Question

A thin wire of mass $M$ and length $L$ is bent to form a circular ring. The moment of inertia of this ring about its axis is

• A. $\frac{1}{4\pi^{2}} \,ML^{2}$
• B. $\frac{1}{12} \, ML^{2}$
• C. $\frac{1}{3\pi^{2}} \,ML^{2}$
• D. $\frac{1}{\pi^{2}} \, ML^{2}$

Answer 1

Answer: (A)

$\frac{1}{4\pi^{2}} \,ML^{2}$

Answer 2

Here, a thin wire of length $L$ is bent to form a circular ring

Then, $2\pi r=L$ [$r$ is the radius of ring]
$\Rightarrow r=\frac{L}{2\pi}$
Hence, the moment of inertia of the ring about its axis
$I=Mr^{2} \Rightarrow I=M \left(\frac{L}{2\pi}\right)^{2}$
$\Rightarrow I=\frac{ML^{2}}{4\pi^{2}}$