A wire gets elongated by 4 mm when a block of mass 2 kg is suspended from it. Now, the block is immersed in water, the wire contracts by 1 mm. Find the density of block when the density of water is `1000kg//m^(3)`.

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Problem
We have a wire that elongates by 4 mm when a 2 kg block is suspended from it. When the block is immersed in water, the wire contracts by 1 mm. We need to find the density of the block, given that the density of water is 1000 kg/m³.

### Step 2: Define Variables
- Let \( m = 2 \, \text{kg} \) (mass of the block)
- Let \( \Delta L = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \) (elongation of the wire)
- Let \( \Delta L' = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \) (contraction of the wire when immersed in water)
- Let \( \rho_w = 1000 \, \text{kg/m}^3 \) (density of water)
- Let \( \rho_b \) be the density of the block (unknown)

### Step 3: Calculate the Weight of the Block
The weight of the block is given by:
\[ W = m \cdot g \]
Assuming \( g \approx 9.81 \, \text{m/s}^2 \):
\[ W = 2 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 19.62 \, \text{N} \]

### Step 4: Apply Young's Modulus for the First Case
Using the formula for Young's modulus:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{W/A}{\Delta L/L} \]
Rearranging gives:
\[ \Delta L = \frac{W \cdot L}{A \cdot Y} \]
This is our first equation.

### Step 5: Apply Young's Modulus for the Second Case
When the block is immersed in water, the effective weight \( W' \) is given by:
\[ W' = W - F_b \]
where \( F_b \) is the buoyant force:
\[ F_b = \rho_w \cdot V \cdot g \]
The volume \( V \) of the block can be expressed as:
\[ V = \frac{m}{\rho_b} \]
Thus, the buoyant force becomes:
\[ F_b = \rho_w \cdot \frac{m}{\rho_b} \cdot g \]
Now, substituting this into the effective weight:
\[ W' = W - \rho_w \cdot \frac{m}{\rho_b} \cdot g \]

### Step 6: Write the Young's Modulus Equation for the Second Case
Using the same Young's modulus formula for the second case:
\[ Y = \frac{W'}{A} \cdot \frac{L}{\Delta L'} \]
This gives us:
\[ \Delta L' = \frac{W' \cdot L}{A \cdot Y} \]
Substituting \( W' \):
\[ \Delta L' = \frac{\left(W - \rho_w \cdot \frac{m}{\rho_b} \cdot g\right) \cdot L}{A \cdot Y} \]

### Step 7: Set Up the Ratio of Extensions
From the two cases, we can set up the ratio:
\[ \frac{\Delta L}{\Delta L'} = \frac{W}{W - \rho_w \cdot \frac{m}{\rho_b} \cdot g} \]
Substituting the known values:
\[ \frac{4 \times 10^{-3}}{1 \times 10^{-3}} = \frac{19.62}{19.62 - 1000 \cdot \frac{2}{\rho_b} \cdot 9.81} \]
This simplifies to:
\[ 4 = \frac{19.62}{19.62 - \frac{19620}{\rho_b}} \]

### Step 8: Solve for Density of the Block
Cross-multiplying gives:
\[ 4(19.62 - \frac{19620}{\rho_b}) = 19.62 \]
Expanding and rearranging:
\[ 78.48 - \frac{78480}{\rho_b} = 19.62 \]
\[ \frac{78480}{\rho_b} = 78.48 - 19.62 \]
\[ \frac{78480}{\rho_b} = 58.86 \]
Thus,
\[ \rho_b = \frac{78480}{58.86} \approx 1333.33 \, \text{kg/m}^3 \]

### Final Answer
The density of the block is approximately \( 1333.33 \, \text{kg/m}^3 \).