In the sonometer experiment , a tuning fork of frequency `256 Hz` is in resonance with `0.4 m` length of the wire when the iron load attached to free end of wire is `2 kg`. If the load is immersed in water , the length of the wire in resonance would be ( specific gravity of iron ` = 8`)

Options

A. `0.37 m`
B. `0.43 m`
C. `0.31 m`
D. `0.2 m`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between tension, length, and frequency
The fundamental frequency \( f_0 \) of a vibrating string is given by the formula:
\[ f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Where:
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the linear mass density of the wire.

Since the frequency \( f_0 \) and the linear mass density \( \mu \) remain constant, we can say that:
\[ f_0 \propto \frac{1}{L} \sqrt{T} \]

### Step 2: Establish the relationship between lengths and tensions
From the above relationship, we can derive that:
\[ \frac{L_2}{L_1} = \sqrt{\frac{T_2}{T_1}} \]
Where:
- \( L_1 \) is the initial length of the wire (0.4 m),
- \( T_1 \) is the initial tension (when the load is 2 kg),
- \( L_2 \) is the new length of the wire when the load is immersed in water,
- \( T_2 \) is the new tension.

### Step 3: Calculate the initial tension \( T_1 \)
The initial tension \( T_1 \) can be calculated using the weight of the load:
\[ T_1 = m \cdot g = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]

### Step 4: Calculate the new tension \( T_2 \) when the load is immersed in water
The buoyant force acting on the load when immersed in water can be calculated using the specific gravity of iron:
- Specific gravity of iron = 8,
- Density of water \( \rho_w = 1000 \, \text{kg/m}^3 \).

The volume \( V \) of the iron can be calculated as:
\[ V = \frac{m}{\rho_{iron}} = \frac{2 \, \text{kg}}{8 \cdot 1000 \, \text{kg/m}^3} = \frac{2}{8000} = 0.00025 \, \text{m}^3 \]

The buoyant force \( F_b \) is given by:
\[ F_b = \rho_w \cdot V \cdot g = 1000 \cdot 0.00025 \cdot 10 = 2.5 \, \text{N} \]

Now, the new tension \( T_2 \) is given by:
\[ T_2 = T_1 - F_b = 20 \, \text{N} - 2.5 \, \text{N} = 17.5 \, \text{N} \]

### Step 5: Calculate the new length \( L_2 \)
Now we can substitute \( T_1 \) and \( T_2 \) into the length ratio equation:
\[ \frac{L_2}{0.4} = \sqrt{\frac{17.5}{20}} \]
Calculating the right side:
\[ \sqrt{\frac{17.5}{20}} = \sqrt{0.875} \approx 0.935 \]
Thus,
\[ L_2 = 0.4 \cdot 0.935 \approx 0.374 \, \text{m} \]

### Step 6: Final result
The new length of the wire in resonance when the load is immersed in water is approximately:
\[ L_2 \approx 0.37 \, \text{m} \]