A stone in hung in air from a wire which is stretched over a sonometer . The bridges of the sonometer are `40 cm` apart when the wire is in unison with a tuning fork of frequency `256 Hz`. When the stone is completely immersed in water , the length between the bridges is `22 cm` for re - establishing unison . The specific gravity of the material of the stone is

Options

A. `((40)^(2))/((40)^(2)+(22)^(2))`
B. `((40)^(2))/((40)^(2)-(22)^(2))`
C. `((40)^(2)+(22)^(2))/((40)^(2))`
D. `((40)^(2)-(22)^(2))/((40)^(2))`

Answer

To solve the problem, we need to find the specific gravity of the stone using the information provided about the frequencies of the wire in two different scenarios: when the stone is in air and when it is immersed in water.

### Step-by-Step Solution:

1. **Understand the Frequency Formula**:
The frequency of a vibrating 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, and \( \mu \) is the mass per unit length of the wire.

2. **Set Up the Equations**:
- When the stone is in air (Length \( L_1 = 40 \, \text{cm} = 0.4 \, \text{m} \)):
\[ f_1 = 256 \, \text{Hz} = \frac{1}{2 \times 0.4} \sqrt{\frac{T}{\mu}} \quad \text{(Equation 1)} \]
- When the stone is immersed in water (Length \( L_2 = 22 \, \text{cm} = 0.22 \, \text{m} \)):
\[ f_2 = 256 \, \text{Hz} = \frac{1}{2 \times 0.22} \sqrt{\frac{T - \rho V g}{\mu}} \quad \text{(Equation 2)} \]
Here, \( \rho \) is the density of water, \( V \) is the volume of the stone, and \( g \) is the acceleration due to gravity.

3. **Equate the Frequencies**:
From Equation 1:
\[ 256 = \frac{1}{0.8} \sqrt{\frac{T}{\mu}} \implies T = 256 \times 0.8 \sqrt{\mu} \]
From Equation 2:
\[ 256 = \frac{1}{0.44} \sqrt{\frac{T - \rho V g}{\mu}} \implies T - \rho V g = 256 \times 0.44 \sqrt{\mu} \]

4. **Substituting Tension**:
Substitute \( T \) from Equation 1 into Equation 2:
\[ 256 \times 0.8 \sqrt{\mu} - \rho V g = 256 \times 0.44 \sqrt{\mu} \]
Rearranging gives:
\[ \rho V g = (256 \times 0.8 - 256 \times 0.44) \sqrt{\mu} \]
Simplifying:
\[ \rho V g = 256 \times (0.8 - 0.44) \sqrt{\mu} \]
\[ \rho V g = 256 \times 0.36 \sqrt{\mu} \]

5. **Relate Volume and Density**:
The volume \( V \) of the stone can be expressed in terms of its mass \( m \) and density \( d \):
\[ V = \frac{m}{d} \]
Substituting this into the equation gives:
\[ \rho \left(\frac{m}{d}\right) g = 256 \times 0.36 \sqrt{\mu} \]

6. **Specific Gravity Calculation**:
The specific gravity \( SG \) is defined as:
\[ SG = \frac{d}{\rho} \]
Rearranging the equation gives:
\[ d = \frac{256 \times 0.36 \sqrt{\mu} \cdot d}{\rho g} \]
Solving for \( SG \):
\[ SG = \frac{22^2}{40^2} = \frac{484}{1600} = 0.3025 \]

7. **Final Result**:
The specific gravity of the material of the stone is approximately:
\[ SG \approx 0.3025 \]