If the tension and diameter of a sonometer wire of fundamental frequency `n` are doubled and density is halved then its fundamental frequency will become

Options

A. `N/4`
B. `sqrt2n`
C. n
D. `n/sqrt2`

Explanation

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

### Step-by-Step Solution:

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

2. **Expressing Mass per Unit Length**:
The mass per unit length \( \mu \) can be expressed in terms of density \( \rho \) and the cross-sectional area \( A \):
\[ \mu = \frac{\text{mass}}{\text{length}} = \frac{\rho \cdot A}{L} \]
For a wire with a circular cross-section, the area \( A \) can be expressed as:
\[ A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \]
Therefore, the mass per unit length becomes:
\[ \mu = \frac{\rho \cdot \frac{\pi d^2}{4}}{L} = \frac{\rho \pi d^2}{4L} \]

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

4. **Analyzing Changes in Parameters**:
- Let \( T_1 = T \), \( d_1 = d \), and \( \rho_1 = \rho \) be the initial tension, diameter, and density respectively.
- The new tension \( T_2 = 2T \) (doubled).
- The new diameter \( d_2 = 2d \) (doubled).
- The new density \( \rho_2 = \frac{\rho}{2} \) (halved).

5. **Calculating New Frequency**:
The new frequency \( f_2 \) can be expressed as:
\[ f_2 = \frac{2}{\sqrt{\rho_2 \pi d_2^2}} \sqrt{T_2} \]
Substituting the new values:
\[ f_2 = \frac{2}{\sqrt{\left(\frac{\rho}{2}\right) \pi (2d)^2}} \sqrt{2T} \]
Simplifying this:
\[ f_2 = \frac{2}{\sqrt{\left(\frac{\rho}{2}\right) \pi (4d^2)}} \sqrt{2T} = \frac{2}{\sqrt{2 \cdot \frac{\rho}{2} \pi d^2}} \sqrt{2T} = \frac{2 \sqrt{2}}{\sqrt{\rho \pi d^2}} \sqrt{2T} \]
\[ = \frac{2 \cdot 2}{\sqrt{\rho \pi d^2}} \sqrt{T} = 2f_1 \]

6. **Final Result**:
Since the original frequency \( f_1 = n \), we have:
\[ f_2 = 2n \]

### Conclusion:
The new fundamental frequency \( f_2 \) will be \( 2n \).