Question

A wire of length \( L \) and radius \( r \) is loaded with a weight \( Mg \). If \( Y \) and \( \sigma \) denote the Young's modulus and Poisson's ratio of the material of the wire respectively, then the decrease in the radius of the wire \( \Delta r \) is given by

• A. \( \frac{MgY}{\pi \sigma} \)
• B. \( \frac{Mg \sigma}{\pi Y} \)
• C. \( \frac{\pi \sigma}{MgY} \)
• D. \( \frac{Mg \sigma}{\pi Y} \)

Hint:

Poisson's ratio helps relate the change in length and radius for materials under tension.

Answer 1

The Correct Option is B

Step 1: Using the formula for longitudinal strain.
The longitudinal strain \( \epsilon \) is given by the formula: \[ \epsilon = \frac{\Delta L}{L} = \frac{F}{A Y} \] where \( F = Mg \) is the force and \( A = \pi r^2 \) is the cross-sectional area. The change in radius \( \Delta r \) is related to the strain and Poisson's ratio \( \sigma \).
Step 2: Using Poisson's ratio.
The decrease in the radius \( \Delta r \) is given by the formula: \[ \Delta r = \frac{\Delta L \sigma}{L} \] Substituting the values, we get the result: \[ \Delta r = \frac{Mg \sigma}{\pi Y} \] Step 3: Conclusion.
Thus, the decrease in the radius of the wire is \( \frac{Mg \sigma}{\pi Y} \), which corresponds to option (B).