The young 's modulus of brass and steel are ` 1.0 xx 10^(11) Nm^(-2)` and `2.0 xx 10^(11) N//m^(2)`. Respectively . A brass wire and steel wire of the same length extend by 1 mm, each under the same forece . It radii of brass and steel wires are `R_(B)` and `R_(S)` respectively, them

Options

A. `R_(s)= sqrt(2)R_(B)`
B. `R_(s)= (R_(B))/sqrt(2)`
C. `R_(s)= 4R_(B)`
D. `R_(S)= (R_(B))/(2)`

Answer

To solve the problem, we need to relate the Young's modulus of the materials (brass and steel) to their respective dimensions and the forces applied. We will use the formula for Young's modulus, which is given by:

\[ Y = \frac{F \cdot L_0}{A \cdot \Delta L} \]

Where:
- \( Y \) is the Young's modulus,
- \( F \) is the force applied,
- \( L_0 \) is the original length of the wire,
- \( A \) is the cross-sectional area of the wire,
- \( \Delta L \) is the change in length (extension).

### Step-by-Step Solution:

1. **Identify the given data:**
- Young's modulus of brass, \( Y_B = 1.0 \times 10^{11} \, \text{N/m}^2 \)
- Young's modulus of steel, \( Y_S = 2.0 \times 10^{11} \, \text{N/m}^2 \)
- Extension of both wires, \( \Delta L = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Length of both wires, \( L_0 \) (same for both)

2. **Write the equations for Young's modulus for both materials:**
- For brass:
\[ Y_B = \frac{F \cdot L_0}{A_B \cdot \Delta L} \]
- For steel:
\[ Y_S = \frac{F \cdot L_0}{A_S \cdot \Delta L} \]

3. **Express the cross-sectional area \( A \) in terms of the radius \( R \):**
- The cross-sectional area \( A \) of a wire is given by:
\[ A = \pi R^2 \]
- Therefore, for brass and steel:
\[ A_B = \pi R_B^2 \quad \text{and} \quad A_S = \pi R_S^2 \]

4. **Substitute the area expressions into the Young's modulus equations:**
- For brass:
\[ Y_B = \frac{F \cdot L_0}{\pi R_B^2 \cdot \Delta L} \]
- For steel:
\[ Y_S = \frac{F \cdot L_0}{\pi R_S^2 \cdot \Delta L} \]

5. **Divide the two equations to eliminate \( F \), \( L_0 \), and \( \Delta L \):**
\[ \frac{Y_B}{Y_S} = \frac{R_S^2}{R_B^2} \]

6. **Rearranging gives us the relationship between the radii:**
\[ \frac{R_B^2}{R_S^2} = \frac{Y_S}{Y_B} \]
\[ R_B^2 = R_S^2 \cdot \frac{Y_S}{Y_B} \]
\[ R_B = R_S \sqrt{\frac{Y_S}{Y_B}} \]

7. **Substituting the values of Young's modulus:**
\[ R_B = R_S \sqrt{\frac{2.0 \times 10^{11}}{1.0 \times 10^{11}}} \]
\[ R_B = R_S \sqrt{2} \]

### Final Result:
The radius of the brass wire is related to the radius of the steel wire by:
\[ R_B = R_S \sqrt{2} \]