Q: A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX’ is

(a) $\large \frac{\rho L^3}{8 \pi^2}$

(b) $\large \frac{\rho L^3}{16 \pi^2}$

(c) $\large \frac{5 \rho L^3}{16 \pi^2}$

(d) $\large \frac{3 \rho L^3}{8\pi^2}$

Ans: (d)

Sol: Mass of the ring = ρ L

Let R = Radius of the ring then L = 2 π R

$\large R = \frac{L}{2\pi}$

Moment of Inertia of the ring about diameter $ I_d = \frac{MR^2}{2}$

According to Parallel axis theorem;

Moment of Inertia of the ring about XX’ is

$\large I_{XX’} = I_d + MR^2$

$\large I_{XX’} = \frac{MR^2}{2} + MR^2$

$\large I_{XX’} = \frac{3}{2}M R^2 $

$\large I_{XX’} = \frac{3}{2}(\rho L) (\frac{L}{2\pi})^2 $

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