Question

A thin wire of length $l$ having linear density $\rho$ is bent into a circular loop with C as its centre, as shown in figure. The moment of inertia of the loop about the line AB is.

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

Answer 1

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

Given : Length of the wire $=l$

Linear mass density $=\rho$

$\therefore$ Mass of the wire $M=\rho l$

Also from $l=2\pi R$ $⟹R=\frac{l}{2\pi }$

Moment of inertia of the loop about an axis passing through C and perpendicular to the plane of loop is given by $I_c=M{R}^2$

MOI of loop about $PQ$ $I_1=\frac{1}{2}M{R}^2$

Let MOI of loop about axis $AB$ be $I_{AB}$

Using parallel axis theorem : $I_{AB}=I_1+M{R}^2=\frac{3}{2}M{R}^2$

$\therefore$ $I_{AB}=\frac{3}{2}\times \rho l\times \frac{l^2}{4{\pi }^2}=\frac{3}{8}\frac{\rho l^3}{{\pi }^2}$