A wire suspended vertically is stretched by a 20 kgf applied to its free end. The increase in length of the wire is 2 mm. The energy stored in the wire is `(g=10ms^(-2))`:

Options

A. 0.1 J
B. 0.2 J
C. 0.4 J
D. 5 J

Explanation

To find the energy stored in the wire when it is stretched, we can follow these steps:

### Step 1: Calculate the Force Applied
The force applied to the wire is given as 20 kgf. We need to convert this to Newtons (N) using the acceleration due to gravity (g = 10 m/s²).

\[ \text{Force (F)} = \text{mass} \times g = 20 \, \text{kg} \times 10 \, \text{m/s}^2 = 200 \, \text{N} \]

### Step 2: Calculate the Stress in the Wire
Stress is defined as the force applied per unit area. However, we do not have the area of the cross-section of the wire. We can denote the area as \(A\).

\[ \text{Stress} (\sigma) = \frac{F}{A} = \frac{200 \, \text{N}}{A} \]

### Step 3: Calculate the Strain in the Wire
Strain is defined as the change in length divided by the original length. The change in length (\(\Delta L\)) is given as 2 mm, which we convert to meters:

\[ \Delta L = 2 \, \text{mm} = 0.002 \, \text{m} \]

Let the original length of the wire be \(L\). Then,

\[ \text{Strain} (\epsilon) = \frac{\Delta L}{L} = \frac{0.002}{L} \]

### Step 4: Calculate the Energy Stored in the Wire
The energy stored per unit volume in a material is given by the formula:

\[ U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume} \]

The volume of the wire can be expressed as \(A \times L\). Therefore, the energy stored (U) can be rewritten as:

\[ U = \frac{1}{2} \times \sigma \times \epsilon \times (A \times L) \]

Substituting the expressions for stress and strain:

\[ U = \frac{1}{2} \times \left(\frac{200}{A}\right) \times \left(\frac{0.002}{L}\right) \times (A \times L) \]

### Step 5: Simplifying the Expression
Notice that \(A\) and \(L\) cancel out:

\[ U = \frac{1}{2} \times 200 \times 0.002 \]

Calculating this gives:

\[ U = \frac{1}{2} \times 200 \times 0.002 = 0.2 \, \text{Joules} \]

### Final Answer
The energy stored in the wire is **0.2 Joules**.

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