A wire of length l and mass M in the form of a circular ring. The moment of inertia of the ring about its axis is

Question

A wire of length l and mass M in the form of a circular ring. The moment of inertia of the ring about its axis is (A) (M l/4 π2) (B) (M l/2 π) (C) (M l/4 π) (D) (M l2/4 π2)

Tardigrade Admin · Accepted Answer

Wire is bent into circular ring, then radius of wire l=2 π R ⇒ R=(l/2 π) ∴ Moment of inertia of ring about its own axis ∴ I=M R2=M((l/2 π))2=(M l2/4 π2)

Q. A wire of length $l$ and mass $M$ in the form of a circular ring. The moment of inertia of the ring about its axis is

$\frac{Ml}{4 π ^{2}}$

$\frac{Ml}{2 π}$

$\frac{Ml}{4 π}$

$\frac{M l ^{2}}{4 π ^{2}}$

Wire is bent into circular ring, then radius of wire
$l = 2 π R$
$\Rightarrow R = \frac{l}{2 π}$
$∴$ Moment of inertia of ring about its own axis
$∴ I = M R^{2} = M (\frac{l}{2 π})^{2} = \frac{M l ^{2}}{4 π ^{2}}$

Q. A wire of length l and mass M in the form of a circular ring. The moment of inertia of the ring about its axis is

4π2Ml​

2πMl​

4πMl​

4π2Ml2​