Question

Two rings of radius \[R\] and \[nR\] made up of the same material have the ratio of moment of inertia about an axis passing through the center\[1:{\text{ }}8\] . The value of \[n\] is
A. \[2\]
B. \[\sqrt 2 \]
C. \[4\]
D. \[\dfrac{1}{2}\]

Answer 1

Hint: Use the formula of the moment of inertia for each ring. Note that the mass can be written as the product of length and linear density of wire.
Since the radius is given, Put the perimeter formula of the circle to find the length of the wire.
The ratio of moment of inertia for two rings are given, so you can easily find the required entity from the relations.

Formula used:
Moment of inertia of ring, $I = M{R^2}$
Where the mass of the ring $M$ can be written as $M = \lambda L$ $\lambda $ is the linear density and $L$ is the length.
$L = 2\pi R$
The ratio of moment of inertia, $\dfrac{{{I_1}}}{{{I_2}}} = \left( {\dfrac{{{M_1}}}{{{M_2}}}} \right) \times {\left( {\dfrac{{{R_1}}}{{{R_2}}}} \right)^2}$

Complete step by step answer:
The quantity which is gained by the result of the torque that is working on an object according to a rotational axis is divided by the acceleration that is occurred by this torque is called the moment of inertia according to the axis.

The M.I of the ring is $I = M{R^2}$
The ratio of moment of inertia of two rings given in the problem, $\dfrac{{{I_1}}}{{{I_2}}} = \left( {\dfrac{{{M_1}}}{{{M_2}}}} \right) \times {\left( {\dfrac{{{R_1}}}{{{R_2}}}} \right)^2}$
The radius of one ring is, ${R_1} = R$
The radius of one ring is, ${R_2} = nR$
The mass of the first ring, ${M_1} = \lambda {L_1}$
The mass of the second ring, ${M_2} = \lambda {L_2}$
$\lambda $ is the linear density,
${L_1}$and ${L_2}$ are the lengths of two rings respectively.
${L_1} = 2\pi R$ and ${L_2} = 2\pi nR$

So, $\dfrac{{{I_1}}}{{{I_2}}} = \left( {\dfrac{{\lambda {L_1}}}{{\lambda {L_2}}}} \right) \times {\left( {\dfrac{R}{{nR}}} \right)^2}$
$ \Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \left( {\dfrac{{2\pi R}}{{2\pi nR}}} \right) \times {\left( {\dfrac{R}{{nR}}} \right)^2}$
$ \Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{1}{{{n^3}}}$

Given, $\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{1}{8}$
$\therefore \dfrac{1}{{{n^3}}} = \dfrac{1}{8}$
$ \Rightarrow {n^3} = {2^3}$
$ \Rightarrow n = 2$
The value of $n = 2$,

Hence, the correct answer is option (A).

Additional information:
The moment of inertia is not only dependent on the mass of an object, but also the distribution of mass.
In a circular motion, the role of the moment of inertia is the same as that of the mass In linear motion.

Note: The torque is the algebraic sum of inertias active on each point of an object. Therefore, the torque,
 $\tau = (\sum m{r^2})\alpha $
 $\tau = I\alpha $
So, torque = moment of inertia × angular acceleration.
It can be said that torque is the axial vector whose direction towards the angular acceleration.