A wire having Young's modulus `2 xx 10^(11)N//m^(2)` is stretched by a force. If the energy stored per unit volume of the wire is `40 "joule"//m^(3)`, then the stress produced in the wire is

Options

A. `2 xx 10^(6) N//m^(2)`
B. `3 xx 10^(6) N//m^(2)`
C. `5 xx 10^(6) N//m^(2)`
D. `4 xx 10^(6) N//m^(2)`

Explanation

To find the stress produced in the wire, we can use the relationship between Young's modulus, stress, and energy stored per unit volume. Here's a step-by-step solution:

### Step 1: Understand the relationship
Young's modulus (Y) is defined as the ratio of stress (σ) to strain (ε):
\[ Y = \frac{\sigma}{\epsilon} \]
The energy stored per unit volume (U) in a material can be expressed in terms of stress and Young's modulus:
\[ U = \frac{1}{2} \frac{\sigma^2}{Y} \]

### Step 2: Rearranging the formula
From the energy stored per unit volume formula, we can rearrange it to find the stress:
\[ U = \frac{1}{2} \frac{\sigma^2}{Y} \]
Multiplying both sides by 2 gives:
\[ 2U = \frac{\sigma^2}{Y} \]
Now, multiplying both sides by Y:
\[ \sigma^2 = 2UY \]

### Step 3: Substitute the known values
We know:
- Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- Energy stored per unit volume, \( U = 40 \, \text{J/m}^3 \)

Substituting these values into the equation:
\[ \sigma^2 = 2 \times 40 \, \text{J/m}^3 \times (2 \times 10^{11} \, \text{N/m}^2) \]

### Step 4: Calculate the stress
Calculating the right-hand side:
\[ \sigma^2 = 2 \times 40 \times 2 \times 10^{11} \]
\[ \sigma^2 = 160 \times 10^{11} \]
Taking the square root to find stress:
\[ \sigma = \sqrt{160 \times 10^{11}} \]

### Step 5: Simplifying the square root
We can simplify \( \sqrt{160} \):
\[ \sqrt{160} = \sqrt{16 \times 10} = 4\sqrt{10} \]
Thus,
\[ \sigma = 4\sqrt{10} \times 10^{5.5} \]

### Step 6: Final calculation
Approximating \( \sqrt{10} \approx 3.16 \):
\[ \sigma \approx 4 \times 3.16 \times 10^{5.5} \]
\[ \sigma \approx 12.64 \times 10^{5.5} \]
\[ \sigma \approx 4 \times 10^{6} \, \text{N/m}^2 \]

### Conclusion
The stress produced in the wire is:
\[ \sigma \approx 4 \times 10^{6} \, \text{N/m}^2 \]