Consider two uniform discs of the same thickness and different radii $\mathrm{R}_{1}=\mathrm{R}$ and $\mathrm{R}_{2}=\alpha \mathrm{R}$ made of the same material. If the ratio of their moments of inertia $\mathrm{I}_{1}$ and $\mathrm{I}_{2}$, respectively, about their axes is $I_{1}: I_{2}=1: 16$ then the value of $\alpha$ is:

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Rotational Mechanics

Consider two uniform discs of the same thickness and different radii

R_{1} = R

and

R_{2} = α R

made of the same material. If the ratio of their moments of inertia

I_{1}

and

I_{2}

, respectively, about their axes is

I_{1} : I_{2} = 1 : 16

then the value of

α

is:

(A)4
(B) $2$
(C)2✓ Correct
(D) $22$

Answer: (C)

Solution

Both discs share the same thickness and material, so a disc's mass scales as its area,

m \propto R^{2}

, and its moment of inertia about the central axis is

I = \frac{1}{2} m R^{2} \propto R^{4}

. Hence

\frac{I _{1}}{I _{2}} = \frac{1}{α ^{4}} = \frac{1}{16}

gives

α^{4} = 16

and

α = 2

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option

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