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Step-by-Step Solution
Step 1: Identify Mass of Each Disc
For a disc made of iron, its mass depends on volume and density. Let the density of the iron be $d$. Then:
$M = d \times \text{Volume} = d \times (\text{Area}) \times (\text{Thickness}).$
• For disc $X$, the radius is $R$ and thickness is $t$. Therefore,
$M_X = d \times \pi R^2 \times t.$
• For disc $Y$, the radius is $4R$ and thickness is $\frac{t}{4}$. Therefore,
$M_Y = d \times \pi (4R)^2 \times \frac{t}{4} = d \times \pi \cdot 16R^2 \times \frac{t}{4} = 4\,d\,\pi\,R^2\,t.$
Step 2: Write Down the Moment of Inertia Formula for a Uniform Disc
The moment of inertia of a circular disc of mass $M$ and radius $R$ (about an axis perpendicular to the plane of the disc and passing through its center) is:
$I = \frac{1}{2}\,M\,R^2.$
Step 3: Calculate $I_X$ and $I_Y$
1) For disc $X$:
$I_X = \frac{1}{2} \times M_X \times R^2
= \frac{1}{2} \times \bigl(d \,\pi R^2 \, t\bigr) \times R^2
= \frac{d\,\pi}{2}\,t\,R^4.$
2) For disc $Y$:
$I_Y = \frac{1}{2} \times M_Y \times (4R)^2
= \frac{1}{2} \times \bigl(4\,d\,\pi\,R^2\,t\bigr) \times (4R)^2.
Note that $(4R)^2 = 16R^2$, so
$I_Y
= \frac{1}{2} \times 4\,d\,\pi\,R^2\,t \times 16R^2
= 2 \times d\,\pi\,R^2\,t \times 16R^2
= 32\,d\,\pi\,R^4\,t.
Step 4: Find the Ratio $\frac{I_Y}{I_X}$
Using the expressions for $I_X$ and $I_Y$:
$\frac{I_Y}{I_X}
= \frac{32\,d\,\pi\,R^4\,t}{\frac{d\,\pi}{2}\,t\,R^4}
= \frac{32}{\frac{1}{2}} = 32 \times 2 = 64.$
Step 5: Conclude the Relationship
Since $\frac{I_Y}{I_X} = 64$, we get:
$I_Y = 64\,I_X.$
Hence, the correct answer is:
$\boxed{I_Y = 64\,I_X.}$
