A `20kg` load is suspended from the lower end of a wire `10cm` long and `1mm^(2)` in cross-sectional area. The upper half of the wire is made of iron and the lower half with aluminium. The total elongation in the wire is
`(Y_("iron") = 20xx10^(10) N//m^(2), Y_(Al) = 7xx10^(10) N//m^(2))`

Options

A. `18.9xx10^(-3)m`
B. `17.8xx10^(-3)m`
C. `1.78xx10^(-3)m`
D. `1.89xx10^(-4)m`

Explanation

To solve the problem of finding the total elongation of a composite wire made of iron and aluminum under a load, we can follow these steps:

### Step 1: Identify the Given Data
- Load (mass) = 20 kg
- Length of the wire (L) = 10 cm = 0.1 m
- Cross-sectional area (A) = 1 mm² = 1 × 10⁻⁶ m²
- Young's modulus for iron (Y_iron) = 2 × 10¹¹ N/m²
- Young's modulus for aluminum (Y_al) = 7 × 10¹⁰ N/m²

### Step 2: Calculate the Force (Weight) Acting on the Wire
The force (F) due to the load can be calculated using the formula:
\[ F = m \cdot g \]
where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).

Calculating the force:
\[ F = 20 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 196.2 \, \text{N} \]

### Step 3: Determine the Length of Each Section of the Wire
Since the wire is 10 cm long and made of two halves:
- Length of the iron section (L_iron) = 0.1 m / 2 = 0.05 m
- Length of the aluminum section (L_al) = 0.1 m / 2 = 0.05 m

### Step 4: Calculate the Elongation of Each Section
Using the formula for elongation:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

#### For the Iron Section:
\[ \Delta L_1 = \frac{F \cdot L_{iron}}{A \cdot Y_{iron}} \]
\[ \Delta L_1 = \frac{196.2 \, \text{N} \cdot 0.05 \, \text{m}}{1 \times 10^{-6} \, \text{m}^2 \cdot 2 \times 10^{11} \, \text{N/m}^2} \]
\[ \Delta L_1 = \frac{9.81 \times 10^{-3}}{2 \times 10^{5}} \]
\[ \Delta L_1 = 4.905 \times 10^{-8} \, \text{m} \]

#### For the Aluminum Section:
\[ \Delta L_2 = \frac{F \cdot L_{al}}{A \cdot Y_{al}} \]
\[ \Delta L_2 = \frac{196.2 \, \text{N} \cdot 0.05 \, \text{m}}{1 \times 10^{-6} \, \text{m}^2 \cdot 7 \times 10^{10} \, \text{N/m}^2} \]
\[ \Delta L_2 = \frac{9.81 \times 10^{-3}}{7 \times 10^{4}} \]
\[ \Delta L_2 = 1.40143 \times 10^{-7} \, \text{m} \]

### Step 5: Calculate the Total Elongation
The total elongation (\( \Delta L \)) of the wire is the sum of the elongations of the two sections:
\[ \Delta L = \Delta L_1 + \Delta L_2 \]
\[ \Delta L = 4.905 \times 10^{-8} + 1.40143 \times 10^{-7} \]
\[ \Delta L = 1.89193 \times 10^{-7} \, \text{m} \]

### Final Answer
The total elongation of the wire is approximately:
\[ \Delta L \approx 1.89 \times 10^{-7} \, \text{m} \]

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