If the work done in strectching 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. (A) `8 J`
B. (B) `16 J`
C. (C) `4 J`
D. (D) `32 J`

Explanation

To solve the problem, we need to find the work done in stretching a wire of the same material but with different dimensions. Here are the steps to find the solution:

### Step 1: Understand the Given Data
- Work done to stretch the first wire (W1) = 2 J
- Elongation of the first wire (ΔL1) = 1 mm = \(1 \times 10^{-3}\) m
- Radius of the second wire (R2) = 2 * R1 (double the radius of the first wire)
- Length of the second wire (L2) = \(L1 / 2\) (half the length of the first wire)
- Elongation of the second wire (ΔL2) = 1 mm = \(1 \times 10^{-3}\) m

### Step 2: Use the Formula for Work Done in Stretching a Wire
The work done (W) in stretching a wire can be expressed as:
\[ W = \frac{1}{2} \cdot F \cdot \Delta L \]
Where:
- F = applied force
- ΔL = elongation

### Step 3: Relate Force to Young's Modulus
The force can be related to Young's modulus (Y) using the formula:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
From this, we can express the force as:
\[ F = \frac{Y \cdot A \cdot \Delta L}{L} \]

### Step 4: Substitute Force into the Work Done Formula
Substituting the expression for F into the work done formula gives:
\[ W = \frac{1}{2} \cdot \left(\frac{Y \cdot A \cdot \Delta L}{L}\right) \cdot \Delta L = \frac{1}{2} \cdot \frac{Y \cdot A \cdot (\Delta L)^2}{L} \]

### Step 5: Set Up the Ratio of Work Done for Both Wires
Let \(W1\) be the work done for the first wire and \(W2\) for the second wire. We can set up the ratio:
\[ \frac{W1}{W2} = \frac{A1 \cdot (\Delta L1)^2 \cdot L2}{A2 \cdot (\Delta L2)^2 \cdot L1} \]

### Step 6: Calculate the Cross-Sectional Areas
The cross-sectional area \(A\) of a wire with radius \(R\) is given by:
\[ A = \pi R^2 \]
Thus:
- \(A1 = \pi R1^2\)
- \(A2 = \pi (2R1)^2 = 4\pi R1^2\)

### Step 7: Substitute Areas and Lengths into the Ratio
Substituting the areas into the ratio:
\[ \frac{W1}{W2} = \frac{\pi R1^2 \cdot (\Delta L1)^2 \cdot (L1/2)}{4\pi R1^2 \cdot (\Delta L2)^2 \cdot L1} \]
This simplifies to:
\[ \frac{W1}{W2} = \frac{1}{4} \cdot \frac{(\Delta L1)^2}{(\Delta L2)^2} \cdot \frac{1}{2} \]

### Step 8: Substitute Known Values
Since \(\Delta L1 = \Delta L2 = 1 \text{ mm}\), we have:
\[ \frac{W1}{W2} = \frac{1}{4} \cdot \frac{1^2}{1^2} \cdot \frac{1}{2} = \frac{1}{8} \]
Thus:
\[ W2 = 8 \cdot W1 = 8 \cdot 2 \text{ J} = 16 \text{ J} \]

### Final Answer
The work necessary for stretching the second wire by 1 mm is **16 J**.

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