A material has Poisson's ratio `0.5`, If a uniform rod of it suffers a longtiudinal strain of `2xx10^(-3)` then the percentage increases in its volume is

Options

A. `0%`
B. `10%`
C. `20%`
D. `5%`

Explanation

To solve the problem, we need to find the percentage increase in the volume of a rod given its longitudinal strain and Poisson's ratio. Let's break down the steps:

### Step 1: Understand the given values
- Poisson's ratio (σ) = 0.5
- Longitudinal strain (ΔL/L) = 2 × 10^(-3)

### Step 2: Relate longitudinal strain to lateral strain
Using the definition of Poisson's ratio:
\[ σ = -\frac{\Delta R / R}{\Delta L / L} \]
Where:
- ΔR/R is the lateral strain
- ΔL/L is the longitudinal strain

We can rearrange this to find ΔR/R:
\[ \Delta R / R = -σ \cdot \Delta L / L \]
Substituting the known values:
\[ \Delta R / R = -0.5 \cdot (2 \times 10^{-3}) = -1 \times 10^{-3} \]

### Step 3: Calculate the total strain in volume
The change in volume (ΔV/V) for a material under strain can be calculated using the formula:
\[ \frac{\Delta V}{V} = \frac{\Delta L}{L} + 2 \cdot \frac{\Delta R}{R} \]
Substituting the values we have:
\[ \frac{\Delta V}{V} = (2 \times 10^{-3}) + 2 \cdot (-1 \times 10^{-3}) \]
\[ \frac{\Delta V}{V} = 2 \times 10^{-3} - 2 \times 10^{-3} = 0 \]

### Step 4: Calculate the percentage increase in volume
Since the change in volume ratio is 0, the percentage increase in volume is:
\[ \text{Percentage increase in volume} = \frac{\Delta V}{V} \times 100\% = 0 \times 100\% = 0\% \]

### Final Answer
The percentage increase in the volume of the rod is **0%**.

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