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 formula for linear expansion and the information given in the question.

### Step 1: Understand the problem and identify the known values
We have three rods:
1. Rod A (metal A): Length = 20 cm, expands by 0.075 cm when temperature changes from 0°C to 100°C.
2. Rod B (metal B): Length = 20 cm, expands by 0.045 cm for the same temperature change.
3. Composite rod (part A and part B): Length = 20 cm, expands by 0.065 cm for the same temperature change.

### Step 2: Calculate the coefficient of linear expansion for metal A
The formula for linear expansion is:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
Where:
- \(\Delta L\) = change in length
- \(L\) = original length
- \(\alpha\) = coefficient of linear expansion
- \(\Delta T\) = change in temperature

For metal A:
\[ 0.075 = 20 \cdot \alpha_A \cdot 100 \]
Rearranging gives:
\[ \alpha_A = \frac{0.075}{2000} = 3.75 \times 10^{-5} \, \text{°C}^{-1} \]

### Step 3: Calculate the coefficient of linear expansion for metal B
For metal B:
\[ 0.045 = 20 \cdot \alpha_B \cdot 100 \]
Rearranging gives:
\[ \alpha_B = \frac{0.045}{2000} = 2.25 \times 10^{-5} \, \text{°C}^{-1} \]

### Step 4: Set up the equation for the composite rod
Let \(L_A\) be the length of the portion made of metal A and \(L_B\) be the length of the portion made of metal B. We know:
\[ L_A + L_B = 20 \, \text{cm} \]
Also, the total expansion of the composite rod is given by:
\[ \Delta L = \Delta L_A + \Delta L_B \]
Where:
\[ \Delta L_A = L_A \cdot \alpha_A \cdot 100 \]
\[ \Delta L_B = L_B \cdot \alpha_B \cdot 100 \]
Substituting these into the total expansion gives:
\[ 0.065 = L_A \cdot (3.75 \times 10^{-5}) \cdot 100 + L_B \cdot (2.25 \times 10^{-5}) \cdot 100 \]
This simplifies to:
\[ 0.065 = L_A \cdot 3.75 \times 10^{-3} + L_B \cdot 2.25 \times 10^{-3} \]

### Step 5: Substitute \(L_B\) in terms of \(L_A\)
From the first equation:
\[ L_B = 20 - L_A \]
Substituting this into the expansion equation:
\[ 0.065 = L_A \cdot 3.75 \times 10^{-3} + (20 - L_A) \cdot 2.25 \times 10^{-3} \]

### Step 6: Solve for \(L_A\)
Expanding and simplifying:
\[ 0.065 = L_A \cdot 3.75 \times 10^{-3} + 20 \cdot 2.25 \times 10^{-3} - L_A \cdot 2.25 \times 10^{-3} \]
Combining like terms:
\[ 0.065 = L_A \cdot (3.75 \times 10^{-3} - 2.25 \times 10^{-3}) + 45 \times 10^{-3} \]
\[ 0.065 = L_A \cdot 1.5 \times 10^{-3} + 45 \times 10^{-3} \]
Subtract \(45 \times 10^{-3}\) from both sides:
\[ 0.065 - 0.045 = L_A \cdot 1.5 \times 10^{-3} \]
\[ 0.020 = L_A \cdot 1.5 \times 10^{-3} \]
Now, solving for \(L_A\):
\[ L_A = \frac{0.020}{1.5 \times 10^{-3}} = \frac{20}{1.5} \approx 13.33 \, \text{cm} \]

### Step 7: Conclusion
The length of the portion made of metal A in the composite rod is approximately **13.33 cm**.