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` force , if its elastic limit is not to be exceeded.

Explanation

To determine 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: Convert the load from kg to Newtons
The load applied to the wire is given as 10 kg. To convert this to Newtons, we use the formula:
\[ \text{Force (F)} = \text{mass (m)} \times \text{acceleration due to gravity (g)} \]
Here, \( g \) is approximately \( 9.8 \, \text{m/s}^2 \).

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

### Step 2: Use the elastic limit to find the stress
The elastic limit of copper is given as \( 1.5 \times 10^8 \, \text{N/m}^2 \). The stress (\( \sigma \)) in the wire due to the applied force can be expressed as:
\[ \sigma = \frac{F}{A} \]
where \( A \) is the cross-sectional area of the wire.

### Step 3: Express the cross-sectional area in terms of diameter
The cross-sectional area \( A \) of a circular wire can be calculated using the formula:
\[ A = \frac{\pi}{4} d^2 \]
where \( d \) is the diameter of the wire.

### Step 4: Set up the equation using the elastic limit
To ensure that the elastic limit is not exceeded, we set the stress equal to the elastic limit:
\[ \sigma = \frac{F}{A} \leq 1.5 \times 10^8 \, \text{N/m}^2 \]
Substituting the expression for \( A \):
\[ \frac{F}{\frac{\pi}{4} d^2} \leq 1.5 \times 10^8 \]

### Step 5: Rearrange the equation to solve for diameter \( d \)
Rearranging gives:
\[ d^2 \geq \frac{4F}{\pi \cdot (1.5 \times 10^8)} \]
Substituting \( F = 98 \, \text{N} \):
\[ d^2 \geq \frac{4 \times 98}{\pi \cdot (1.5 \times 10^8)} \]

### Step 6: Calculate \( d^2 \)
Calculating the right-hand side:
\[ d^2 \geq \frac{392}{\pi \cdot (1.5 \times 10^8)} \approx \frac{392}{4.712 \times 10^8} \approx 8.33 \times 10^{-7} \, \text{m}^2 \]

### Step 7: Find \( d \)
Taking the square root to find \( d \):
\[ d \geq \sqrt{8.33 \times 10^{-7}} \approx 9.13 \times 10^{-4} \, \text{m} \]

### Step 8: Convert \( d \) to mm
To convert meters to millimeters:
\[ d \geq 9.13 \times 10^{-4} \, \text{m} \times 1000 = 0.913 \, \text{mm} \]

### Final Answer
The minimum diameter of the copper wire that can support a load of 10 kg without exceeding the elastic limit is approximately **0.913 mm**.

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