If the elastic limit of copper is `1.5xx10^(8) N//m^(2)`,determine the minimum diameter a copper wire can have under a load of` 10.0 kg`find , if its elastic limit is not to be exceeded.

Explanation

To solve the problem of determining the minimum diameter of a copper wire that can support a load of 10 kg without exceeding its elastic limit, we can follow these steps:

### Step 1: Write down the given data
- Elastic limit of copper (\( \sigma_m \)): \( 1.5 \times 10^8 \, \text{N/m}^2 \)
- Load (\( F \)): \( 10 \, \text{kg} \)

### Step 2: Convert the load into force
The force due to the load can be calculated using the formula:
\[ F = m \cdot g \]
where \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \).

Calculating the force:
\[ F = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N} \]

### Step 3: Relate stress to force and area
The stress (\( \sigma \)) in the wire can be defined as:
\[ \sigma = \frac{F}{A} \]
where \( A \) is the cross-sectional area of the wire.

### Step 4: Express the area in terms of diameter
For a circular cross-section, the area \( A \) can be expressed in terms of the diameter \( d \):
\[ A = \frac{\pi}{4} d^2 \]

### Step 5: Set up the equation using the elastic limit
Since we want to ensure that the stress does not exceed the elastic limit, we can set up the equation:
\[ \sigma_m = \frac{F}{A} \]
Substituting the expression for area:
\[ \sigma_m = \frac{F}{\frac{\pi}{4} d^2} \]

### Step 6: Rearrange the equation to solve for diameter
Rearranging the equation gives:
\[ d^2 = \frac{4F}{\pi \sigma_m} \]
Taking the square root to find \( d \):
\[ d = \sqrt{\frac{4F}{\pi \sigma_m}} \]

### Step 7: Substitute the values into the equation
Substituting \( F = 98 \, \text{N} \) and \( \sigma_m = 1.5 \times 10^8 \, \text{N/m}^2 \):
\[ d = \sqrt{\frac{4 \times 98}{\pi \times 1.5 \times 10^8}} \]

### Step 8: Calculate the diameter
Calculating the above expression:
\[ d = \sqrt{\frac{392}{\pi \times 1.5 \times 10^8}} \]
\[ d \approx \sqrt{\frac{392}{4.712 \times 10^8}} \]
\[ d \approx \sqrt{8.32 \times 10^{-7}} \]
\[ d \approx 0.00091 \, \text{m} \]
Converting to millimeters:
\[ d \approx 0.91 \, \text{mm} \]

### Final Answer
The minimum diameter of the copper wire that can support a load of 10 kg without exceeding the elastic limit is approximately **0.91 mm**.

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