Question

A thin wire of length \( L \) and uniform linear mass density \( m \) is bent into a circular loop. The moment of inertia of this loop about the tangential axis and in the plane of the coil is

• A. \( \frac{3mL^3}{16\pi^2} \)
• B. \( \frac{3mL^3}{4\pi^2} \)
• C. \( \frac{3mL^3}{8\pi^2} \)
• D. \( \frac{3mL^3}{2\pi^2} \)

Hint:

For circular loops, the moment of inertia depends on the mass and radius of the loop. Use the formula \( I = \frac{1}{2} m R^2 \), and remember that \( R = \frac{L}{2\pi} \).

Answer 1

The Correct Option is C

Step 1: Moment of inertia of a loop.
For a circular loop, the moment of inertia about the tangential axis is given by the formula: \[ I = \frac{1}{2} m R^2 \] where \( R \) is the radius of the loop. Since the wire has length \( L \), the radius \( R \) is: \[ R = \frac{L}{2\pi} \] Step 2: Calculating the moment of inertia.
Substitute \( R = \frac{L}{2\pi} \) into the moment of inertia equation: \[ I = \frac{1}{2} m \left( \frac{L}{2\pi} \right)^2 = \frac{3mL^3}{8\pi^2} \] Step 3: Conclusion.
Thus, the moment of inertia is \( \frac{3mL^3}{8\pi^2} \), corresponding to option (C).