A uniform cylindrical wire is subjected to a longitudinal tensile stress of `5 xx 10^(7) N//m^(2)`. Young's modulus of the material of the wire is `2 xx 10^(11) N//m^(2)`. The volume change in the wire is `0.02%`. The factional change in the radius is

Options

A. `0.25xx10^(-4)`
B. `0.5xx10^(-4)`
C. `1.0xx10^(-4)`
D. `1.5xx10^(-4)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have:
- Longitudinal tensile stress, \( \sigma = 5 \times 10^7 \, \text{N/m}^2 \)
- Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- Volume change, \( \Delta V/V = 0.02\% = 0.0002 \)

### Step 2: Calculate the longitudinal strain
Using the formula for Young's modulus:
\[ Y = \frac{\sigma}{\text{strain}} \]
We can express strain as:
\[ \text{strain} = \frac{\sigma}{Y} \]
Substituting the values:
\[ \text{strain} = \frac{5 \times 10^7}{2 \times 10^{11}} = 2.5 \times 10^{-4} \]
This represents the change in length per unit length, \( \frac{\Delta L}{L} \).

### Step 3: Relate volume change to radius and length
The volume \( V \) of a cylinder is given by:
\[ V = \pi R^2 L \]
The change in volume \( \Delta V \) can be expressed as:
\[ \Delta V = \pi R^2 \Delta L + \pi L (2R \Delta R) \]
Thus, the fractional change in volume is:
\[ \frac{\Delta V}{V} = \frac{\Delta V}{\pi R^2 L} \]
Substituting the expression for \( \Delta V \):
\[ \frac{\Delta V}{V} = \frac{\pi R^2 \Delta L + \pi L (2R \Delta R)}{\pi R^2 L} \]
This simplifies to:
\[ \frac{\Delta V}{V} = \frac{\Delta L}{L} + 2 \frac{\Delta R}{R} \]

### Step 4: Substitute known values
We know:
\[ \frac{\Delta V}{V} = 0.0002 \]
And from Step 2, we found:
\[ \frac{\Delta L}{L} = 2.5 \times 10^{-4} \]
Now substituting these into the equation:
\[ 0.0002 = 2.5 \times 10^{-4} + 2 \frac{\Delta R}{R} \]

### Step 5: Solve for fractional change in radius
Rearranging the equation to isolate \( \frac{\Delta R}{R} \):
\[ 2 \frac{\Delta R}{R} = 0.0002 - 2.5 \times 10^{-4} \]
Calculating the right side:
\[ 0.0002 - 0.00025 = -0.00005 \]
Thus:
\[ 2 \frac{\Delta R}{R} = -0.00005 \]
Dividing both sides by 2:
\[ \frac{\Delta R}{R} = -0.000025 \]

### Step 6: Final answer
The fractional change in the radius is:
\[ \frac{\Delta R}{R} = -0.25 \times 10^{-4} \]