A 5 metre long wire is fixed to the ceiling. A weight of `10kg` is hung at the lower end and is `1metre` above the floor. The wire was elongated by `1mm`. The energy stored in the wire duw to streching is

Options

A. 0.01J
B. 0.05 J
C. 0.02 J
D. 0.04J

Explanation

To find the energy stored in the wire due to stretching, we can use the formula for the elastic potential energy stored in a stretched wire:

\[ \text{Energy} (E) = \frac{1}{2} \times F \times L \]

where:
- \( F \) is the force applied to the wire,
- \( L \) is the original length of the wire.

### Step 1: Calculate the Force (F)
The force applied to the wire is due to the weight hung at the end. The weight can be calculated using the formula:

\[ F = m \times g \]

where:
- \( m = 10 \, \text{kg} \) (mass of the weight),
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).

Calculating the force:

\[ F = 10 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 98.1 \, \text{N} \]

### Step 2: Determine the Original Length (L)
The original length of the wire is given as:

\[ L = 5 \, \text{m} \]

### Step 3: Convert the Elongation to Meters
The elongation of the wire is given as \( 1 \, \text{mm} \). We need to convert this to meters:

\[ \text{Elongation} = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \]

### Step 4: Calculate the Energy Stored (E)
Now we can substitute the values of \( F \) and \( L \) into the energy formula:

\[ E = \frac{1}{2} \times F \times L \]

Substituting the values:

\[ E = \frac{1}{2} \times 98.1 \, \text{N} \times 5 \, \text{m} \]

Calculating the energy:

\[ E = \frac{1}{2} \times 98.1 \times 5 = \frac{490.5}{2} = 245.25 \, \text{J} \]

### Step 5: Final Calculation
However, since we need the energy stored due to the elongation, we need to consider the elongation in the calculation:

Using the elongation in the formula for energy stored:

\[ E = \frac{1}{2} \times F \times \text{Elongation} \]

Substituting the values:

\[ E = \frac{1}{2} \times 98.1 \, \text{N} \times 1 \times 10^{-3} \, \text{m} \]

Calculating the energy:

\[ E = \frac{1}{2} \times 98.1 \times 0.001 = 0.04905 \, \text{J} \approx 0.05 \, \text{J} \]

Thus, the energy stored in the wire due to stretching is approximately:

\[ \boxed{0.05 \, \text{J}} \]