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

Options

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

Explanation

To solve the problem, we will use the concept of Young's modulus and the relationship between force, extension, and density. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have a steel wire that expands by \( L_a \) when a weight is suspended from it in air. When the same weight is immersed in water, the extension reduces to \( L_w \). We need to find the relative density (specific gravity) of the weight.

### Step 2: Apply Young's Modulus in Air
Young's modulus \( Y \) is defined as:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
Where:
- \( F \) is the force applied (weight of the object)
- \( L \) is the original length of the wire
- \( A \) is the cross-sectional area of the wire
- \( \Delta L \) is the extension (change in length)

In air, the force \( F \) is equal to the weight of the object, which can be expressed as:
\[ F = V \cdot \rho \cdot g \]
Where:
- \( V \) is the volume of the weight
- \( \rho \) is the density of the weight
- \( g \) is the acceleration due to gravity

Thus, we can write:
\[ Y = \frac{V \cdot \rho \cdot g \cdot L}{A \cdot L_a} \]

### Step 3: Apply Young's Modulus in Water
When the weight is immersed in water, the effective force acting on the wire is reduced due to buoyancy. The buoyant force is equal to the weight of the water displaced, which is:
\[ F_{buoyant} = V \cdot \rho_{water} \cdot g \]
Where \( \rho_{water} \) is the density of water (approximately 1 g/cm³).

The effective force when the weight is submerged becomes:
\[ F_{effective} = V \cdot \rho \cdot g - V \cdot \rho_{water} \cdot g \]

Using Young's modulus again, we have:
\[ Y = \frac{(V \cdot \rho \cdot g - V \cdot \rho_{water} \cdot g) \cdot L}{A \cdot L_w} \]

### Step 4: Set the Two Equations Equal
Since Young's modulus remains constant, we can set the two equations equal to each other:
\[ \frac{V \cdot \rho \cdot g \cdot L}{A \cdot L_a} = \frac{(V \cdot \rho \cdot g - V \cdot \rho_{water} \cdot g) \cdot L}{A \cdot L_w} \]

### Step 5: Simplify the Equation
By canceling out common terms (like \( V \), \( g \), and \( L \)), we get:
\[ \frac{\rho}{L_a} = \frac{\rho - \rho_{water}}{L_w} \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ \rho \cdot L_w = (\rho - \rho_{water}) \cdot L_a \]
Expanding this:
\[ \rho \cdot L_w = \rho \cdot L_a - \rho_{water} \cdot L_a \]

### Step 7: Solve for Relative Density
Rearranging to isolate \( \rho \):
\[ \rho \cdot (L_a - L_w) = \rho_{water} \cdot L_a \]
Thus,
\[ \rho = \frac{\rho_{water} \cdot L_a}{L_a - L_w} \]

### Step 8: Calculate Relative Density
Since the relative density (specific gravity) is the ratio of the density of the material to the density of water, we can express it as:
\[ \text{Relative Density} = \frac{\rho}{\rho_{water}} = \frac{L_a}{L_a - L_w} \]

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