A wire is suspended vertically from a rigid support. When loaded with a steel weight in air, the wire extends by `16cm`. When the weight is completely immersed in Water, the extension is reduced to `14 cm`. The relative density of the material of the weight is

Options

A. `2g//cm^(3)`
B. `6g//cm^(3)`
C. `8g//cm^(3)`
D. `16 g//cm^(3)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
We have a wire suspended vertically from a rigid support, and a steel weight is attached to it. The wire extends by 16 cm when the weight is in air and by 14 cm when the weight is immersed in water. We need to find the relative density of the material of the weight.

### Step 2: Set up the equations for the two scenarios
1. **In air**:
- The force acting on the wire is the weight of the steel weight, \( F = mg \).
- The extension of the wire is given as \( \Delta L_1 = 16 \, \text{cm} = 0.16 \, \text{m} \).
- Using Young's modulus, we have:
\[ \frac{mg}{\Delta L_1} = \frac{AY}{L} \]
- Rearranging gives us:
\[ mg = \frac{AY}{L} \Delta L_1 \]

2. **In water**:
- The effective force acting on the wire is the weight minus the buoyant force:
\[ F' = mg - F_b \]
- The buoyant force \( F_b \) can be expressed as \( F_b = \rho_{water} \cdot g \cdot V \), where \( V \) is the volume of the steel weight.
- The extension of the wire in this case is \( \Delta L_2 = 14 \, \text{cm} = 0.14 \, \text{m} \).
- Using Young's modulus again, we have:
\[ \frac{F'}{\Delta L_2} = \frac{AY}{L} \]
- Rearranging gives us:
\[ F' = \frac{AY}{L} \Delta L_2 \]

### Step 3: Substitute and simplify
From the first scenario, we have:
\[ mg = \frac{AY}{L} \cdot 0.16 \]
From the second scenario, substituting \( F' \):
\[ mg - F_b = \frac{AY}{L} \cdot 0.14 \]
Substituting \( F_b \):
\[ mg - \rho_{water} \cdot g \cdot V = \frac{AY}{L} \cdot 0.14 \]

### Step 4: Relate the two equations
We can equate the two expressions for \( mg \):
\[ \frac{AY}{L} \cdot 0.16 = \frac{AY}{L} \cdot 0.14 + \rho_{water} \cdot g \cdot V \]
Rearranging gives:
\[ \frac{AY}{L} \cdot (0.16 - 0.14) = \rho_{water} \cdot g \cdot V \]
\[ \frac{AY}{L} \cdot 0.02 = \rho_{water} \cdot g \cdot V \]

### Step 5: Express the volume and density
The volume \( V \) can be expressed in terms of density:
\[ V = \frac{m}{\rho_{steel}} \]
Substituting this into the equation gives:
\[ \frac{AY}{L} \cdot 0.02 = \rho_{water} \cdot g \cdot \frac{m}{\rho_{steel}} \]

### Step 6: Solve for relative density
Now we can express the relative density:
\[ \frac{\rho_{steel}}{\rho_{water}} = \frac{0.02 \cdot \rho_{water} \cdot g}{0.02 \cdot g} \]
This simplifies to:
\[ \rho_{steel} = 8 \cdot \rho_{water} \]
Given that \( \rho_{water} = 1 \, \text{g/cm}^3 \), we find:
\[ \rho_{steel} = 8 \, \text{g/cm}^3 \]

### Step 7: Calculate relative density
The relative density is:
\[ \text{Relative Density} = \frac{\rho_{steel}}{\rho_{water}} = \frac{8}{1} = 8 \]

### Final Answer
The relative density of the material of the weight is **8**.

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