The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of

Options

A. `1:2`
B. `2:1`
C. `4:1`
D. `1:1`

Answer

To solve the problem, we need to understand the relationship between Young's modulus, stress, and strain in the two wires made of steel and brass.

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
- Young's modulus (Y) is defined as the ratio of stress (force per unit area) to strain (relative change in length).
- For two materials, we have:
\[ Y_s = \frac{F_s L}{A \Delta L_s} \quad \text{(for steel)} \]
\[ Y_b = \frac{F_b L}{A \Delta L_b} \quad \text{(for brass)} \]
- Given that \( Y_s = 2 Y_b \).

2. **Setting Up the Ratio**:
- From the definition, we can express the ratio of Young's moduli:
\[ \frac{Y_s}{Y_b} = \frac{F_s \Delta L_b}{F_b \Delta L_s} \]
- Since the lengths (L) and areas (A) of both wires are the same, they cancel out.

3. **Substituting the Given Relation**:
- We know that \( Y_s = 2 Y_b \), thus:
\[ \frac{Y_s}{Y_b} = 2 \quad \Rightarrow \quad \frac{F_s \Delta L_b}{F_b \Delta L_s} = 2 \]

4. **Equal Elongation Condition**:
- For the ends of the wires to be at the same level after elongation, the elongations must be equal:
\[ \Delta L_s = \Delta L_b \]
- Therefore, we can simplify the equation:
\[ \frac{F_s}{F_b} = 2 \]

5. **Interpreting Forces as Weights**:
- The forces \( F_s \) and \( F_b \) correspond to the weights added to the steel and brass wires respectively:
\[ F_s = W_s \quad \text{and} \quad F_b = W_b \]
- Thus, we have:
\[ \frac{W_s}{W_b} = 2 \]

6. **Final Ratio**:
- This means the weight added to the steel wire is twice that added to the brass wire:
\[ W_s : W_b = 2 : 1 \]

### Conclusion:
The ratio of the weights added to the steel and brass wires must be \( 2 : 1 \).