Two wires A and B are of same material. Their lengths are in the ratio 1:2 and diameters are in the ratio 2:1 when stretched by force `F_A` and `F_B` respectively they get equal increase in their lengths. Then the ratio `(F_A)/(F_B)` should be

Options

A. `1:2`
B. `1:1`
C. `2:1`
D. `8:1`

Answer

To solve the problem, we need to find the ratio of the forces \( F_A \) and \( F_B \) applied to two wires A and B, which are made of the same material and have specific length and diameter ratios.

### Step-by-Step Solution:

1. **Identify the Given Ratios:**
- The lengths of the wires A and B are in the ratio \( L_A : L_B = 1 : 2 \).
- The diameters of the wires A and B are in the ratio \( d_A : d_B = 2 : 1 \).

2. **Express Lengths and Diameters:**
- Let \( L_A = L \) and \( L_B = 2L \) (since the ratio is 1:2).
- Let \( d_A = 2d \) and \( d_B = d \) (since the ratio is 2:1).

3. **Calculate the Radii:**
- The radius \( r \) is half of the diameter, so:
- \( r_A = \frac{d_A}{2} = \frac{2d}{2} = d \)
- \( r_B = \frac{d_B}{2} = \frac{d}{2} \)
- Thus, the ratio of the radii is \( r_A : r_B = d : \frac{d}{2} = 2 : 1 \).

4. **Use the Formula for Force:**
- The force applied to stretch a wire is given by:
\[ F = \frac{Y \cdot A \cdot \Delta L}{L} \]
- Where \( Y \) is the Young's modulus, \( A \) is the cross-sectional area, \( \Delta L \) is the change in length, and \( L \) is the original length of the wire.

5. **Calculate the Cross-Sectional Areas:**
- The area \( A \) of a wire is given by:
\[ A = \pi r^2 \]
- Therefore:
- \( A_A = \pi r_A^2 = \pi d^2 \)
- \( A_B = \pi r_B^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \)

6. **Write the Forces for Each Wire:**
- For wire A:
\[ F_A = \frac{Y \cdot A_A \cdot \Delta L}{L_A} = \frac{Y \cdot \pi d^2 \cdot \Delta L}{L} \]
- For wire B:
\[ F_B = \frac{Y \cdot A_B \cdot \Delta L}{L_B} = \frac{Y \cdot \frac{\pi d^2}{4} \cdot \Delta L}{2L} \]

7. **Set the Forces Equal:**
- Since both wires experience the same increase in length (\( \Delta L \)) and are made of the same material (same \( Y \)):
\[ F_A = \frac{Y \cdot \pi d^2 \cdot \Delta L}{L} \]
\[ F_B = \frac{Y \cdot \frac{\pi d^2}{4} \cdot \Delta L}{2L} \]

8. **Calculate the Ratio of Forces:**
- Now, we can find the ratio \( \frac{F_A}{F_B} \):
\[ \frac{F_A}{F_B} = \frac{\frac{Y \cdot \pi d^2 \cdot \Delta L}{L}}{\frac{Y \cdot \frac{\pi d^2}{4} \cdot \Delta L}{2L}} = \frac{1}{\frac{1}{4} \cdot \frac{1}{2}} = 8 \]

9. **Final Ratio:**
- Therefore, the ratio \( \frac{F_A}{F_B} = 8 : 1 \).

### Conclusion:
The required ratio of the forces \( \frac{F_A}{F_B} \) is \( 8 : 1 \).