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 step by step, we will analyze how the fundamental frequency of a sonometer wire changes when the tension and diameter are doubled, and the density is halved.

### Step 1: Understand the formula for fundamental frequency
The fundamental frequency \( f \) of a vibrating string (or sonometer 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.

### Step 2: Express mass per unit length in terms of density
The mass per unit length \( \mu \) can be expressed as:
\[ \mu = \rho \cdot A \]
where:
- \( \rho \) = density of the material,
- \( A \) = cross-sectional area of the wire.

For a cylindrical wire, the area \( A \) can be expressed in terms of diameter \( d \):
\[ A = \frac{\pi d^2}{4} \]
Thus,
\[ \mu = \rho \cdot \frac{\pi d^2}{4} \]

### Step 3: Substitute \( \mu \) into the frequency formula
Substituting \( \mu \) into the frequency formula gives:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\rho \cdot \frac{\pi d^2}{4}}} \]
This simplifies to:
\[ f = \frac{1}{2L} \sqrt{\frac{4T}{\rho \pi d^2}} \]

### Step 4: Analyze the changes in parameters
According to the problem:
- The tension \( T \) is doubled: \( T_2 = 2T_1 \)
- The diameter \( d \) is doubled: \( d_2 = 2d_1 \)
- The density \( \rho \) is halved: \( \rho_2 = \frac{1}{2} \rho_1 \)

### Step 5: Substitute the new values into the frequency formula
Now we can find the new frequency \( f_2 \):
\[ f_2 = \frac{1}{2L} \sqrt{\frac{4T_2}{\rho_2 \pi d_2^2}} \]
Substituting the changes:
\[ f_2 = \frac{1}{2L} \sqrt{\frac{4(2T_1)}{\frac{1}{2} \rho_1 \pi (2d_1)^2}} \]
This simplifies to:
\[ f_2 = \frac{1}{2L} \sqrt{\frac{8T_1}{\frac{1}{2} \rho_1 \pi (4d_1^2)}} \]
\[ f_2 = \frac{1}{2L} \sqrt{\frac{8T_1 \cdot 2}{\rho_1 \pi (4d_1^2)}} \]
\[ f_2 = \frac{1}{2L} \sqrt{\frac{16T_1}{\rho_1 \pi d_1^2}} \]

### Step 6: Relate \( f_2 \) to \( f_1 \)
From the original frequency \( f_1 \):
\[ f_1 = \frac{1}{2L} \sqrt{\frac{4T_1}{\rho_1 \pi d_1^2}} \]
Now, comparing \( f_2 \) and \( f_1 \):
\[ f_2 = 2f_1 \]

### Conclusion
Thus, the new fundamental frequency \( f_2 \) when the tension and diameter are doubled and the density is halved will be:
\[ f_2 = 2n \]

### Final Answer
The fundamental frequency will become \( 2n \).

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