To solve the problem step by step, we will use the concepts of tension, frequency, and the relationship between the lengths of the sonometer wire in air and water.
### Step 1: Understand the relationship between frequency, tension, and length
The frequency of a vibrating string (sonometer wire) is given by the formula:
\[
f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}
\]
where:
- \( f \) is the frequency,
- \( L \) 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 the initial conditions
Given that the tuning fork frequency is 256 Hz and it resonates with \( \frac{1}{\sqrt{7}} \) times the length of the sonometer wire, we can denote:
- \( L_1 = \frac{1}{\sqrt{7}} L \)
- \( T_1 \) is the tension when the load is in air.
### Step 3: Calculate the tension in air
The tension \( T_1 \) in the wire when the load is in air is equal to the weight of the iron load:
\[
T_1 = mg = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N}
\]
### Step 4: Calculate the specific gravity and tension in water
The specific gravity of iron is given as 8. This means that the density of iron is 8 times the density of water. The effective weight of the iron load when immersed in water is given by:
\[
T_2 = T_1 - \text{buoyant force}
\]
The buoyant force can be calculated using the volume of the iron and the density of water:
\[
\text{Volume of iron} = \frac{mass}{density} = \frac{2 \, \text{kg}}{8 \times 1000 \, \text{kg/m}^3} = \frac{2}{8000} \, \text{m}^3 = 0.00025 \, \text{m}^3
\]
The buoyant force is:
\[
\text{Buoyant force} = \text{Volume} \times \text{Density of water} \times g = 0.00025 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 = 2.45 \, \text{N}
\]
Thus, the tension in water is:
\[
T_2 = T_1 - \text{Buoyant force} = 19.6 \, \text{N} - 2.45 \, \text{N} = 17.15 \, \text{N}
\]
### Step 5: Set up the ratio of tensions and lengths
Since the frequency remains constant, we can set up the ratio:
\[
\frac{\sqrt{T_1}}{\sqrt{T_2}} = \frac{L_1}{L_2}
\]
Substituting the values:
\[
\frac{\sqrt{19.6}}{\sqrt{17.15}} = \frac{\frac{1}{\sqrt{7}} L}{L_2}
\]
### Step 6: Solve for \( L_2 \)
Squaring both sides gives:
\[
\frac{19.6}{17.15} = \frac{\frac{1}{7} L^2}{L_2^2}
\]
Rearranging gives:
\[
L_2^2 = \frac{17.15}{19.6} \times \frac{1}{7} L^2
\]
Taking the square root:
\[
L_2 = L \sqrt{\frac{17.15}{19.6 \times 7}} = 1 \times \sqrt{\frac{17.15}{137.2}} \approx 0.29 \, \text{m}
\]
### Final Answer
The length of the wire in water that will be in resonance with the same tuning fork is approximately \( 0.29 \, \text{m} \).
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