Two wires A and B of same material have radii in the ratio `2:1` and lengths in the ratio `4:1`. The ratio of the normal forces required to produce the same change in the lengths of these two wires is

Options

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

Answer

To solve the problem, we need to find the ratio of the normal forces required to produce the same change in length of two wires A and B, given their radii and lengths.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two wires (A and B) made of the same material. The radii of the wires are in the ratio \( r_A : r_B = 2 : 1 \) and their lengths are in the ratio \( L_A : L_B = 4 : 1 \). We need to find the ratio of the forces \( F_A : F_B \) required to produce the same change in length \( \Delta L \).

2. **Using Young's Modulus**:
The relationship between stress and strain is given by:
\[ \text{Stress} = \text{Young's Modulus} \times \text{Strain} \]
Stress can also be defined as:
\[ \text{Stress} = \frac{F}{A} \]
where \( F \) is the force applied and \( A \) is the cross-sectional area of the wire.

3. **Calculating Cross-Sectional Area**:
The cross-sectional area \( A \) of a wire with radius \( r \) is given by:
\[ A = \pi r^2 \]
Therefore, for wires A and B:
\[ A_A = \pi r_A^2 \quad \text{and} \quad A_B = \pi r_B^2 \]

4. **Expressing Forces**:
From the relationship of stress and strain, we can express the force for each wire as:
\[ F_A = Y \cdot A_A \cdot \frac{\Delta L}{L_A} \quad \text{and} \quad F_B = Y \cdot A_B \cdot \frac{\Delta L}{L_B} \]
where \( Y \) is Young's modulus, which is the same for both wires since they are made of the same material.

5. **Substituting Areas**:
Substituting the areas into the force equations:
\[ F_A = Y \cdot \pi r_A^2 \cdot \frac{\Delta L}{L_A} \quad \text{and} \quad F_B = Y \cdot \pi r_B^2 \cdot \frac{\Delta L}{L_B} \]

6. **Finding the Ratio of Forces**:
Now, we can find the ratio of the forces:
\[ \frac{F_A}{F_B} = \frac{Y \cdot \pi r_A^2 \cdot \frac{\Delta L}{L_A}}{Y \cdot \pi r_B^2 \cdot \frac{\Delta L}{L_B}} = \frac{r_A^2 \cdot L_B}{r_B^2 \cdot L_A} \]

7. **Substituting the Ratios**:
Given the ratios \( r_A : r_B = 2 : 1 \) and \( L_A : L_B = 4 : 1 \), we have:
\[ r_A = 2k, \quad r_B = k \quad (k \text{ is a constant}) \]
\[ L_A = 4m, \quad L_B = m \quad (m \text{ is a constant}) \]

8. **Calculating the Ratio**:
Now substituting these values into the force ratio:
\[ \frac{F_A}{F_B} = \frac{(2k)^2 \cdot m}{(k)^2 \cdot (4m)} = \frac{4k^2 \cdot m}{k^2 \cdot 4m} = \frac{4}{4} = 1 \]

9. **Final Result**:
Therefore, the ratio of the normal forces required to produce the same change in lengths of the two wires is:
\[ \frac{F_A}{F_B} = 1 : 1 \]