An aluminum wire and a steel wire of the same length and cross-section are joined end to end. The compoiste wire is hung from a rigid support and a load is suspended from the free end. If the length of the composite wire is `2.7mm` then the increases in the length of wire is (in `mm`)
`(Y_(Al) = 2xx10^(11)Nm^(2), Y_("Steel") = 7xx10^(11) Nm^(-2))`

Options

A. `1.7,1`
B. `1.3,4`
C. `1.5,1.2`
D. `2.1,0.6`

Explanation

To solve the problem, we need to find the increase in length of both the aluminum and steel wires when a load is applied to the composite wire. We will use the Young's modulus and the relationship between the elongations of the two wires.

### Step-by-step Solution:

1. **Understand the Problem:**
We have two wires (aluminum and steel) of the same length and cross-section joined end to end. The total elongation of the composite wire is given as \(2.7 \, \text{mm}\).

2. **Identify Young's Modulus Values:**
- Young's modulus for aluminum, \(Y_{Al} = 2 \times 10^{11} \, \text{N/m}^2\)
- Young's modulus for steel, \(Y_{Steel} = 7 \times 10^{11} \, \text{N/m}^2\)

3. **Use the Relationship Between Elongations:**
The relationship between the elongations of the two wires can be expressed as:
\[ \frac{\Delta L_{Al}}{\Delta L_{Steel}} = \frac{Y_{Steel}}{Y_{Al}} \]
This means that the elongation in aluminum is to the elongation in steel as the Young's modulus of steel is to that of aluminum.

4. **Substitute the Values:**
\[ \frac{\Delta L_{Al}}{\Delta L_{Steel}} = \frac{7 \times 10^{11}}{2 \times 10^{11}} = \frac{7}{2} \]
Let \(\Delta L_{Steel} = x\). Then, \(\Delta L_{Al} = \frac{7}{2} x\).

5. **Set Up the Total Elongation Equation:**
The total elongation of the composite wire is the sum of the elongations of both wires:
\[ \Delta L_{Al} + \Delta L_{Steel} = 2.7 \, \text{mm} \]
Substituting the elongations:
\[ \frac{7}{2} x + x = 2.7 \]

6. **Combine Like Terms:**
\[ \frac{7x + 2x}{2} = 2.7 \]
\[ \frac{9x}{2} = 2.7 \]

7. **Solve for \(x\):**
Multiply both sides by 2:
\[ 9x = 5.4 \]
\[ x = \frac{5.4}{9} = 0.6 \, \text{mm} \]
Thus, \(\Delta L_{Steel} = 0.6 \, \text{mm}\).

8. **Find \(\Delta L_{Al}\):**
Now substitute back to find \(\Delta L_{Al}\):
\[ \Delta L_{Al} = 2.7 - \Delta L_{Steel} = 2.7 - 0.6 = 2.1 \, \text{mm} \]

### Final Results:
- Increase in length of aluminum wire, \(\Delta L_{Al} = 2.1 \, \text{mm}\)
- Increase in length of steel wire, \(\Delta L_{Steel} = 0.6 \, \text{mm}\)