Question

Moment of inertia of the rod about a perpendicular axis through its centre is given by I.
If the same rod is bent in the form of ring and its Moment of inertia of a ring about its diameter is I|. then the ratio $\dfrac{{\text{I}}}{{{{\text{I}}^{\text{1}}}}}$is
$
  A.\dfrac{{3{\pi ^2}}}{2} \\
  B.\dfrac{{8{\pi ^2}}}{3} \\
  C.\dfrac{{2{\pi ^2}}}{3} \\
  D.\dfrac{{5{\pi ^2}}}{3} \\
 $

Answer 1

Hint: Here we will be applying a formula to calculate moment of inertia of a given body.

Formula used:
Moment of inertia of the rod about a perpendicular axis through its centre is given by
$I = \dfrac{{M{L^2}}}{{12}}$
Moment of inertia of a ring about its diameter is given by, ${I^1} = \dfrac{{M{R^2}}}{2}$
where, M is the mass of the uniform rod or ring, l is the length of the uniform rod and R is the radius of the ring.

Complete step-by-step answer:
Moment of inertia I is the rotational analogue of mass. In rotation of a body about a fixed axis, the moment of inertia plays a similar role as mass does in linear motion. It is a measure of inertia of the body for rotational motion (rotational inertia). The moment of inertia of a body about a given axis of rotation is defined as the sum of the product of the masses of the particles in the body and square of the distance from the axis of rotation.
$I = M{R^2}$
Let us consider uniform rod of mass M and length L. then Moment of inertia of the rod about a perpendicular axis through its centre is given by
$I = \dfrac{{M{L^2}}}{{12}}$ ……….. (1)
It is given in the question that, if the same rod is bent in the form of a ring, then there is no change in length and mass. Hence, circumference of a ring is equal to the length of the rod.
$\therefore L = 2\pi R$
Where r is the radius of the ring
We can now consider the Moment of inertia of a ring about its diameter is given by,
 ${{\text{I}}^{\text{1}}} = \dfrac{{M{R^2}}}{2}$ …………… (2)
We shall now divide the equation (1) by (2), we get,
$\dfrac{{\text{I}}}{{{{\text{I}}^{\text{1}}}}} = \dfrac{{\dfrac{{M{L^2}}}{{12}}}}{{\dfrac{{M{R^2}}}{2}}}$
$\dfrac{{\text{I}}}{{{{\text{I}}^{\text{1}}}}} = \dfrac{{M{L^2}}}{{12}} \times \dfrac{2}{{M{R^2}}}$
$\dfrac{{\text{I}}}{{{{\text{I}}^{\text{1}}}}} = \dfrac{{{L^2}}}{{6{R^2}}} = \dfrac{{4{\pi ^2}{R^2}}}{{6{R^2}}}$
$\dfrac{{\text{I}}}{{{{\text{I}}^{\text{1}}}}} = \dfrac{{2{\pi ^2}}}{3}$
From the above equations we found the value for the moment of inertia.
Therefore, the correct option is (C).

Note:In almost all physics problems that involve rotational motions it is important to solve those problems with the rotational inertia formula.