Question

A thin uniform wire of mass \( m \) and linear mass density \( \rho \) is bent in the form of a circular loop. The moment of inertia of the loop about its diameter is:

• A. \( \frac{m^2}{4\pi^2 p^2} \)
• B. \( \frac{m^3}{4p^2} \)
• C. \( \frac{m^3}{8\pi^2 p^2} \)
• D. \( \frac{m^3}{8p} \)

Hint:

Use mass density to determine the radius of the circular loop. - Use the standard formula for moment of inertia of a ring about its diameter.

Answer 1

The Correct Option is C

Step 1: Find the radius of the loop
Since \( \rho \) is linear mass density: \[ m = \rho \times 2\pi R. \] Solving for \( R \): \[ R = \frac{m}{2\pi \rho}. \] Step 2: Compute moment of inertia
For a circular ring about its diameter: \[ I = \frac{1}{2} m R^2. \] Substituting \( R = \frac{m}{2\pi \rho} \), \[ I = \frac{1}{2} m \left(\frac{m}{2\pi \rho}\right)^2. \] \[ I = \frac{m^3}{8\pi^2 p^2}. \] Thus, the correct answer is \( \boxed{\frac{m^3}{8\pi^2 p^2}} \).