An object of specific gravity `rho` is hung from a thin steel wire. The fundamental frequency for transverse standing waves in wire is `300 Hz`. The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency in `Hz` is

Explanation

To solve the problem step by step, we will analyze the situation of the object hanging from the steel wire and how its immersion in water affects the fundamental frequency of the wire.

### Step 1: Understand the Initial Conditions
The object has a specific gravity of \( \rho \) and is hanging from a thin steel wire. The fundamental frequency of transverse standing waves in the wire is given as \( f_1 = 300 \, \text{Hz} \).

### Step 2: Determine the Initial Tension in the Wire
The tension \( T \) in the wire when the object is hanging in air can be expressed as:
\[ T = mg \]
where \( m \) is the mass of the object. The mass can be related to the specific gravity:
\[ m = \rho \cdot V \cdot g \]
where \( V \) is the volume of the object. Thus, the tension becomes:
\[ T = \rho \cdot V \cdot g \]

### Step 3: Calculate the Mass per Unit Length of the Wire
The mass per unit length \( \mu \) of the wire is a constant that does not change with the immersion of the object. It remains the same in both scenarios.

### Step 4: Write the Expression for the Fundamental Frequency
The fundamental frequency \( f \) of the wire is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Substituting for \( T \):
\[ f_1 = \frac{1}{2L} \sqrt{\frac{\rho V g}{\mu}} = 300 \, \text{Hz} \]

### Step 5: Analyze the New Conditions When the Object is Immersed
When the object is immersed in water with half of its volume submerged, the upthrust (buoyant force) acting on the object is given by:
\[ \text{Upthrust} = \frac{1}{2} V \cdot \rho_w \cdot g \]
where \( \rho_w \) is the density of water.

### Step 6: Calculate the New Tension in the Wire
The new tension \( T' \) in the wire when the object is half submerged is:
\[ T' = mg - \text{Upthrust} = \rho V g - \frac{1}{2} V \cdot \rho_w \cdot g \]
This simplifies to:
\[ T' = Vg \left( \rho - \frac{1}{2} \rho_w \right) \]

### Step 7: Write the Expression for the New Fundamental Frequency
The new fundamental frequency \( f' \) can be expressed as:
\[ f' = \frac{1}{2L} \sqrt{\frac{T'}{\mu}} \]
Substituting for \( T' \):
\[ f' = \frac{1}{2L} \sqrt{\frac{Vg \left( \rho - \frac{1}{2} \rho_w \right)}{\mu}} \]

### Step 8: Relate the New Frequency to the Original Frequency
We can relate the new frequency \( f' \) to the original frequency \( f_1 \):
\[ \frac{f'}{f_1} = \sqrt{\frac{\rho - \frac{1}{2} \rho_w}{\rho}} \]
Thus, we can write:
\[ f' = f_1 \sqrt{\frac{\rho - \frac{1}{2} \rho_w}{\rho}} \]
Substituting \( f_1 = 300 \, \text{Hz} \):
\[ f' = 300 \sqrt{\frac{\rho - \frac{1}{2} \cdot 1}{\rho}} \]

### Step 9: Final Calculation
Assuming specific gravity \( \rho \) is given in terms of water (where \( \rho_w = 1 \)), we can compute the final frequency.

### Conclusion
The new fundamental frequency when the object is half submerged in water is:
\[ f' = 300 \sqrt{\frac{\rho - 0.5}{\rho}} \]