A wire when stretched by the weight of a solid,' gives a fundamental )frequaicy `v_(1)` when the solid is immersed in water it gives a frequency v, and when immersed in liquid it gives a frequency of `v_(2)` Calculate the specific gravityofthe solidand that ofthe liquid.

Answer

To solve the problem, we need to calculate the specific gravity of the solid and the liquid based on the frequencies given when the solid is suspended in air, water, and another liquid. Let's break down the solution step by step.

### Step 1: Understand the Problem
We have a wire that vibrates with a fundamental frequency \( v_1 \) when a solid is hung from it. When the solid is immersed in water, the frequency changes to \( v \), and when immersed in another liquid, the frequency changes to \( v_2 \). We need to find the specific gravity of the solid and the liquid.

### Step 2: Establish the Relationship for Frequency
The frequency of a vibrating wire is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire, \( L \) is the length of the wire, and \( \mu \) is the mass per unit length of the wire.

### Step 3: Case 1 - Solid in Air
When the solid is in air, the tension \( T_1 \) is simply the weight of the solid:
\[ T_1 = mg \]
Thus, the frequency \( v_1 \) can be expressed as:
\[ v_1 = \frac{1}{2L} \sqrt{\frac{mg}{\mu}} \]

### Step 4: Case 2 - Solid in Water
When the solid is immersed in water, the buoyant force reduces the effective weight:
\[ T_2 = mg - V \rho_{water} g \]
where \( V \) is the volume of the solid and \( \rho_{water} \) is the density of water. The frequency \( v \) can be expressed as:
\[ v = \frac{1}{2L} \sqrt{\frac{mg - V \rho_{water} g}{\mu}} \]

### Step 5: Case 3 - Solid in Liquid
When the solid is immersed in another liquid, the tension becomes:
\[ T_3 = mg - V \rho_{liquid} g \]
The frequency \( v_2 \) can be expressed as:
\[ v_2 = \frac{1}{2L} \sqrt{\frac{mg - V \rho_{liquid} g}{\mu}} \]

### Step 6: Formulate the Equations
From the above expressions, we can write:
1. For air:
\[ v_1^2 = \frac{mg}{4L^2 \mu} \]
2. For water:
\[ v^2 = \frac{mg - V \rho_{water} g}{4L^2 \mu} \]
3. For liquid:
\[ v_2^2 = \frac{mg - V \rho_{liquid} g}{4L^2 \mu} \]

### Step 7: Relate the Frequencies
From the equations for \( v \) and \( v_1 \):
\[ \frac{v^2}{v_1^2} = \frac{mg - V \rho_{water} g}{mg} \]
This simplifies to:
\[ 1 - \frac{V \rho_{water}}{m} = \frac{v^2}{v_1^2} \]
From this, we can express the specific gravity of the solid \( S_s \) as:
\[ S_s = \frac{v_1^2}{v_1^2 - v^2} \]

### Step 8: Relate the Liquid's Specific Gravity
Using the same method for the liquid:
\[ \frac{v_2^2}{v_1^2} = \frac{mg - V \rho_{liquid} g}{mg} \]
This simplifies to:
\[ 1 - \frac{V \rho_{liquid}}{m} = \frac{v_2^2}{v_1^2} \]
Thus, the specific gravity of the liquid \( S_l \) can be expressed as:
\[ S_l = \frac{v_1^2 - v_2^2}{v_1^2} \]

### Final Expressions
1. Specific Gravity of Solid:
\[ S_s = \frac{v_1^2}{v_1^2 - v^2} \]
2. Specific Gravity of Liquid:
\[ S_l = \frac{v_1^2 - v_2^2}{v_1^2} \]