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

Options

A. `(L^(2)+e^(2))/(L^(2))`
B. `(L^(2)-e^(2))/(L^(2))`
C. `(L^(2))/(L^(2)-e^(2))`
D. `(L^(2))/(L^(2)+e^(2))`

Answer

To find the specific gravity of the stone, we can follow these steps:

### Step 1: Understand the relationship between frequency, tension, and length
The frequency of a vibrating string (or 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: Analyze the situation in air and water
When the stone is hung in air, the tension in the wire is \( T_{air} \). When the stone is immersed in water, the tension changes to \( T_{water} \). The frequency remains constant in both cases since it is in unison with the tuning fork.

### Step 3: Set up the ratio of lengths and tensions
Since the frequency does not change, we can set up the following ratio:
\[ \frac{L}{l} = \sqrt{\frac{T_{air}}{T_{water}}} \]

### Step 4: Express tensions in terms of densities
The tension in the wire when the stone is in air can be expressed as:
\[ T_{air} = V \rho g \]
where \( V \) is the volume of the stone, \( \rho \) is the density of the stone, and \( g \) is the acceleration due to gravity.

When the stone is immersed in water, the effective weight of the stone is reduced due to the buoyant force. The tension can be expressed as:
\[ T_{water} = V \rho g - V \rho_{water} g \]
where \( \rho_{water} \) is the density of water.

### Step 5: Substitute the tensions into the ratio
Substituting the expressions for tensions into the ratio gives:
\[ \frac{L}{l} = \sqrt{\frac{V \rho g}{V \rho g - V \rho_{water} g}} \]
This simplifies to:
\[ \frac{L}{l} = \sqrt{\frac{\rho}{\rho - \rho_{water}}} \]

### Step 6: Square both sides and rearrange
Squaring both sides results in:
\[ \left(\frac{L}{l}\right)^2 = \frac{\rho}{\rho - \rho_{water}} \]

### Step 7: Solve for the density of the stone
Rearranging gives:
\[ \rho = \left(\frac{L}{l}\right)^2 (\rho - \rho_{water}) \]
This can be further rearranged to find the specific gravity (which is the ratio of the density of the stone to the density of water):
\[ \rho_{specific} = \frac{\rho}{\rho_{water}} = \frac{L^2}{L^2 - l^2} \]

### Final Answer
Thus, the specific gravity of the material of the stone is:
\[ \text{Specific Gravity} = \frac{L^2}{L^2 - l^2} \]