
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