Question

To get 300 MW electric power for half an hour, how much mass is to be completely converted into energy?

• A. \( 6 \times 10^{-2} \, \text{kg} \)
• B. \( 3 \times 10^{-6} \, \text{kg} \)
• C. \( 6 \times 10^{-3} \, \text{kg} \)
• D. \( 6 \times 10^{-6} \, \text{kg} \)

Hint:

Always convert power into total energy using \( E = Pt \) before applying \( E = mc^2 \).

Answer 1

The Correct Option is D

Step 1: Write the relation between energy and mass.
Energy is related to mass by Einstein’s equation:
\[ E = mc^2 \]

Step 2: Write the given power.
\[ P = 300 \, \text{MW} = 300 \times 10^6 \, W \]

Step 3: Convert time into seconds.
\[ t = 30 \, \text{minutes} = 1800 \, \text{s} \]

Step 4: Calculate total energy required.
\[ E = Pt \]
\[ E = 300 \times 10^6 \times 1800 \]
\[ E = 5.4 \times 10^{11} \, J \]

Step 5: Use Einstein’s equation to find mass.
\[ m = \frac{E}{c^2} \]

Step 6: Substitute the value of speed of light.
\[ c = 3 \times 10^8 \, \text{m/s} \]
\[ c^2 = 9 \times 10^{16} \]
\[ m = \frac{5.4 \times 10^{11}}{9 \times 10^{16}} \]

Step 7: Calculate final value.
\[ m = 0.6 \times 10^{-5} \]
\[ m = 6 \times 10^{-6} \, \text{kg} \]
Final Answer:
\[ \boxed{6 \times 10^{-6} \, \text{kg}} \]