Question

Two discs of the same material and thickness have radii $0.2\,m$ and $0.6\,m$.Their moments of inertia about their axes will be in the ratio
A. $1:81$
B. $1:27$
C. $1:9$
D. $1:3$

Answer 1

Applying the moment of inertia of a disc about an axis formula we can solve this question where the discs are made of the same material and thickness but of different radii. Using this information and putting in the moment of inertia formula we can find the required ratio.

Formula used:
Moment of inertia of a disc about it axis is:
$I = \dfrac{1}{2}M{R^2}$
Where, $I$ = Moment of Inertia, $M$ = Mass of the disc and $R$ = radius of the disc.

Complete step by step answer:
As per the given problem we know that, two discs are made up of the same material and the thickness and radii of the two discs is given as $0.2m$ and $0.6m$ respectively.We need to find the ratio of the moment of inertia of the two dist with the help of given information.

Let $I_1$ be the moment of inertia of the disc one having radius $0.2m$ and $I_2$ be the moment of inertia of the disc two having radius $0.6m$. As the disc is made up of the same material and thickness we can say that the density of the two dice must be the same.
Mass = Volume of the disc multiply by density
Volume of a disc $= \pi {R^2}t$
Thickness of the disc = t

Now putting in moment of inertia formula we will get,
Disc 1: $M_1 = \pi R{1^2}t \times \rho$ and $R_1 = 0.2\,m$
Putting this in $I_1$ we will get,
$I_1 = \dfrac{1}{2}M1R{1^2}$
$\Rightarrow I_1 = \dfrac{1}{2}\pi R{1^2}t \times \rho R{1^2}$
$\Rightarrow I_1 = \dfrac{1}{2}\pi t \times \rho R{1^4}$
Putting radius value we will get,
$I_1 = \dfrac{1}{2}\pi t \times \rho {\left( {0.2} \right)^4} \ldots \ldots \left( 1 \right)$

Disc 2: $M_2 = \pi R{2^2}t \times \rho$ and $R_2 = 0.6\,m$
Putting this in $I_1$ we will get,
$I_2 = \dfrac{1}{2}M2R{2^2}$
$\Rightarrow I_2 = \dfrac{1}{2}\pi R{2^2}t \times \rho R{2^2}$
$\Rightarrow I_2 = \dfrac{1}{2}\pi t \times \rho R{2^4}$
Putting radius value we will get,
$I_2 = \dfrac{1}{2}\pi t \times \rho {\left( {0.6} \right)^4} \ldots \ldots \left( 2 \right)$
Taking ratio of equation $\left( 1 \right)$ is to $\left( 2 \right)$ we will get
$\dfrac{{I_1}}{{I_2}} = \dfrac{{\dfrac{1}{2}\pi t \times \rho {{\left( {0.2} \right)}^4}}}{{\dfrac{1}{2}\pi t \times \rho {{\left( {0.6} \right)}^4}}}$

As the disc are of same material and thickness $\rho$ and $t$ remain same for two,
Cancelling the common terms we will get,
$\dfrac{{I_1}}{{I_2}} = \dfrac{{{{\left( {0.2} \right)}^4}}}{{{{\left( {0.6} \right)}^4}}}$
Dividing numerator and denominator by 2 we will get,
$\dfrac{{I_1}}{{I_2}} = \dfrac{{{{\left( {0.1} \right)}^4}}}{{{{\left( {0.3} \right)}^4}}}$
Now opening the bracket we will get,
$\dfrac{{I_1}}{{I_2}} = \dfrac{{0.0001}}{{0.0081}}$
$\Rightarrow \dfrac{{I_1}}{{I_2}} = \dfrac{{\dfrac{1}{{10000}}}}{{\dfrac{{81}}{{10000}}}}$
Cancelling the common terms we will get,
$\dfrac{{I_1}}{{I_2}} = \dfrac{1}{{81}}$
$\therefore I_1:I_2 = 1:81$

Therefore the correct option is $\left( A \right)$.

Note: Main point to learn from this problem is that whenever a body is made up of the same material then their density will remain the same. As we are calculating the ratio of the two bodies, their units cancel out and become unitless. Note that different bodies have different moments of inertia.