Question

A thin wire of length $L$ and uniform linear mass density $\rho$ is bent into a circular loop with centre at $O$ as shown. The moment of inertia of the loop about the axis $X{X}^{\prime }$ is:

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

Answer 1

The correct option is D $\frac{3\pi L^3}{8{\pi }^2}$

Moment of inertia about axis passing through center$=M{R}^2$

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

About $I{I}^{\prime }$ moment of inertia$=\frac{1}{2}M{R}^2$

About $X{X}^{\prime }=\frac{1}{2}M{R}^2+M{R}^2$

$=\frac{3}{2}M{R}^2$

Substituting value of R

Moment of inertia$=\frac{3}{2}M{\left(\frac{L}{2\pi }\right)}^2$

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

$\rho =\frac{mass}{volume}\Rightarrow M=\rho \times L$

Linear mass density$=\frac{mass}{length}$

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

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