A material has Poisson's ratio `0.50`. If a uniform rod of it suffers a longitudinal strain of `2 xx 10^(-3)`, then the percentage change in volume is

Options

A. 0.6
B. 0.4
C. 0.2
D. 0.4

Explanation

To solve the problem, we need to find the percentage change in volume of a rod when it undergoes a longitudinal strain, given its Poisson's ratio.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Poisson's ratio (σ) = 0.50
- Longitudinal strain (dL/L) = 2 × 10^(-3)

2. **Relate Longitudinal Strain to Lateral Strain:**
- The relationship between longitudinal strain and lateral strain is given by the formula:
\[ \sigma = -\frac{dR/R}{dL/L} \]
- Rearranging this gives:
\[ dR/R = -\sigma \cdot \frac{dL}{L} \]

3. **Substitute Values:**
- Substituting the known values into the equation:
\[ dR/R = -0.50 \cdot (2 \times 10^{-3}) = -10^{-3} \]

4. **Volume of the Rod:**
- The volume (V) of a cylindrical rod is given by:
\[ V = \pi R^2 L \]

5. **Differentiate Volume:**
- To find the change in volume (dV), we differentiate V:
\[ dV = \pi R^2 dL + 2\pi R L dR \]

6. **Express dV/V:**
- Dividing dV by V gives:
\[ \frac{dV}{V} = \frac{\pi R^2 dL + 2\pi R L dR}{\pi R^2 L} \]
- This simplifies to:
\[ \frac{dV}{V} = \frac{dL}{L} + 2\frac{dR}{R} \]

7. **Substitute Strains:**
- Now substitute the values of dL/L and dR/R:
\[ \frac{dV}{V} = (2 \times 10^{-3}) + 2(-10^{-3}) = 2 \times 10^{-3} - 2 \times 10^{-3} = 0 \]

8. **Calculate Percentage Change in Volume:**
- The percentage change in volume is given by:
\[ \frac{dV}{V} \times 100 = 0 \times 100 = 0\% \]

### Final Answer:
The percentage change in volume is **0%**.