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 step by step, we need to analyze the situation involving the brass wire and the mass hanging from it.

### Step 1: Understand the problem
We have a brass wire that is initially at a temperature of \(40^\circ C\) and has a radius of \(1 \, \text{mm}\). A mass \(M\) is hung from the wire. When the temperature drops to \(20^\circ C\), the wire regains its original length of \(0.2 \, \text{m}\). We need to find the value of \(M\).

### Step 2: Identify the relevant formulas
1. **Change in length due to temperature change**:
\[ \Delta L = \alpha L \Delta T \]
where:
- \(\Delta L\) = change in length
- \(\alpha\) = coefficient of linear expansion
- \(L\) = original length
- \(\Delta T\) = change in temperature

2. **Stress and strain relationship**:
\[ \text{Stress} = \frac{F}{A} = \frac{mg}{A} \]
where:
- \(F\) = force due to the mass \(M\)
- \(A\) = cross-sectional area of the wire

3. **Young's Modulus**:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]

### Step 3: Calculate the change in length due to cooling
Given:
- \(\Delta T = 40^\circ C - 20^\circ C = 20^\circ C\)
- \(\alpha = 10^{-5} \, \text{°C}^{-1}\)
- \(L = 0.2 \, \text{m}\)

Substituting into the formula for change in length:
\[ \Delta L = \alpha L \Delta T = (10^{-5})(0.2)(20) = 4 \times 10^{-5} \, \text{m} \]

### Step 4: Calculate the cross-sectional area of the wire
The radius \(r\) of the wire is \(1 \, \text{mm} = 1 \times 10^{-3} \, \text{m}\).
The area \(A\) is given by:
\[ A = \pi r^2 = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]

### Step 5: Relate stress and strain
From Young's modulus:
\[ Y = \frac{mg/A}{\Delta L/L} \]
Rearranging gives:
\[ mg = Y \cdot A \cdot \frac{\Delta L}{L} \]

### Step 6: Substitute known values
Substituting \(Y = 10^{11} \, \text{N/m}^2\), \(A = \pi \times 10^{-6} \, \text{m}^2\), \(\Delta L = 4 \times 10^{-5} \, \text{m}\), and \(L = 0.2 \, \text{m}\):
\[ mg = (10^{11}) \cdot (\pi \times 10^{-6}) \cdot \frac{4 \times 10^{-5}}{0.2} \]

### Step 7: Calculate \(mg\)
Calculating the right-hand side:
\[ mg = (10^{11}) \cdot (\pi \times 10^{-6}) \cdot (2 \times 10^{-4}) = 2\pi \times 10^{1} \, \text{N} \]
Thus,
\[ M = \frac{2\pi \times 10^{1}}{g} = \frac{2\pi \times 10^{1}}{10} = 2\pi \, \text{kg} \]

### Step 8: Final calculation
Using \(\pi \approx 3.14\):
\[ M \approx 2 \times 3.14 = 6.28 \, \text{kg} \]

### Final Answer:
The value of \(M\) is approximately \(6.28 \, \text{kg}\).