At `40^(@)C` a brass wire of 1mm radius is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from` 40^(@)C` to `20^(@)C` it regains its original length of 0.2 m the value of M is close to (coefficient of linear expansion and young's modulus of brass are `10^(-5)//.^(@)C` and `10^(11)//N//m^(2)` respectively `g=10ms^(-2))`

Options

A. 9kg
B. 0.5kg
C. 1.5 kg
D. 0.9 kg

Answer

To solve the problem, we need to find the mass \( M \) that is hung from the brass wire. We will use the concepts of thermal expansion and Hooke's law.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a brass wire with a radius of \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \).
- The wire is initially at \( 40^\circ C \) and is cooled to \( 20^\circ C \).
- The original length of the wire is \( L_0 = 0.2 \, \text{m} \).
- The coefficient of linear expansion of brass is \( \alpha = 10^{-5} \, \text{°C}^{-1} \).
- The Young's modulus of brass is \( Y = 10^{11} \, \text{N/m}^2 \).
- The acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \).

2. **Calculate the Change in Temperature**:
\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 20^\circ C - 40^\circ C = -20^\circ C \]

3. **Calculate the Change in Length Due to Thermal Contraction**:
The change in length \( \Delta L \) due to thermal contraction can be calculated using the formula:
\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]
Substituting the values:
\[ \Delta L = 0.2 \, \text{m} \cdot (10^{-5} \, \text{°C}^{-1}) \cdot (-20) = -4 \times 10^{-5} \, \text{m} \]

4. **Using Hooke's Law**:
According to Hooke's law, the extension \( \Delta L \) due to the mass \( M \) hanging from the wire is given by:
\[ \Delta L = \frac{M g L_0}{A Y} \]
where \( A \) is the cross-sectional area of the wire:
\[ A = \pi r^2 = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]

5. **Equating the Two Expressions for \( \Delta L \)**:
Since the wire regains its original length, we can equate the two expressions for \( \Delta L \):
\[ -4 \times 10^{-5} = \frac{M g L_0}{A Y} \]
Rearranging gives:
\[ M = -4 \times 10^{-5} \cdot \frac{A Y}{g L_0} \]

6. **Substituting the Values**:
Now substituting the values into the equation:
\[ M = -4 \times 10^{-5} \cdot \frac{\pi \times 10^{-6} \cdot 10^{11}}{10 \cdot 0.2} \]
Simplifying:
\[ M = -4 \times 10^{-5} \cdot \frac{\pi \times 10^{5}}{2} = -2 \times 10^{-5} \cdot \pi \times 10^{5} = -2\pi \, \text{kg} \]

7. **Final Result**:
The mass \( M \) is approximately \( 2\pi \, \text{kg} \).

### Conclusion:
The value of \( M \) is close to \( 6.28 \, \text{kg} \).