A wire elongates 2mm when a stone is suspended from it. When the stone is completely immersed in water, the wire contracts by 0.6 mm. Find the density of stone.

Answer

To solve the problem, we will use the concepts of elasticity, specifically the relationship between force, stress, and strain, along with the principle of buoyancy.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- When the stone is suspended from the wire, it causes an elongation of 2 mm. This elongation can be attributed to the weight of the stone acting on the wire.
- When the stone is immersed in water, the wire contracts by 0.6 mm due to the buoyant force acting on the stone.

2. **Calculating the Force on the Wire:**
- Let the density of the stone be \( D \) kg/m³, and the volume of the stone be \( V \) m³.
- The weight of the stone when suspended in air is given by:
\[ F_{\text{air}} = mg = D \cdot V \cdot g \]
- Here, \( g \) is the acceleration due to gravity.

3. **Calculating the Buoyant Force:**
- When the stone is immersed in water, the buoyant force acting on it is equal to the weight of the water displaced by the stone:
\[ F_{\text{buoyant}} = \text{density of water} \cdot V \cdot g = 1000 \cdot V \cdot g \]
- The effective force on the wire when the stone is immersed in water is:
\[ F_{\text{water}} = mg - F_{\text{buoyant}} = D \cdot V \cdot g - 1000 \cdot V \cdot g \]

4. **Relating the Forces to Elongation and Contraction:**
- The elongation of the wire due to the weight of the stone is given as 2 mm (or 0.002 m).
- The contraction of the wire when the stone is immersed is given as 0.6 mm (or 0.0006 m).
- Using the relationship between force, stress, and strain, we can express these elongations in terms of the forces:
\[ \text{For elongation: } \frac{D \cdot V \cdot g}{A} = \frac{Y \cdot 0.002}{L_0} \]
\[ \text{For contraction: } \frac{D \cdot V \cdot g - 1000 \cdot V \cdot g}{A} = \frac{Y \cdot 0.0006}{L_0} \]
- Here, \( A \) is the cross-sectional area of the wire, \( Y \) is Young's modulus, and \( L_0 \) is the original length of the wire.

5. **Setting Up the Equation:**
- From the elongation:
\[ \frac{D \cdot g}{A} = \frac{Y \cdot 0.002}{L_0} \]
- From the contraction:
\[ \frac{(D - 1000) \cdot g}{A} = \frac{Y \cdot 0.0006}{L_0} \]

6. **Dividing the Two Equations:**
- Dividing the elongation equation by the contraction equation:
\[ \frac{D \cdot g}{(D - 1000) \cdot g} = \frac{0.002}{0.0006} \]
- Simplifying gives:
\[ \frac{D}{D - 1000} = \frac{2}{0.6} = \frac{10}{3} \]

7. **Cross-Multiplying and Solving for D:**
- Cross-multiplying:
\[ 3D = 10(D - 1000) \]
- Expanding and rearranging:
\[ 3D = 10D - 10000 \]
\[ 10000 = 10D - 3D \]
\[ 10000 = 7D \]
\[ D = \frac{10000}{7} \approx 1428.57 \text{ kg/m}^3 \]

### Final Answer:
The density of the stone is approximately \( 1428.57 \, \text{kg/m}^3 \).