When a rod of length 2.5 m and radius 10 mm is subjected to a force the elongation is 3mm and decrease of radius is 0.0025 mm. Find poisson.s ratio of the material of the rod.

Answer

To find the Poisson's ratio of the material of the rod, we can use the formula for Poisson's ratio (ν):

\[ \nu = -\frac{\Delta r / r_0}{\Delta L / L_0} \]

Where:
- \(\Delta r\) = change in radius
- \(r_0\) = original radius
- \(\Delta L\) = change in length (elongation)
- \(L_0\) = original length

### Step-by-Step Solution:

1. **Identify the given values:**
- Original length of the rod, \(L_0 = 2.5 \, \text{m} = 2500 \, \text{mm}\) (since 1 m = 1000 mm)
- Original radius of the rod, \(r_0 = 10 \, \text{mm}\)
- Change in length (elongation), \(\Delta L = 3 \, \text{mm}\)
- Change in radius, \(\Delta r = -0.0025 \, \text{mm}\) (negative because it is a decrease)

2. **Substitute the values into the Poisson's ratio formula:**
\[ \nu = -\frac{\Delta r / r_0}{\Delta L / L_0} \]
\[ \nu = -\frac{-0.0025 \, \text{mm} / 10 \, \text{mm}}{3 \, \text{mm} / 2500 \, \text{mm}} \]

3. **Calculate the numerator:**
\[ \Delta r / r_0 = -0.0025 / 10 = -0.00025 \]

4. **Calculate the denominator:**
\[ \Delta L / L_0 = 3 / 2500 = 0.0012 \]

5. **Substituting back into the equation:**
\[ \nu = -\frac{-0.00025}{0.0012} \]

6. **Calculate the Poisson's ratio:**
\[ \nu = \frac{0.00025}{0.0012} \approx 0.2083 \]

7. **Round to two decimal places:**
\[ \nu \approx 0.2 \]

### Final Answer:
The Poisson's ratio of the material of the rod is approximately \(0.2\).