In the process of nuclear fission of `1 g` uranium, the mass lost is `0.92 mg`. The efficiency of power house run by the fission reactor is `10%`.To obtain `400` megawatt power from the power house, how much uranium will be required per hour? `(c=3xx10^(8) m s^(-1))`.

Explanation

To solve the problem step by step, we will follow the outlined process to determine how much uranium is needed per hour to generate 400 megawatts of power with a 10% efficiency.

### Step 1: Calculate the actual power needed
Given that the efficiency of the power house is 10%, to produce 400 megawatts (MW) of output power, we need to find the input power required.

\[ \text{Input Power} = \frac{\text{Output Power}}{\text{Efficiency}} = \frac{400 \text{ MW}}{0.10} = 4000 \text{ MW} \]
...