If the work done in stretching a wire by `1mm` is `2J`, then work necessary for stretching another wire of same material but with double radius of cross -section and half of the length by `1 mm` is

Options

A. 16J
B. 8J
C. 4J
D. `(1)/(4)J`

Explanation

To solve the problem, we need to determine the work necessary to stretch a wire of the same material, but with different dimensions. We will use the relationship between work done, force, and the properties of the wire.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Work done to stretch the first wire (W1) = 2 J
- The first wire has a certain radius (r) and length (l).
- The second wire has double the radius (2r) and half the length (l/2).

2. **Understand the Relationship Between Work, Force, and Dimensions:**
- The work done (W) in stretching a wire can be expressed as:
\[ W \propto \text{Force} \times \text{Extension} \]
- The force (F) required to stretch a wire is given by:
\[ F = \frac{Y \cdot A \cdot \Delta l}{L} \]
where:
- \(Y\) = Young's modulus (constant for the same material)
- \(A\) = Cross-sectional area
- \(\Delta l\) = Extension (1 mm in both cases)
- \(L\) = Original length of the wire

3. **Calculate the Cross-Sectional Areas:**
- For the first wire:
\[ A_1 = \pi r^2 \]
- For the second wire (with radius 2r):
\[ A_2 = \pi (2r)^2 = 4\pi r^2 \]

4. **Set Up the Ratio of Forces:**
- Since the extension (\(\Delta l\)) is the same for both wires, we can write the ratio of the forces:
\[ \frac{F_1}{F_2} = \frac{A_1/L_1}{A_2/L_2} \]
- Substituting the areas and lengths:
\[ \frac{F_1}{F_2} = \frac{\pi r^2 / l}{4\pi r^2 / (l/2)} = \frac{\pi r^2 \cdot 2}{4\pi r^2 \cdot l} = \frac{2}{4} = \frac{1}{2} \]

5. **Relate the Work Done:**
- Since work done is proportional to force, we can write:
\[ \frac{W_1}{W_2} = \frac{F_1}{F_2} \]
- Therefore:
\[ \frac{W_1}{W_2} = \frac{1}{2} \]
- Rearranging gives:
\[ W_2 = 2 \cdot W_1 \]

6. **Substitute the Value of W1:**
- Given \(W_1 = 2 J\):
\[ W_2 = 2 \cdot 2 J = 4 J \]

### Final Answer:
The work necessary for stretching the second wire by 1 mm is **4 Joules**.