Young's modulus of brass and steel are `10 xx 10^(10) N//m` and `2 xx 10^(11) N//m^(2)`, respectively. A brass wire and a steel wire of the same length are extended by `1 mm` under the same force. The radii of the brass and steel wires are `R_(B)` and `R_(S)`. respectively. Then

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))//4`

Explanation

To solve the problem, we need to relate the radii of the brass and steel wires given their Young's moduli and the fact that they are extended by the same amount under the same force.

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
Young's modulus \( Y \) is defined as:
\[ Y = \frac{F \cdot L_0}{A \cdot \Delta L} \]
where:
- \( F \) is the force applied,
- \( L_0 \) is the original length,
- \( A \) is the cross-sectional area,
- \( \Delta L \) is the change in length.

2. **Setting Up the Equations for Brass and Steel**:
Let:
- \( Y_B = 10 \times 10^{10} \, \text{N/m}^2 \) (Young's modulus for brass),
- \( Y_S = 2 \times 10^{11} \, \text{N/m}^2 \) (Young's modulus for steel),
- \( R_B \) = radius of the brass wire,
- \( R_S \) = radius of the steel wire.

Since both wires are extended by the same amount (\( \Delta L = 1 \, \text{mm} = 0.001 \, \text{m} \)), we can write the equations for the two wires:
\[ \Delta L = \frac{F \cdot L_0}{A_B \cdot Y_B} \quad \text{(for brass)} \]
\[ \Delta L = \frac{F \cdot L_0}{A_S \cdot Y_S} \quad \text{(for steel)} \]

3. **Cross-Sectional Area**:
The cross-sectional area \( A \) for a wire with radius \( R \) is given by:
\[ A = \pi R^2 \]
Therefore, we can rewrite the equations as:
\[ \Delta L = \frac{F \cdot L_0}{\pi R_B^2 \cdot Y_B} \]
\[ \Delta L = \frac{F \cdot L_0}{\pi R_S^2 \cdot Y_S} \]

4. **Equating the Two Expressions**:
Since \( \Delta L \) is the same for both wires, we can set the two expressions equal to each other:
\[ \frac{F \cdot L_0}{\pi R_B^2 \cdot Y_B} = \frac{F \cdot L_0}{\pi R_S^2 \cdot Y_S} \]

5. **Simplifying the Equation**:
Canceling \( F \) and \( L_0 \) from both sides and \( \pi \) as well, we get:
\[ \frac{1}{R_B^2 Y_B} = \frac{1}{R_S^2 Y_S} \]
Rearranging gives:
\[ \frac{R_S^2}{R_B^2} = \frac{Y_B}{Y_S} \]

6. **Substituting Values**:
Substitute the values of \( Y_B \) and \( Y_S \):
\[ \frac{R_S^2}{R_B^2} = \frac{10 \times 10^{10}}{2 \times 10^{11}} = \frac{10}{20} = \frac{1}{2} \]

7. **Finding the Relationship Between Radii**:
Taking the square root of both sides:
\[ \frac{R_S}{R_B} = \frac{1}{\sqrt{2}} \]
Therefore, we can express \( R_S \) in terms of \( R_B \):
\[ R_S = \frac{R_B}{\sqrt{2}} \]

### Final Answer:
The radius of the steel wire \( R_S \) is equal to \( \frac{R_B}{\sqrt{2}} \).