Two wires A and B have the same length and area of cross-secton. But Young's modulus of A is two times the Young's modulus of B. Then the ratio of force constant of A to that of B is

Options

A. 1
B. 2
C. `1//2`
D. `sqrt2`

Answer

To solve the problem, we need to find the ratio of the force constants of two wires A and B, given that they have the same length and area of cross-section, but the Young's modulus of wire A is twice that of wire B.

### Step-by-Step Solution:

1. **Understand the Definitions**:
- Young's modulus (Y) is a measure of the stiffness of a material. It is defined as the ratio of stress (force per unit area) to strain (relative change in length).
- The force constant (k) of a wire can be expressed in terms of Young's modulus (Y), the area of cross-section (A), and the original length (L) of the wire.

2. **Formula for Force Constant**:
- The force constant (k) can be expressed as:
\[ k = \frac{Y \cdot A}{L} \]
- Here, \(Y\) is Young's modulus, \(A\) is the area of cross-section, and \(L\) is the original length of the wire.

3. **Set Up the Ratio**:
- For wire A, the force constant \(k_A\) is:
\[ k_A = \frac{Y_A \cdot A}{L} \]
- For wire B, the force constant \(k_B\) is:
\[ k_B = \frac{Y_B \cdot A}{L} \]
- Now, we can find the ratio of the force constants:
\[ \frac{k_A}{k_B} = \frac{Y_A \cdot A / L}{Y_B \cdot A / L} \]

4. **Simplify the Ratio**:
- Since the area \(A\) and length \(L\) are the same for both wires, they cancel out:
\[ \frac{k_A}{k_B} = \frac{Y_A}{Y_B} \]

5. **Substitute the Given Values**:
- We are given that \(Y_A = 2Y_B\). Therefore, substituting this into the ratio gives:
\[ \frac{k_A}{k_B} = \frac{2Y_B}{Y_B} = 2 \]

6. **Final Ratio**:
- Thus, the ratio of the force constant of wire A to that of wire B is:
\[ \frac{k_A}{k_B} = 2:1 \]

### Conclusion:
The ratio of the force constant of wire A to that of wire B is \(2:1\).