Question

Moment of inertia of a straight wire about an axis perpendicular to the wire and passing through one of its end is $I$. This wire is now framed into a circle (a ring) of single turn. The moment of inertia of this ring about an axis passing through centre and perpendicular to its plane would be:

• A. $\left(\frac{3}{{\pi }^2}\right)I$
• B. $\left(\frac{3}{{4\pi }^2}\right)I$
• C. $\left(\frac{{\pi }^2}{3}\right)I$
• D. $\left(\frac{{4\pi }^2}{3}\right)I$

Answer 1

The correct option is B $\left(\frac{3}{{4\pi }^2}\right)I$

For a wire of length l and mass m its moment of inertia about an axis passing through its one end and perpendicular to its length is $m{l}^2/3$ so $I=m{l}^2/3$ or $m{l}^2=3I$

Now when this wire is changed in a ring then the circumference of the circle is $2\pi r=length=l$ so radius $r=l/2\pi$

Now Moment of Inertia of a ring of radius r and mass m about an axis passing through its centre and perpendicular to its plane is $m{r}^2$

putting value of r in terms of l we get new M.O.I as $\left(m\right)l^2/4{\pi }^2=m{l}^2/4{\pi }^2$

or $3I/4{\pi }^2$

Option B is correct.