Fundamental frequency of a sonometer wire is n. If
the length and diameter of the wire are doubled
keeping the tension same, then the new fundamental
frequency is

Options

A. `(2n)/sqrt2`
B. `n/(2sqrt(2))`
C. `sqrt(2)n`
D. `n/4`

Answer

To solve the problem, we need to analyze how the fundamental frequency of a sonometer wire changes when the length and diameter of the wire are doubled while keeping the tension constant.

### Step-by-Step Solution:

1. **Understanding the Fundamental Frequency**:
The fundamental frequency (f) of a vibrating string or wire is given by the formula:
\[ f = \frac{V}{2L} \]
where \( V \) is the velocity of the wave on the wire and \( L \) is the length of the wire.

2. **Velocity of the Wave**:
The velocity \( V \) is related to the tension \( T \) and the linear mass density \( \mu \) of the wire:
\[ V = \sqrt{\frac{T}{\mu}} \]

3. **Linear Mass Density**:
The linear mass density \( \mu \) is defined as:
\[ \mu = \frac{m}{L} \]
where \( m \) is the mass of the wire. The mass can be expressed in terms of volume and density:
\[ m = \text{Density} \times \text{Volume} = \rho \times A \times L \]
where \( A \) is the cross-sectional area of the wire. For a circular wire, the area \( A \) can be expressed in terms of the diameter \( d \):
\[ A = \frac{\pi d^2}{4} \]

4. **Substituting for Linear Mass Density**:
Thus, we can write:
\[ \mu = \frac{\rho \cdot \frac{\pi d^2}{4} \cdot L}{L} = \frac{\rho \pi d^2}{4} \]

5. **Substituting into the Frequency Formula**:
Now substituting \( \mu \) back into the frequency formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\frac{\rho \pi d^2}{4}}} = \frac{1}{2L} \sqrt{\frac{4T}{\rho \pi d^2}} = \frac{2}{\sqrt{\rho \pi}} \cdot \frac{\sqrt{T}}{L d} \]

6. **Effect of Doubling Length and Diameter**:
If the length \( L \) and diameter \( d \) are both doubled:
- New length \( L' = 2L \)
- New diameter \( d' = 2d \)

Substitute these into the frequency formula:
\[ f' = \frac{2}{\sqrt{\rho \pi}} \cdot \frac{\sqrt{T}}{L' d'} = \frac{2}{\sqrt{\rho \pi}} \cdot \frac{\sqrt{T}}{(2L)(2d)} = \frac{2}{\sqrt{\rho \pi}} \cdot \frac{\sqrt{T}}{4Ld} = \frac{1}{4} \cdot \frac{2}{\sqrt{\rho \pi}} \cdot \frac{\sqrt{T}}{Ld} = \frac{f}{4} \]

7. **Conclusion**:
Therefore, the new fundamental frequency \( f' \) is:
\[ f' = \frac{f}{4} \]
If the original fundamental frequency is \( n \), then the new fundamental frequency is:
\[ n' = \frac{n}{4} \]

### Final Answer:
The new fundamental frequency is \( \frac{n}{4} \).