Fundamental frequency of sonometer wire is `n`. If the length, tension and diameter of wire are tripled. The new fundamental frequency is

Options

A. `n//sqrt3`
B. `n//2sqrt3`
C. `nsqrt3`
D. `n//3sqrt3`

Explanation

To find the new fundamental frequency of a sonometer wire when its length, tension, and diameter are all tripled, we can follow these steps:

### Step 1: Understand the relationship of frequency with length, tension, and mass per unit length.
The fundamental frequency \( f \) of a vibrating string (or wire) is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the mass per unit length of the wire.

### Step 2: Determine how each parameter changes.
Given that the length \( L \), tension \( T \), and diameter \( d \) of the wire are all tripled:
- New length \( L' = 3L \)
- New tension \( T' = 3T \)

The mass per unit length \( \mu \) depends on the diameter and density of the wire. The mass per unit length is given by:
\[ \mu = \rho \cdot A \]
where \( A \) is the cross-sectional area. For a wire with a circular cross-section, the area \( A \) is given by:
\[ A = \pi r^2 \]
If the diameter is tripled, the radius \( r \) is also tripled:
- New radius \( r' = 3r \)
- New area \( A' = \pi (3r)^2 = 9\pi r^2 \)

Thus, the new mass per unit length becomes:
\[ \mu' = \rho \cdot A' = \rho \cdot 9\pi r^2 = 9\mu \]

### Step 3: Substitute the new values into the frequency formula.
Now we can substitute the new values into the frequency formula:
\[ f' = \frac{1}{2L'} \sqrt{\frac{T'}{\mu'}} \]
Substituting the new values:
\[ f' = \frac{1}{2(3L)} \sqrt{\frac{3T}{9\mu}} \]
This simplifies to:
\[ f' = \frac{1}{6L} \sqrt{\frac{3T}{9\mu}} = \frac{1}{6L} \cdot \frac{1}{3} \sqrt{\frac{T}{\mu}} = \frac{1}{18L} \sqrt{\frac{T}{\mu}} \]

### Step 4: Relate the new frequency to the original frequency.
From the original frequency formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
We can see that:
\[ f' = \frac{1}{18} \cdot 3f = \frac{f}{6} \]

### Conclusion
The new fundamental frequency \( f' \) when the length, tension, and diameter of the wire are tripled is:
\[ f' = \frac{n}{6} \]