An iron load of 2 kg is suspended in the air from the free end of a Sonometer wire of length 1 m. A tuning fork of frequency 256 Hz is in resonance with time the length of the sonometer wire. If the load is immersed in water that will be in resonance with the same tuning fork is (specified gravity of iron=8)

Options

A. `sqrt8`
B. `sqrt6`
C. `1/sqrt6`
D. `sqrt(7/8)`

Answer

To solve the problem step by step, we will analyze the situation where an iron load is suspended from a sonometer wire and how its resonance frequency changes when the load is immersed in water.

### Step 1: Understand the Initial Condition
Initially, we have an iron load of mass \( m = 2 \, \text{kg} \) suspended from a sonometer wire of length \( L = 1 \, \text{m} \). The frequency of the tuning fork is \( f = 256 \, \text{Hz} \).

### Step 2: Calculate the Tension in the Wire
The tension \( T \) in the wire due to the weight of the iron load is given by:
\[ T = m \cdot g \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). Thus,
\[ T = 2 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 19.62 \, \text{N} \]

### Step 3: Relate Frequency to Tension and Length
The frequency of the vibrating wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( \mu \) is the mass per unit length of the wire. Since the frequency is constant, we can express the relationship between tension and length as:
\[ f \propto \sqrt{\frac{T}{L}} \]

### Step 4: Condition When the Load is Immersed in Water
When the iron load is immersed in water, the apparent weight of the load is reduced due to the buoyant force. The buoyant force \( F_b \) is given by:
\[ F_b = \rho_{water} \cdot V \cdot g \]
where \( \rho_{water} = 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 \) and \( V \) is the volume of the iron load.

### Step 5: Calculate the Volume of the Iron Load
The density of iron is given by:
\[ \text{Density of Iron} = \text{Specific Gravity} \times \text{Density of Water} = 8 \times 1000 \, \text{kg/m}^3 = 8000 \, \text{kg/m}^3 \]
The volume \( V \) of the iron load can be calculated as:
\[ V = \frac{m}{\text{Density}} = \frac{2 \, \text{kg}}{8000 \, \text{kg/m}^3} = 0.00025 \, \text{m}^3 \]

### Step 6: Calculate the New Tension in the Wire
The new tension \( T' \) when the load is submerged in water is given by:
\[ T' = mg - F_b = mg - \rho_{water} \cdot V \cdot g \]
Substituting the values:
\[ T' = 2 \cdot 9.81 - 1000 \cdot 0.00025 \cdot 9.81 \]
Calculating \( F_b \):
\[ F_b = 1000 \cdot 0.00025 \cdot 9.81 = 2.4525 \, \text{N} \]
Thus,
\[ T' = 19.62 - 2.4525 = 17.1675 \, \text{N} \]

### Step 7: Set Up the New Frequency Condition
Using the same frequency relationship:
\[ \frac{T}{L} = \frac{T'}{L'} \]
We know \( L = 1 \, \text{m} \) and \( T = 19.62 \, \text{N} \), and we want to find \( L' \):
\[ \frac{19.62}{1} = \frac{17.1675}{L'} \]
Solving for \( L' \):
\[ L' = \frac{17.1675}{19.62} \approx 0.874 \, \text{m} \]

### Step 8: Conclusion
The new length of the sonometer wire when the load is immersed in water is approximately \( 0.874 \, \text{m} \) or \( \frac{7}{8} \, \text{m} \).