Question

A thin wire of length L and uniform linear mass density $\rho$ Is bent into a circular loop with centre at 0 as shown in the figure. The moment of inertia of the loop about the axis XX’ is

• A. $\frac{\rho L^3}{8{\pi }^2}$
• B. $\frac{\rho L^3}{16{\pi }^2}$
• C. $\frac{5\rho L^3}{16{\pi }^2}$
• D. $\frac{3\rho L^3}{8{\pi }^2}$

Answer 1

The correct option is D

$\frac{3\rho L^3}{8{\pi }^2}$

Mass of ring, $m=\rho L$
Now, $2\pi R=L\Rightarrow R=\frac{L}{2\pi }$
Moment of inertia about XX' = $\frac{m{R}^2}{2}+m{R}^2=\frac{3}{2}m{R}^2=\frac{3\rho L^3}{8{\pi }^2}$