Question

A wire of weight W and area of cross-section A elongates under its own weight. If Y is the Young's modulus and \(\sigma\) is the Poisson's ratio of the material of the wire, then the fractional change in the radius of the wire is:

• A. \( \frac{2\sigma W}{AY} \)
• B. \( \frac{\sigma W}{2AY} \)
• C. \( \frac{\sigma W}{3AY} \)
• D. \( \frac{\sigma W}{AY} \)

Hint:

For vertical wires, the effective stress for elongation is half the weight divided by area, because the tension varies linearly along the length.

Answer 1

The Correct Option is B

Step 1: Find the effective stretching force.
A wire hanging under its own weight $W$ behaves as though the whole weight acts at its centre of mass, halfway down. So the effective force producing the average stretch is $\dfrac{W}{2}$.
Step 2: Write the longitudinal stress.
Stress is force over area: \[ \text{stress} = \frac{W/2}{A} = \frac{W}{2A}. \]
Step 3: Get the longitudinal strain.
Young's modulus is $Y = \dfrac{\text{stress}}{\text{longitudinal strain}}$, so the longitudinal strain (the fractional change in length) is \[ \varepsilon_L = \frac{\text{stress}}{Y} = \frac{W}{2AY}. \]
Step 4: Bring in Poisson's ratio.
Poisson's ratio $\sigma$ links the sideways strain to the lengthwise strain: \[ \sigma = \frac{\text{lateral strain}}{\text{longitudinal strain}}. \] The lateral strain here is the fractional change in radius, $\dfrac{\Delta r}{r}$.
Step 5: Solve for the fractional radius change.
Rearranging, \[ \frac{\Delta r}{r} = \sigma\,\varepsilon_L = \sigma\cdot\frac{W}{2AY}. \]
Step 6: State the result.
So the fractional change in the radius of the wire is $\dfrac{\sigma W}{2AY}$.
\[ \boxed{\dfrac{\sigma W}{2AY}} \]