A mass of 5 kg is hung from a copper wire of 5 mm diameter and 2 m in length. Calculate the extension produced. What should be the minimum diameter of the wire so that its elastic limit is not exceeded ? Elastic limit for copper `=1.5xx10^(9)` dyne `cm^(-2)` and Y for copper `=1.1xx10^(12)` dyne `cm^(-2)`

Answer

To solve the problem step by step, we will first calculate the extension produced in the copper wire when a mass of 5 kg is hung from it. Then we will determine the minimum diameter of the wire so that its elastic limit is not exceeded.

### Step 1: Calculate the Force (F)
The force exerted by the mass can be calculated using the formula:
\[ F = m \cdot g \]
Where:
- \( m = 5 \, \text{kg} = 5000 \, \text{grams} \)
- \( g = 980 \, \text{cm/s}^2 \) (acceleration due to gravity in CGS units)

Calculating the force:
\[ F = 5000 \, \text{g} \cdot 980 \, \text{cm/s}^2 = 4900000 \, \text{dyne} \]

### Step 2: Calculate the Area (A) of the Wire
The area of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Where:
- Diameter \( d = 5 \, \text{mm} = 0.5 \, \text{cm} \)
- Radius \( r = \frac{d}{2} = \frac{0.5}{2} = 0.25 \, \text{cm} \)

Calculating the area:
\[ A = \pi (0.25 \, \text{cm})^2 = \pi (0.0625 \, \text{cm}^2) \approx 0.19635 \, \text{cm}^2 \]

### Step 3: Calculate the Young's Modulus (Y)
The Young's modulus for copper is given as:
\[ Y = 1.1 \times 10^{12} \, \text{dyne/cm}^2 \]

### Step 4: Calculate the Extension (ΔL)
Using the formula for extension:
\[ \Delta L = \frac{F L}{A Y} \]
Where:
- \( L = 200 \, \text{cm} \)

Substituting the values:
\[ \Delta L = \frac{4900000 \, \text{dyne} \cdot 200 \, \text{cm}}{0.19635 \, \text{cm}^2 \cdot 1.1 \times 10^{12} \, \text{dyne/cm}^2} \]

Calculating:
\[ \Delta L = \frac{980000000}{0.216985 \times 10^{12}} \]
\[ \Delta L \approx 4.54 \times 10^{-3} \, \text{cm} \]

### Step 5: Calculate the Minimum Diameter to Avoid Exceeding Elastic Limit
The elastic limit for copper is given as:
\[ \sigma = 1.5 \times 10^{9} \, \text{dyne/cm}^2 \]

Using the formula for stress:
\[ \sigma = \frac{F}{A} \]

Rearranging to find the area:
\[ A = \frac{F}{\sigma} \]

Substituting the values:
\[ A = \frac{4900000 \, \text{dyne}}{1.5 \times 10^{9} \, \text{dyne/cm}^2} \]
\[ A \approx 0.003267 \, \text{cm}^2 \]

### Step 6: Calculate the Minimum Radius (r)
Using the area to find the radius:
\[ A = \pi r^2 \]
\[ r^2 = \frac{A}{\pi} \]
\[ r = \sqrt{\frac{A}{\pi}} \]

Calculating:
\[ r = \sqrt{\frac{0.003267}{\pi}} \approx 0.032 \, \text{cm} \]

### Step 7: Calculate the Minimum Diameter (d)
The diameter is twice the radius:
\[ d = 2r = 2 \times 0.032 \, \text{cm} = 0.064 \, \text{cm} \]

### Final Answers
- The extension produced in the wire is approximately \( \Delta L \approx 4.54 \times 10^{-3} \, \text{cm} \).
- The minimum diameter of the wire to avoid exceeding the elastic limit is \( d \approx 0.064 \, \text{cm} \).