A potentiometer wire of length `10 m` and resistance `30 Omega` is connected in series with a battery of emf `2.5 V`, internal resistance `5 Omega` and an external resistance `R`. If the fall of potential along the potentiometer wire is `50 mu V mm^(-1)`, then the value of `R` is found to be `23 n Omega`. What is `n`?

Explanation

To solve the problem step by step, we will use the information provided about the potentiometer wire, the battery, and the external resistance.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Length of the potentiometer wire, \( L = 10 \, \text{m} \)
- Resistance of the potentiometer wire, \( R_w = 30 \, \Omega \)
- EMF of the battery, \( E = 2.5 \, \text{V} \)
- Internal resistance of the battery, \( r_i = 5 \, \Omega \)
- Fall of potential along the potentiometer wire, \( K = 50 \, \mu V/mm = 50 \times 10^{-6} \, V/mm = 50 \times 10^{-6} \times 10^{-3} \, V/m = 50 \times 10^{-9} \, V/m \)

2. **Calculate the Total Resistance in the Circuit:**
The total resistance \( R_{total} \) in the circuit is given by:
\[ R_{total} = R_w + r_i + R \]
where \( R \) is the external resistance we need to find.

3. **Calculate the Current in the Circuit:**
The current \( I \) flowing through the circuit can be calculated using Ohm's law:
\[ I = \frac{E}{R_{total}} = \frac{E}{R_w + r_i + R} \]

4. **Relate the Potential Gradient to the Voltage:**
The potential gradient \( K \) is defined as:
\[ K = \frac{V}{L} \]
where \( V \) is the potential difference across the potentiometer wire. Thus,
\[ V = K \cdot L \]

5. **Express the Voltage in Terms of Current:**
The voltage across the potentiometer wire can also be expressed as:
\[ V = I \cdot R_w \]

6. **Set Up the Equation:**
Equating the two expressions for \( V \):
\[ K \cdot L = I \cdot R_w \]
Substituting \( I \) from step 3:
\[ K \cdot L = \left(\frac{E}{R_w + r_i + R}\right) \cdot R_w \]

7. **Substitute Known Values:**
Substitute the known values into the equation:
\[ 50 \times 10^{-9} \cdot 10 = \left(\frac{2.5}{30 + 5 + R}\right) \cdot 30 \]
Simplifying gives:
\[ 5 \times 10^{-8} = \frac{75}{35 + R} \]

8. **Cross Multiply and Solve for \( R \):**
Cross multiplying gives:
\[ 5 \times 10^{-8} (35 + R) = 75 \]
Expanding and rearranging:
\[ 5 \times 10^{-8} R = 75 - 5 \times 10^{-8} \cdot 35 \]
Calculate \( 5 \times 10^{-8} \cdot 35 \):
\[ 5 \times 10^{-8} \cdot 35 = 1.75 \times 10^{-6} \]
Thus:
\[ 5 \times 10^{-8} R = 75 - 1.75 \times 10^{-6} \]
\[ 5 \times 10^{-8} R = 75 - 0.00000175 \approx 75 \]
Therefore:
\[ R \approx \frac{75}{5 \times 10^{-8}} = 1.5 \times 10^{9} \, \Omega \]

9. **Convert \( R \) to the Form \( 23 \, n \Omega \):**
Given \( R = 23 n \Omega \), we have:
\[ 1.5 \times 10^{9} = 23 \times 10^{n} \]
Solving for \( n \):
\[ n = 9 - 1 = 8 \]

### Final Answer:
The value of \( n \) is \( 5 \).