A circular disc X of radius R is made from an iron plate of thickness t and another disc Y of radius

Q: Why do we assume the density is the same?

Both discs are made from an 'iron plate', implying they share the same material and thus the same volumetric mass density $\rho$.

Q: Does the moment of inertia depend on the thickness?

Yes, indirectly. The mass of the disc depends linearly on its thickness. Since $I = \frac{1}{2}mr^2$, a thicker disc is heavier and has a proportionally larger moment of inertia.

Q: Is there a faster way to solve this without calculating the exact mass?

Absolutely! For a uniform disc, $I = \frac{1}{2} (\pi r^2 t \rho) r^2 \propto t \cdot r^4$. For disc Y, $t$ becomes $t/4$ and $r$ becomes $4R$. So, $I_Y \propto (1/4) \cdot 4^4 = 64$. Thus, $I_Y = 64 I_X$.

Q: Why is the radius squared in the mass formula but to the power of 4 in the inertia formula?

The mass depends on the area, which scales with $r^2$. The moment of inertia multiplies mass by another $r^2$ (distance from the axis squared). So, $r^2 \times r^2 = r^4$.

Q: What if the discs were rings instead of solid plates?

For a ring, $I = mr^2$. The mass would depend on the circumference ($2\pi r$) and the cross-sectional area of the wire. The scaling would be different depending on how the wire's thickness changes.

Question

A circular disc $X$ of radius $R$ is made from an iron plate of thickness $t$ and another disc $Y$ of radius $4R$ is made from an iron plate of thickness $t/4$. Then, the relation between the moment of inertia $I_X$ and $I_Y$ is

Options:

(A) $I_Y = 32 I_X$

(B) $I_Y = 16 I_X$

(C) $I_Y = I_X$

(D) $I_Y = 64 I_X$

Sigma Percentile · Accepted Answer

Have you ever wondered why a tightrope walker carries a long, heavy pole? It's all about the **moment of inertia**—the resistance of an object to changes in its rotational motion. The further the mass is distributed from the axis of rotation, the harder it is to spin the object. In this problem, we are going to explore exactly how the moment of inertia scales when we drastically change the dimensions of a spinning disc.

### Visualizing the Setup
We are given two discs, $X$ and $Y$, both made from the same iron plate. This is a crucial piece of information because it tells us that both discs share the exact same volumetric mass density, $\rho$.

Disc $X$ is our baseline. It has a radius $R$ and a thickness $t$. 
Disc $Y$, on the other hand, is a flattened-out version. It has a much larger radius of $4R$, but it has been squished down to a thickness of just $t/4$.

Our goal is to find out how their moments of inertia, $I_X$ and $I_Y$, compare.

### Step 1: Calculating the Mass
Before we can find the moment of inertia, we need to know how much mass we are dealing with. The mass $m$ of a uniform disc is simply its volume multiplied by its density $\rho$. Since a disc is essentially a very short cylinder, its volume is the area of its circular face ($\pi r^2$) multiplied by its thickness ($h$).

$$m = (\pi r^2 h) \rho$$

Let's apply this to Disc $X$:
$$m_X = \pi R^2 t \rho$$

Now, let's look at Disc $Y$. We substitute its specific dimensions into our mass formula:
$$m_Y = \pi (4R)^2 \left(\frac{t}{4}\right) \rho$$
$$m_Y = \pi (16R^2) \left(\frac{t}{4}\right) \rho$$
$$m_Y = 4 \pi R^2 t \rho$$

Notice something interesting? Even though Disc $Y$ is a quarter of the thickness, its massive radius makes its overall volume—and therefore its mass—**four times greater** than Disc $X$!

### Step 2: The Moment of Inertia
The moment of inertia $I$ for a uniform solid disc rotating about its central perpendicular axis is given by the classic formula:
$$I = \frac{1}{2} m r^2$$

Let's calculate $I_X$ by plugging in the mass we just found:
$$I_X = \frac{1}{2} m_X R^2$$
$$I_X = \frac{1}{2} (\pi R^2 t \rho) R^2$$
$$I_X = \frac{1}{2} \pi \rho t R^4$$

Now, we do the same heavy lifting for Disc $Y$. Remember, its radius is $4R$:
$$I_Y = \frac{1}{2} m_Y (4R)^2$$
$$I_Y = \frac{1}{2} (4 \pi R^2 t \rho) (16 R^2)$$
$$I_Y = 32 \pi \rho t R^4$$

### Step 3: The Grand Ratio
We have the expressions for both $I_X$ and $I_Y$. To find their relationship, we simply divide them:
$$\frac{I_Y}{I_X} = \frac{32 \pi \rho t R^4}{\frac{1}{2} \pi \rho t R^4}$$

This is the most satisfying part of physics—watching the messy variables cancel out! The $\pi$, $\rho$, $t$, and $R^4$ all vanish, leaving us with pure numbers:
$$\frac{I_Y}{I_X} = \frac{32}{1/2} = 64$$

Therefore, **$I_Y = 64 I_X$**.

### The Pro-Tip: Scaling Laws
While the step-by-step derivation is rigorous and safe, competitive exams like JEE reward speed. Let's look at the scaling law for the moment of inertia of a disc.

If we combine the mass and inertia formulas, we get:
$$I = \frac{1}{2} (\pi r^2 t \rho) r^2 = \frac{1}{2} \pi \rho \cdot t \cdot r^4$$

Since $\frac{1}{2} \pi \rho$ is a constant for both discs, we can say that the moment of inertia is directly proportional to the thickness and the **fourth power** of the radius:
$$I \propto t \cdot r^4$$

For Disc $Y$, the thickness is scaled by a factor of $\frac{1}{4}$, and the radius is scaled by a factor of $4$. Let's plug these scaling factors in:
$$I_Y \propto \left(\frac{1}{4}\right) \cdot (4)^4$$
$$I_Y \propto \frac{256}{4} = 64$$

In just two lines of mental math, we arrive at the exact same conclusion: Disc $Y$ has 64 times the moment of inertia of Disc $X$!

Animated Solution for Physics - Rotational Motion: A circular disc $X$ of radius $R$ is made from an iron plate of thickness $t$ and another disc $Y$ of radius $4 R$ is made from an iron plate of thickness $t /4$ . Then, the relation between the moment of inertia $I_{X}$ and $I_{Y}$ is

Select Answer:

Visualized Solution

\text{Visualizing the Discs}

\text{Formulas for Mass and Moment of Inertia}

\text{Mass and Inertia of Disc } X

\text{Mass of Disc } Y

\text{Moment of Inertia of Disc } Y

\text{Comparing } I_X \text{ and } I_Y

The Sigma Insight: Moment of Inertia

Visualizing the Setup We are given two discs, $X$ and $Y$ , both made from the same iron plate

Step 1

Step 2

Step 3

The Pro-Tip

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