A rod of length 20 cm made of metal A expands by 0.075 when its temperature is raised from `0^(@)`C to `100^(@)C`. Another rod of a different metal B having the same length expands by 0.045 cm for the same change in temperature. A third rod of the same length is composed of two parts, one of metal A and other of metal B. This rod expands by 0.065 cm for the same change in temperature. The portion made of metal A has the length :

Options

A. 13.33 cm
B. 10 cm
C. 15 cm
D. 18 cm

Explanation

To solve the problem step by step, we will use the concept of linear expansion of materials. The formula for linear expansion is given by:

\[ \Delta L = \alpha L_0 \Delta T \]

where:
- \(\Delta L\) = change in length
- \(\alpha\) = coefficient of linear expansion
- \(L_0\) = original length
- \(\Delta T\) = change in temperature

### Step 1: Define the Variables
Let:
- \(L_A\) = length of the portion made of metal A
- \(L_B\) = length of the portion made of metal B

Given that the total length of the rod is 20 cm, we have:
\[ L_A + L_B = 20 \quad \text{(1)} \]

### Step 2: Write the Expansion Equations
For metal A, the change in length when heated from \(0^\circ C\) to \(100^\circ C\) is:
\[ \Delta L_A = 0.075 \text{ cm} \]

For metal B, the change in length is:
\[ \Delta L_B = 0.045 \text{ cm} \]

Using the linear expansion formula, we can express the expansions in terms of the lengths:
\[ \Delta L_A = \alpha_A L_A \Delta T \quad \text{(2)} \]
\[ \Delta L_B = \alpha_B L_B \Delta T \quad \text{(3)} \]

### Step 3: Substitute the Values
Since \(\Delta T = 100^\circ C\), we can write:
\[ 0.075 = \alpha_A L_A \cdot 100 \quad \Rightarrow \quad \alpha_A L_A = \frac{0.075}{100} = 0.00075 \quad \text{(4)} \]
\[ 0.045 = \alpha_B L_B \cdot 100 \quad \Rightarrow \quad \alpha_B L_B = \frac{0.045}{100} = 0.00045 \quad \text{(5)} \]

### Step 4: Write the Total Expansion Equation
The total expansion of the combined rod is given as:
\[ \Delta L_A + \Delta L_B = 0.065 \text{ cm} \]
Using the expansions from equations (2) and (3):
\[ 0.075 + 0.045 = 0.065 \quad \Rightarrow \quad \alpha_A L_A + \alpha_B L_B = \frac{0.065}{100} = 0.00065 \quad \text{(6)} \]

### Step 5: Substitute \(L_B\) from Equation (1)
From equation (1), we can express \(L_B\) in terms of \(L_A\):
\[ L_B = 20 - L_A \]
Substituting this into equation (6):
\[ 0.075 L_A + 0.045(20 - L_A) = 0.00065 \]

### Step 6: Simplify and Solve for \(L_A\)
Expanding and simplifying:
\[ 0.075 L_A + 0.9 - 0.045 L_A = 0.00065 \]
\[ (0.075 - 0.045) L_A + 0.9 = 0.00065 \]
\[ 0.03 L_A = 0.00065 - 0.9 \]
\[ 0.03 L_A = -0.89935 \]
\[ L_A = \frac{-0.89935}{0.03} \approx 13.33 \text{ cm} \]

### Conclusion
The length of the portion made of metal A is approximately \(13.33\) cm.

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