Two wires A & B of same material having elongation 2mm and 4mm resp. on applying 2N take. If radius of B is 4 times the radius of A and ratio of length of A is to B in the form of 1/x then the value of x is.

Answer

To solve the problem, we need to analyze the elongation of two wires A and B under the same force and relate their dimensions to find the ratio of their lengths.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two wires A and B made of the same material, with elongations of 2 mm and 4 mm respectively when a force of 2 N is applied to each.

2. **Young's Modulus Relation**:
The Young's modulus (Y) for a material is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
where:
- \( F \) is the force applied,
- \( A \) is the cross-sectional area,
- \( \Delta L \) is the elongation,
- \( L \) is the original length of the wire.

3. **Cross-Sectional Area**:
The cross-sectional area \( A \) of a wire can be expressed in terms of its radius \( r \):
\[ A = \pi r^2 \]

4. **Setting Up the Equations**:
For wire A:
\[ Y = \frac{F \cdot L_A}{\pi r_A^2 \cdot \Delta L_A} \]
For wire B:
\[ Y = \frac{F \cdot L_B}{\pi r_B^2 \cdot \Delta L_B} \]
Since both wires are made of the same material and subjected to the same force, we can equate the Young's moduli:
\[ \frac{F \cdot L_A}{\pi r_A^2 \cdot \Delta L_A} = \frac{F \cdot L_B}{\pi r_B^2 \cdot \Delta L_B} \]

5. **Canceling Common Terms**:
Cancel \( F \) and \( \pi \) from both sides:
\[ \frac{L_A}{r_A^2 \cdot \Delta L_A} = \frac{L_B}{r_B^2 \cdot \Delta L_B} \]

6. **Substituting Known Values**:
We know:
- \( \Delta L_A = 2 \text{ mm} \)
- \( \Delta L_B = 4 \text{ mm} \)
- \( r_B = 4r_A \) (given that the radius of B is 4 times that of A)

Substitute these values into the equation:
\[ \frac{L_A}{r_A^2 \cdot 2} = \frac{L_B}{(4r_A)^2 \cdot 4} \]

7. **Simplifying the Equation**:
The equation becomes:
\[ \frac{L_A}{2r_A^2} = \frac{L_B}{16r_A^2 \cdot 4} \]
Simplifying further:
\[ \frac{L_A}{2r_A^2} = \frac{L_B}{64r_A^2} \]

8. **Cross-Multiplying**:
Cross-multiply to eliminate the fractions:
\[ 64L_A = 2L_B \]
Thus:
\[ \frac{L_A}{L_B} = \frac{2}{64} = \frac{1}{32} \]

9. **Finding the Ratio**:
The ratio of lengths \( \frac{L_A}{L_B} \) is given as \( \frac{1}{x} \). Therefore, we have:
\[ \frac{1}{x} = \frac{1}{32} \implies x = 32 \]

### Final Answer:
The value of \( x \) is **32**.