Question

A nuclear reactor delivers a power of \(10^9\) W, the amount of fuel consumed by the reactor in one hour is

• A. 0.72 g
• B. 0.04 g
• C. 0.96 g
• D. 0.08 g

Answer 1

The Correct Option is B

Using the equation \( P = \frac{mc^2}{t} \), we can solve for mass \( m \): \[ m = \frac{P \cdot t}{c^2} \] 

Given values:

• Power \( P = 10^9 \, \text{W} \)
• Speed of light \( c = 3 \times 10^8 \, \text{m/s} \)
• Time \( t = 60 \times 60 = 3600 \, \text{s} \)

Substituting the given values into the equation:

\[ m = \frac{(3 \times 10^8)^2 \cdot 10^9}{3600 \cdot 10^{16}} \]

Simplifying the equation:

\[ m = \frac{9 \times 10^{16}}{36 \times 10^{11}} = 4 \times 10^{-5} \, \text{kg} \]

Therefore, the mass \( m \) is: 0.04 g

Answer 2

The power equation is given by:

\( P = \frac{E}{t} = \frac{1}{2} mv^2 \)  

Substitute the given values: \( P = 10^9 \), \( t = 3600 \, \text{sec} \), and \( v = 9 \times 10^{16} \, \text{m/s} \):

\( 10^9 = \frac{1}{2} m \times (9 \times 10^{16})^2 \)

Now, solve for the mass \( m \):

\( m = \frac{2 \times 10^9}{(9 \times 10^{16})^2} \)

After calculating:

\( m = 4 \times 10^{-5} \, \text{kg} \)

Converting the mass to grams:

\( m = 4 \times 10^{-5} \times 10^3 = 4 \times 10^{-2} \, \text{g} = 0.04 \, \text{g} \)

Therefore, the mass \( m \) is \( 0.04 \, \text{g} \).