When a force is applied on a wire of uniform cross-sectional area `3xx10^-6m^2` and length 4 m, the increase in length is 1mm. Energy stored in it will be `(Y=2xx10^(11)N//m^2`)

Options

A. 6270 J
B. 0.177 J
C. 0.075 J
D. 0.150 J

Explanation

To find the energy stored in the wire when a force is applied, we can use the formula for elastic potential energy stored in a stretched wire:

\[ U = \frac{1}{2} F \Delta L \]

where:
- \( U \) is the energy stored,
- \( F \) is the force applied,
- \( \Delta L \) is the increase in length of the wire.

First, we need to find the force \( F \) applied to the wire using Young's modulus \( Y \):

\[ Y = \frac{F/A}{\Delta L/L} \]

Rearranging this, we can express the force \( F \) as:

\[ F = Y \cdot \frac{A \cdot \Delta L}{L} \]

### Step 1: Substitute the given values into the formula for force

Given:
- \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- \( A = 3 \times 10^{-6} \, \text{m}^2 \)
- \( \Delta L = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- \( L = 4 \, \text{m} \)

Substituting these values into the equation for force:

\[ F = 2 \times 10^{11} \, \text{N/m}^2 \cdot \frac{3 \times 10^{-6} \, \text{m}^2 \cdot 1 \times 10^{-3} \, \text{m}}{4 \, \text{m}} \]

### Step 2: Calculate the force \( F \)

Calculating the numerator:

\[ 3 \times 10^{-6} \, \text{m}^2 \cdot 1 \times 10^{-3} \, \text{m} = 3 \times 10^{-9} \, \text{m}^3 \]

Now substituting this back into the force equation:

\[ F = 2 \times 10^{11} \cdot \frac{3 \times 10^{-9}}{4} \]

Calculating \( \frac{3 \times 10^{-9}}{4} \):

\[ \frac{3 \times 10^{-9}}{4} = 0.75 \times 10^{-9} = 7.5 \times 10^{-10} \]

Now substituting this into the force equation:

\[ F = 2 \times 10^{11} \cdot 7.5 \times 10^{-10} = 1.5 \times 10^{2} \, \text{N} = 150 \, \text{N} \]

### Step 3: Calculate the energy stored \( U \)

Now we can calculate the energy stored using the force we just calculated:

\[ U = \frac{1}{2} F \Delta L \]

Substituting the values of \( F \) and \( \Delta L \):

\[ U = \frac{1}{2} \cdot 150 \, \text{N} \cdot 1 \times 10^{-3} \, \text{m} \]

Calculating this:

\[ U = \frac{1}{2} \cdot 150 \cdot 1 \times 10^{-3} = 75 \times 10^{-3} = 0.075 \, \text{J} \]

### Final Answer

The energy stored in the wire is \( 0.075 \, \text{J} \).

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