Question

10¹⁸ fissions per second is required for producing power of 300 MW in a nuclear power station. To increase the power output to 360 MW the additional number of fissions required per second is

• A. 2×10⁸
• B. 5×10¹⁸
• C. 5×10¹⁷
• D. 6×10¹⁷
• E. 2×10¹⁷

Answer 1

Answer: (E)

Nuclear Fission and Power Output 

\(10^{18}\) fissions per second produce 300 MW of power in a nuclear power station. We need to find the additional number of fissions required per second to increase the power output to 360 MW.

Step 1: Power per Fission

First, we determine the power produced per fission:

\(\text{Power per fission} = \frac{\text{Total Power}}{\text{Number of fissions per second}}\)

\(\text{Power per fission} = \frac{300 \text{ MW}}{10^{18} \text{ fissions/s}}\)

\(\text{Power per fission} = 3 \times 10^{-16} \text{ MW/fission}\)

Step 2: Additional Power

The additional power required is:

\(\Delta \text{Power} = 360 \text{ MW} - 300 \text{ MW} = 60 \text{ MW}\)

Step 3: Additional Fissions Required

Now we can calculate the additional number of fissions needed per second to produce the additional power:

\(\Delta \text{Fissions} = \frac{\Delta \text{Power}}{\text{Power per fission}}\)

\(\Delta \text{Fissions} = \frac{60 \text{ MW}}{3 \times 10^{-16} \text{ MW/fission}}\)

\(\Delta \text{Fissions} = 2 \times 10^{17} \text{ fissions/s}\)

Conclusion

The additional number of fissions required per second to increase the power output to 360 MW is \(2 \times 10^{17}\).

Answer 2

Given:
Power at 10¹⁸ fissions/s = 300 MW 
We are to find additional number of fissions per second required for 360 MW.

Step 1: Let number of fissions per second be proportional to power:
300 MW → 10¹⁸ fissions/s
1 MW → \( \frac{10^{18}}{300} \)
So, 360 MW → \( \frac{10^{18}}{300} \times 360 = 1.2 \times 10^{18} \) fissions/s

Step 2: Additional number of fissions required:
\[ 1.2 \times 10^{18} - 1.0 \times 10^{18} = 0.2 \times 10^{18} = 2 \times 10^{17} \]

Final Answer: \( 2 \times 10^{17} \)