A metal rod (A) of `25 cm` length expands by `0.050 cm` when its temperature is raised from `0^@ C` to `100^@ C`. Another rod (B) of a different metal of length `40 cm` expamds by `0.040 cm` for the same rise in temperature. A third rod ( C) of `50 cm` length is made up of pieces of rods (A) and (B) placed end to end expands by `0.03 cm` on heating from `0^@ C`. Find the lengths of each portion of the composite rod.

Explanation

To solve the problem, we need to find the lengths of the portions of rods A and B that make up the composite rod C. We will use the concept of linear thermal expansion to derive the necessary equations.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Length of rod A, \( L_A = 25 \, \text{cm} \)
- Expansion of rod A, \( \Delta L_A = 0.050 \, \text{cm} \)
- Length of rod B, \( L_B = 40 \, \text{cm} \)
- Expansion of rod B, \( \Delta L_B = 0.040 \, \text{cm} \)
- Total length of composite rod C, \( L_C = 50 \, \text{cm} \)
- Expansion of rod C, \( \Delta L_C = 0.03 \, \text{cm} \)

2. **Calculate the Coefficient of Linear Expansion for Rod A:**
\[ \Delta L_A = L_A \cdot \alpha_A \cdot \Delta T \]
Here, \( \Delta T = 100^\circ C \).
\[ 0.050 = 25 \cdot \alpha_A \cdot 100 \]
\[ \alpha_A = \frac{0.050}{25 \cdot 100} = 2 \times 10^{-5} \, \text{per } ^\circ C \]

3. **Calculate the Coefficient of Linear Expansion for Rod B:**
\[ \Delta L_B = L_B \cdot \alpha_B \cdot \Delta T \]
\[ 0.040 = 40 \cdot \alpha_B \cdot 100 \]
\[ \alpha_B = \frac{0.040}{40 \cdot 100} = 1 \times 10^{-5} \, \text{per } ^\circ C \]

4. **Set Up the Equation for Composite Rod C:**
Let \( L \) be the length of rod A in the composite rod C. Then the length of rod B will be \( 50 - L \).
The total expansion of rod C can be expressed as:
\[ \Delta L_C = \Delta L_A + \Delta L_B \]
Where:
\[ \Delta L_A = L \cdot \alpha_A \cdot \Delta T \]
\[ \Delta L_B = (50 - L) \cdot \alpha_B \cdot \Delta T \]
Substituting these into the equation gives:
\[ 0.03 = L \cdot (2 \times 10^{-5}) \cdot 50 + (50 - L) \cdot (1 \times 10^{-5}) \cdot 50 \]

5. **Simplify the Equation:**
\[ 0.03 = L \cdot (1 \times 10^{-3}) + (50 - L) \cdot (0.5 \times 10^{-3}) \]
\[ 0.03 = 1 \times 10^{-3} L + 0.5 \times 10^{-3} (50 - L) \]
\[ 0.03 = 1 \times 10^{-3} L + 25 \times 10^{-3} - 0.5 \times 10^{-3} L \]
\[ 0.03 = (0.5 \times 10^{-3}) L + 25 \times 10^{-3} \]
\[ 0.03 - 25 \times 10^{-3} = 0.5 \times 10^{-3} L \]
\[ -0.020 = 0.5 \times 10^{-3} L \]
\[ L = \frac{-0.020}{0.5 \times 10^{-3}} = 40 \, \text{cm} \]

6. **Calculate the Length of Rod B:**
\[ L_B = 50 - L = 50 - 40 = 10 \, \text{cm} \]

### Final Answer:
- Length of rod A, \( L_A = 40 \, \text{cm} \)
- Length of rod B, \( L_B = 10 \, \text{cm} \)