A copper wire of length l and radius 2 mm is suspended from a fixed support vertically and a mass m is hung from its other end. The wire initially at `30^@C` is cooled down to `20^@C` to bring it back to its original length l . The coefficient of linear thermal expansion of the copper is `1.7 xx 10^(-5) "'^@C^(-1)` . If l >> the diameter of wire,then find the value of m/5 in kg close to nearest integer. Take, `g = 10 m//s^2 , Y = 1.1 xx 10^11 N//m^2`

Explanation

To solve the problem, we need to find the mass \( m \) that, when hung from a copper wire, will cause it to contract back to its original length after being cooled from \( 30^\circ C \) to \( 20^\circ C \). We will use the concepts of thermal expansion and Young's modulus to derive the solution.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Length of the wire: \( l \)
- Radius of the wire: \( r = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} \)
- Change in temperature: \( \Delta T = 30^\circ C - 20^\circ C = 10^\circ C \)
- Coefficient of linear thermal expansion for copper: \( \alpha = 1.7 \times 10^{-5} \, ^\circ C^{-1} \)
- Young's modulus for copper: \( Y = 1.1 \times 10^{11} \, \text{N/m}^2 \)
- Acceleration due to gravity: \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Cross-Sectional Area \( A \) of the Wire:**
\[ A = \pi r^2 = \pi (2 \times 10^{-3})^2 = \pi (4 \times 10^{-6}) = 4\pi \times 10^{-6} \, \text{m}^2 \]

3. **Calculate the Thermal Contraction \( \Delta L \):**
The formula for linear contraction due to temperature change is:
\[ \Delta L = L \alpha \Delta T \]

4. **Relate the Extension \( \Delta L \) to the Mass \( m \):**
When mass \( m \) is hung from the wire, it causes an extension \( \Delta L \) due to the weight:
\[ \Delta L = \frac{mgL}{A Y} \]

5. **Set the Two Expressions for \( \Delta L \) Equal:**
Since the wire returns to its original length when cooled, we equate the two expressions:
\[ L \alpha \Delta T = \frac{mgL}{A Y} \]

6. **Cancel \( L \) from Both Sides:**
\[ \alpha \Delta T = \frac{mg}{A Y} \]

7. **Solve for Mass \( m \):**
Rearranging gives:
\[ m = \frac{A Y \alpha \Delta T}{g} \]

8. **Substitute the Known Values:**
\[ m = \frac{(4\pi \times 10^{-6}) (1.1 \times 10^{11}) (1.7 \times 10^{-5}) (10)}{10} \]

9. **Calculate \( m \):**
\[ m = (4\pi \times 10^{-6}) (1.1 \times 10^{11}) (1.7 \times 10^{-5}) \]
\[ m \approx 4 \times 3.14 \times 1.1 \times 1.7 \times 10^{0} \approx 23.487 \, \text{kg} \]

10. **Find \( \frac{m}{5} \):**
\[ \frac{m}{5} \approx \frac{23.487}{5} \approx 4.6974 \]

11. **Round to the Nearest Integer:**
\[ \frac{m}{5} \approx 5 \]

### Final Answer:
The value of \( \frac{m}{5} \) in kg, rounded to the nearest integer, is \( 5 \).