A nuclear power supplying electrical power to a villages uses a radioactive material of half life `T` year as the fuel . The amount of fuel at the beginning is such that the total power requirement of the village is `12.5%` of the electrical power available from the plant at that time. If the plant is able to meet the total power needs of the village for a maximum period of `n T` years , then the value of `n ` is

Explanation

To solve the problem step by step, we need to analyze the situation involving the radioactive material and its half-life.

### Step-by-Step Solution:

1. **Understanding the Half-Life**:
The half-life \( T \) of the radioactive material means that after every \( T \) years, the amount of the radioactive material reduces to half of its previous amount.

2. **Initial Amount of Fuel**:
Let's denote the initial amount of fuel as \( A_0 \). At time \( t = 0 \), the power available from the plant is proportional to \( A_0 \).

3. **Power Requirement of the Village**:
The village requires \( 12.5\% \) of the power available from the plant. This means that the power requirement of the village is:
\[ P_{\text{village}} = 0.125 \times P_{\text{plant}} \]
where \( P_{\text{plant}} \) is the power generated by the initial amount of fuel \( A_0 \).

4. **Amount of Fuel Over Time**:
- After \( T \) years, the amount of fuel left is \( \frac{A_0}{2} \).
- After \( 2T \) years, the amount of fuel left is \( \frac{A_0}{4} \).
- After \( 3T \) years, the amount of fuel left is \( \frac{A_0}{8} \).

5. **Power Generation Over Time**:
- Initially, the power generated is proportional to \( A_0 \).
- After \( T \) years, the power generated is proportional to \( \frac{A_0}{2} \).
- After \( 2T \) years, the power generated is proportional to \( \frac{A_0}{4} \).
- After \( 3T \) years, the power generated is proportional to \( \frac{A_0}{8} \).

6. **Meeting Village Power Requirement**:
The power generated after \( 3T \) years is \( \frac{A_0}{8} \), which corresponds to \( 12.5\% \) of the initial power \( A_0 \):
\[ P_{\text{village}} = 0.125 \times P_{\text{plant}} = 0.125 \times A_0 \]
This means that the plant can meet the village's power needs for \( 3T \) years.

7. **Finding \( n \)**:
Since the plant can meet the village's power needs for a maximum period of \( nT \) years, we can equate:
\[ nT = 3T \]
Dividing both sides by \( T \):
\[ n = 3 \]

### Final Answer:
The value of \( n \) is \( 3 \).

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