A composite rod consists of a steel rod of length `25 cm` and area `2A` and a copper rod of length `50 cm` and area `A`. The composite rod is subjected to an axial load `F`. If the Young's moduli of steel and copper are in the ratio `2: 1` then

Options

A. the extension produced in copper rod will be more
B. the extension in copper and steel parts will be in the ratio `1:2`
C. the stress applied to copper rod will be more
D. no extension will be produced in the steel rod

Explanation

To solve the problem, we need to analyze the composite rod made of steel and copper under an axial load \( F \). We will determine the extensions in both rods and compare them.

### Given Data:
- Length of steel rod, \( L_s = 25 \, \text{cm} = 0.25 \, \text{m} \)
- Area of steel rod, \( A_s = 2A \)
- Length of copper rod, \( L_c = 50 \, \text{cm} = 0.50 \, \text{m} \)
- Area of copper rod, \( A_c = A \)
- Ratio of Young's moduli, \( \frac{E_s}{E_c} = \frac{2}{1} \)

### Step 1: Express Young's Moduli
Let the Young's modulus of copper be \( E_c \). Then, the Young's modulus of steel can be expressed as:
\[ E_s = 2E_c \]

### Step 2: Calculate Stress in Each Rod
Stress is defined as force per unit area:
- Stress in the steel rod:
\[ \sigma_s = \frac{F}{A_s} = \frac{F}{2A} \]
- Stress in the copper rod:
\[ \sigma_c = \frac{F}{A_c} = \frac{F}{A} \]

### Step 3: Relate Strain to Stress and Young's Modulus
Using the relationship \( \epsilon = \frac{\sigma}{E} \):
- Strain in the steel rod:
\[ \epsilon_s = \frac{\sigma_s}{E_s} = \frac{\frac{F}{2A}}{2E_c} = \frac{F}{4AE_c} \]
- Strain in the copper rod:
\[ \epsilon_c = \frac{\sigma_c}{E_c} = \frac{\frac{F}{A}}{E_c} = \frac{F}{AE_c} \]

### Step 4: Calculate Extensions
The extension \( \Delta L \) in each rod can be expressed as:
- Extension in the steel rod:
\[ \Delta L_s = \epsilon_s \cdot L_s = \left(\frac{F}{4AE_c}\right) \cdot 0.25 \]
\[ \Delta L_s = \frac{F \cdot 0.25}{4AE_c} = \frac{F}{16AE_c} \]
- Extension in the copper rod:
\[ \Delta L_c = \epsilon_c \cdot L_c = \left(\frac{F}{AE_c}\right) \cdot 0.50 \]
\[ \Delta L_c = \frac{F \cdot 0.50}{AE_c} = \frac{F}{2AE_c} \]

### Step 5: Compare Extensions
Now, we can find the ratio of the extensions:
\[ \frac{\Delta L_s}{\Delta L_c} = \frac{\frac{F}{16AE_c}}{\frac{F}{2AE_c}} = \frac{1/16}{1/2} = \frac{1}{8} \]
This implies:
\[ \Delta L_c = 8 \Delta L_s \]

### Conclusion
The extension produced in the copper rod is greater than that in the steel rod. Thus, the correct answer is that the extension produced in the copper rod will be more.