Q88 / 180 · AIIMS 2019 (25 May) · Physics

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
    $\frac{1}{4} M L^2$
  • B
    $\frac{1}{\pi^2} M L^2$
  • Correct answer
    C
    $\frac{1}{4 \pi^2} M L^2$
  • D
    $\frac{1}{12} M L^2$

Solution

A wire of length $L$ is bent to form a circular ring of radius $R$, $\therefore \quad 2 \pi R=L$ or $R=\frac{L}{2 \pi}$ Moment of inertia of the circular ring about its axis is $I=M R^2=\frac{1}{4 \pi^2} M L^2$

Hence, the correct answer is option: 'C'.

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