Question

Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is $I$. If the same rod is bent into a ring and its moment of inertia about its diameter is $I'$, then the ratio ${\frac {I}{I'}}$ is

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

Answer 1

Answer: (A)

$\frac{2}{3} \,\pi^{2}$

Answer 2

We know that, radius of ring,
$R=\frac{L}{2 \pi}\,\,\,\,\,\dots(i)$
Moment of inertia of thin uniform rod,
$I=\frac{M L^{2}}{12}\,\,\,\,\,\dots(ii)$
and same rod is bent into a ring, then its moment of inertia,
$I'=\frac{1}{2} \,M R^{2}$
From E (i).
$I'=\frac{1}{2} \frac{M L^{2}}{4 \pi^{2}}$
$I'=\frac{M L^{2}}{8 \pi^{2}}$
On dividing E (ii) by E (iii), we get
$\frac{I}{I'}=\frac{M L^{2}}{12} \times \frac{8 \pi^{2}}{M L^{2}}$
$\frac{I}{I'}=\frac{8 \pi^{2}}{12}$
$\frac{I}{I'}=\frac{2 \pi^{2}}{3}$