Question

A circular disc A of radius $r$ is made from an iron plate of thickness t and another circular disc B of radius $4r$ is made from an iron plate of thickness $\dfrac{t}{4}$. The relation between the moments of inertia ${I_A}$ and ${I_B}$ is:
A. ${I_A} > {I_B}$
B. ${I_A} = {I_B}$
C. ${I_A} < {I_B}$
D. the conditions are insufficient to predict the relation between ${I_A}$ and ${I_B}$ .

Answer 1

Hint-By finding the value of moment of inertia of disc $A$ and $B$ about an axis passing through the centre perpendicular to the plane of the disc and on comparing these values we can arrive at the correct answer.

Complete step by step answer:
It is given that a circular disc A and B are made from an iron plate
Radius of disc A ,
${R_A} = r$
Thickness of disc A,
${t_A} = t$
Disc B has radius
 ${R_B} = 4r$
Thickness of disc B is
${t_B} = \dfrac{t}{4}$
We know that the moment of inertia of disc about an axis passing through the centre and perpendicular to the disc is given as,
$I = \dfrac{1}{2}M{R^2}$
Where, $M$ is the mass and $R$ is the radius,
Now we need to find the mass of disc $A$ and $B$
We know that density is the ratio of mass to volume,
$\rho = \dfrac{m}{V}$
Where $m$ is the mass and $V$ is the volume.
Volume of disc will be area multiplied by thickness.
$V = At$
Thus, for disc A
${M_A} = \rho \times {V_A}$
$ \Rightarrow {M_A} = \rho \times {A_A} \times {t_A}$
Area is the area of a circle $\pi {r^2}$. Where, r is the radius.
$ \Rightarrow {M_A} = \rho \times \pi {r^2} \times t$
Similarly, mass of disc B
${M_B} = \rho \times {V_B}$
$ \Rightarrow {M_b} = \rho \times {A_B} \times {t_B}$
$ \Rightarrow {M_B} = \rho \times \pi {\left( {4r} \right)^2} \times \left( {\dfrac{t}{4}} \right)$
$ \Rightarrow {M_B} = \rho \times 4\pi {r^2}t$
Now we can calculate the moment of inertia of disc A
${I_A} = \dfrac{{{M_A}{R_A}^2}}{2}$
On substituting the values, we get
$ \Rightarrow {I_A} = \dfrac{{\left( {\rho \pi {r^2}t} \right) \times {r^2}}}{2}$
Moment of inertia of disc $B$
${I_B} = \dfrac{{{M_B}{R_B}^2}}{2}$
On substituting the values we get,
$ \Rightarrow {I_B} = \dfrac{{\left( {4\rho \pi {r^2}t} \right) \times {{\left( {4r} \right)}^2}}}{2}$
$ \Rightarrow {I_B} = \dfrac{{64 \times \rho \pi {r^2}t \times {r^2}}}{2}$
Let us divide equation 1 by equation 2. Then we get
$ \Rightarrow \dfrac{{{I_A}}}{{{I_B}}} = \dfrac{{\dfrac{{\left( {\rho \pi {r^2}t} \right) \times {r^2}}}{2}}}{{\dfrac{{64 \times \rho \pi {r^2}t \times {r^2}}}{2}}}$
$ \Rightarrow \dfrac{{{I_A}}}{{{I_B}}} = \dfrac{1}{{64}}$
$ \Rightarrow {I_B} = 64{I_A}$
We can see that moment of inertia of B is 64 times the moment of inertia of A.
Hence ${I_B} > {I_A}$

So the correct answer is option C.

Note:The moment of inertia of the disc varies according to the axis of rotation. The moment of inertia about axis passing through centre and perpendicular to the plane of disc is given as
 $I = \dfrac{{MR}}{2}$
Where, $M$ is the mass and $R$ is the radius.
If any other axis of rotation is to be taken then the moment of inertia about that axis will be different from this equation.