A steel wire is suspended vertically from a rigid support. When loaded with a weight in air, it expands by and when the weight is immersed completely in water, the extension is reduced to `L_(w)`. Then relative done of the material of weight is

Options

A. `(L_(a))/(L_(a) - L_(w))`
B. `(L_(w))/(L_(a))`
C. `(L_(a))/(L_(w))`
D. `(L_(w))/(L_(a) - L_(w))`

Explanation

To solve the problem, we need to analyze the situation of the steel wire under two conditions: when the weight is in air and when it is immersed in water. We will use the principles of elasticity and buoyancy to derive the relative density of the material of the weight.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( L_a \) be the extension of the wire when the weight is in air.
- Let \( L_w \) be the extension of the wire when the weight is immersed in water.
- Let \( V \) be the volume of the weight.
- Let \( \rho \) be the relative density of the material of the weight.
- Let \( g \) be the acceleration due to gravity.
- Let \( A \) be the cross-sectional area of the wire.

2. **Extension in Air**:
- When the weight is in air, the force acting on the wire is equal to the weight of the load:
\[ F = V \cdot \rho \cdot g \]
- According to Hooke's law, the extension \( L_a \) of the wire can be expressed as:
\[ L_a = \frac{F \cdot L}{A \cdot Y} \]
- Substituting for \( F \):
\[ L_a = \frac{(V \cdot \rho \cdot g) \cdot L}{A \cdot Y} \tag{1} \]

3. **Extension in Water**:
- When the weight is immersed in water, the effective weight of the load is reduced due to the upthrust (buoyant force). The buoyant force is equal to the weight of the water displaced, which is given by:
\[ \text{Buoyant Force} = V \cdot \rho_w \cdot g \]
- Here, \( \rho_w \) is the density of water (approximately \( 1 \) g/cm³ or \( 1000 \) kg/m³).
- The effective weight when immersed in water becomes:
\[ F' = V \cdot \rho \cdot g - V \cdot \rho_w \cdot g \]
- The extension \( L_w \) in this case is:
\[ L_w = \frac{F' \cdot L}{A \cdot Y} \]
- Substituting for \( F' \):
\[ L_w = \frac{(V \cdot \rho \cdot g - V \cdot \rho_w \cdot g) \cdot L}{A \cdot Y} \tag{2} \]

4. **Setting Up the Equations**:
- From equations (1) and (2), we can set up the relationship between the two extensions:
\[ \frac{L_a}{L_w} = \frac{(V \cdot \rho \cdot g)}{(V \cdot (\rho - \rho_w) \cdot g)} \]
- Simplifying this gives:
\[ \frac{L_a}{L_w} = \frac{\rho}{\rho - \rho_w} \]

5. **Solving for Relative Density**:
- Rearranging the above equation gives:
\[ \rho = \frac{L_a}{L_a - L_w} \cdot \rho_w \]
- Since we are looking for the relative density, we can express it as:
\[ \rho = \frac{L_a}{L_a - L_w} \]

### Conclusion:
The relative density of the material of the weight is given by:
\[ \text{Relative Density} = \frac{L_a}{L_a - L_w} \]