The fundamental frequency of a sonometre wire is n . If its radius is doubled and its tension becomes half, the material of the wire remains same, the new fundamental frequency will be

Answer

To solve the problem, we need to analyze how the fundamental frequency of a sonometer wire changes when its radius is doubled and its tension is halved.

### Step-by-Step Solution:

1. **Understand the Formula for Fundamental Frequency**:
The fundamental frequency \( n \) of a sonometer wire is given by the formula:
\[ n = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire, \( L \) is the length of the wire, and \( \mu \) is the mass per unit length.

2. **Express Mass per Unit Length**:
The mass per unit length \( \mu \) can be expressed in terms of the radius \( r \) and the density \( \rho \) of the material:
\[ \mu = \frac{m}{L} = \frac{\text{Volume} \times \text{Density}}{L} = \frac{\pi r^2 L \rho}{L} = \pi r^2 \rho \]

3. **Substituting \( \mu \) into the Frequency Formula**:
Substitute \( \mu \) back into the frequency formula:
\[ n = \frac{1}{2L} \sqrt{\frac{T}{\pi r^2 \rho}} \]

4. **Determine New Conditions**:
- New radius \( r' = 2r \) (radius is doubled)
- New tension \( T' = \frac{T}{2} \) (tension is halved)

5. **Calculate New Mass per Unit Length**:
Substitute the new radius into the mass per unit length:
\[ \mu' = \pi (r')^2 \rho = \pi (2r)^2 \rho = 4\pi r^2 \rho \]

6. **Substituting New Values into the Frequency Formula**:
The new fundamental frequency \( n' \) can be expressed as:
\[ n' = \frac{1}{2L} \sqrt{\frac{T'}{\mu'}} = \frac{1}{2L} \sqrt{\frac{\frac{T}{2}}{4\pi r^2 \rho}} \]

7. **Simplifying the Expression**:
Simplifying the expression for \( n' \):
\[ n' = \frac{1}{2L} \sqrt{\frac{T}{2 \cdot 4\pi r^2 \rho}} = \frac{1}{2L} \sqrt{\frac{T}{8\pi r^2 \rho}} = \frac{1}{2} \cdot \frac{1}{2L} \sqrt{\frac{T}{\pi r^2 \rho}} \cdot \frac{1}{\sqrt{2}} \]
\[ n' = \frac{n}{2\sqrt{2}} \]

8. **Final Result**:
Thus, the new fundamental frequency \( n' \) is:
\[ n' = \frac{n}{2\sqrt{2}} \]

### Conclusion:
The new fundamental frequency of the sonometer wire, when the radius is doubled and the tension is halved, is \( \frac{n}{2\sqrt{2}} \).