Two wires 'A' and 'B' of the 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. `1:2`
C. `1:4`
D. `1:2`

Answer

To solve the problem, we need to find the ratio of the normal forces required to produce the same change in the lengths of two wires 'A' and 'B'. Let's denote the following:

- \( r_1 \) and \( r_2 \) are the radii of wires A and B, respectively.
- \( L_1 \) and \( L_2 \) are the lengths of wires A and B, respectively.
- \( F_1 \) and \( F_2 \) are the forces applied on wires A and B, respectively.

### Step 1: Understand the relationship between force, area, and length
Using Young's modulus, we have the relationship:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
Where:
- \( Y \) is Young's modulus (constant for the same material),
- \( F \) is the force applied,
- \( L \) is the original length of the wire,
- \( A \) is the cross-sectional area,
- \( \Delta L \) is the change in length.

Since we want the same change in length (\( \Delta L \)) for both wires, we can rearrange the equation to find the force:
\[ F = \frac{Y \cdot A \cdot \Delta L}{L} \]

### Step 2: Calculate the areas of the wires
The cross-sectional area \( A \) of a wire is given by:
\[ A = \pi r^2 \]
Thus, for wires A and B:
\[ A_1 = \pi r_1^2 \quad \text{and} \quad A_2 = \pi r_2^2 \]

### Step 3: Substitute the areas into the force equation
Now, substituting the areas into the force equation for both wires:
\[ F_1 = \frac{Y \cdot \pi r_1^2 \cdot \Delta L}{L_1} \]
\[ F_2 = \frac{Y \cdot \pi r_2^2 \cdot \Delta L}{L_2} \]

### Step 4: Find the ratio of the forces
Now, we can find the ratio of the forces:
\[ \frac{F_1}{F_2} = \frac{Y \cdot \pi r_1^2 \cdot \Delta L / L_1}{Y \cdot \pi r_2^2 \cdot \Delta L / L_2} \]
This simplifies to:
\[ \frac{F_1}{F_2} = \frac{r_1^2 \cdot L_2}{r_2^2 \cdot L_1} \]

### Step 5: Substitute the given ratios
Given that:
- The ratio of the radii \( r_1 : r_2 = 2 : 1 \) implies \( r_1 = 2r \) and \( r_2 = r \).
- The ratio of the lengths \( L_1 : L_2 = 4 : 1 \) implies \( L_1 = 4L \) and \( L_2 = L \).

Now substituting these values into the ratio:
\[ \frac{F_1}{F_2} = \frac{(2r)^2 \cdot L}{r^2 \cdot 4L} = \frac{4r^2 \cdot L}{r^2 \cdot 4L} = \frac{4}{4} = 1 \]

### Final Answer
Thus, the ratio of the normal forces required to produce the same change in the lengths of these two wires is:
\[ F_1 : F_2 = 1 : 1 \]