Question

A uniform metal wire of length \( L \), mass \( M \) and density \( \rho \) is under a tension \( T \). If the speed of transverse wave along the wire is \( V \), then area of cross-section of the wire is

• A. \( \frac{V}{T \rho} \)
• B. \( \frac{T}{V^2 \rho} \)
• C. \( \frac{T^2}{V \rho} \)
• D. \( \frac{V^2}{T \rho} \)

Hint:

To find the area of cross-section of a wire, use the relation between wave speed, tension, and mass per unit length.

Answer 1

The Correct Option is B

Step 1: Using the wave speed formula.
The speed of transverse wave \( V \) along a string is given by the formula: \[ V = \sqrt{\frac{T}{\mu}} \] where \( T \) is the tension and \( \mu \) is the mass per unit length, which is given by \( \mu = \frac{M}{L} = \rho A \), where \( A \) is the area of cross-section.
Step 2: Solving for \( A \).
By substituting \( \mu = \rho A \) into the wave speed formula, we get: \[ V = \sqrt{\frac{T}{\rho A}} \Rightarrow A = \frac{T}{V^2 \rho} \] Step 3: Conclusion.
The correct answer is (B), \( \frac{T}{V^2 \rho} \).