Question

Two discs have the same mass and thickness. Their materials are of densities ${\pi _1}$ and ${\pi _2}$. The ratio of their moment of inertia about central axis will be
A. ${\pi _1}:{\pi _2}$
B. ${\pi _1}{\pi _2}:1$
C. $1:{\pi _1}{\pi _2}$
D. ${\pi _2}:{\pi _1}$

Answer 1

Hint: Through the condition of equal mass and thickness find the relation between densities and radii of the discs. Then use the definition of moment of inertia for disc to find the required ratio.

Complete answer:
Let the two discs be A and B with material densities, ${\pi _1}$ and ${\pi _2}$ and radii ${r_1}$ and ${r_2}$ respectively. Let the constant mass and thickness of the discs be $M$ and $h$.
Mass of disc A = Density of material times the volume
$\Rightarrow M = {\pi _1} \times \pi r_1^2 \times h$
Similarly for disc B, $M = {\pi _2} \times \pi r_2^2 \times h$
Since both the masses are equal,
$\Rightarrow {\pi _1} \times r_1^2 = {\pi _2} \times r_2^2$, cancelling common terms on both sides.
$\Rightarrow \dfrac{{{\pi _1}}}{{{\pi _2}}} = \dfrac{{r_2^2}}{{r_1^2}}$ ... (1)
Moment of inertia of disc A:
${I_A} = \dfrac{{Mr_1^2}}{2} = \dfrac{{{\pi _1} \times \pi r_1^2 \times h \times r_1^2}}{2}$
Moment of inertia of disc B:
${I_B} = \dfrac{{Mr_2^2}}{2} = \dfrac{{{\pi _2} \times \pi r_2^2 \times h \times r_2^2}}{2}$
We will find the ratio of above two expressions and reduce it using earlier results in (1),
$\dfrac{{{I_A}}}{{{I_B}}} = \dfrac{{{\pi _1} \times \pi r_1^2 \times h \times r_1^2}}{{{\pi _2} \times \pi r_2^2 \times h \times r_2^2}} = \dfrac{{{\pi _2}}}{{{\pi _1}}}$
Comparing this result with the given choices we can say that option D is correct.

Note: It is preferable to remember the moment of inertia of some standard bodies like, disc, ring, sphere, solid cylinder. Better practice more similar problems to absorb these relations. Deriving these relations during examinations is not worth spending time on.