A circular disc `X` of radius `R` is made from an iron 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 mertia `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)=64I_(x)`

Explanation

To find the relation between the moments of inertia \( I_x \) and \( I_y \) for the two discs, we will follow these steps:

### Step 1: Understand the Moment of Inertia Formula
The moment of inertia \( I \) of a disc about an axis passing through its center and perpendicular to its plane is given by the formula:
\[ I = \frac{1}{2} M R^2 \]
where \( M \) is the mass of the disc and \( R \) is its radius.

### Step 2: Calculate the Mass of Disc \( X \)
For disc \( X \):
- Radius \( R_x = R \)
- Thickness \( t_x = t \)

The volume \( V_x \) of disc \( X \) is given by:
\[ V_x = \text{Area} \times \text{Thickness} = \pi R_x^2 t_x = \pi R^2 t \]

The mass \( M_x \) of disc \( X \) can be calculated using the density \( \rho \):
\[ M_x = \rho V_x = \rho \pi R^2 t \]

### Step 3: Calculate the Moment of Inertia for Disc \( X \)
Now, substituting \( M_x \) into the moment of inertia formula:
\[ I_x = \frac{1}{2} M_x R_x^2 = \frac{1}{2} (\rho \pi R^2 t) R^2 = \frac{1}{2} \rho \pi R^4 t \]

### Step 4: Calculate the Mass of Disc \( Y \)
For disc \( Y \):
- Radius \( R_y = 4R \)
- Thickness \( t_y = \frac{t}{4} \)

The volume \( V_y \) of disc \( Y \) is:
\[ V_y = \pi R_y^2 t_y = \pi (4R)^2 \left(\frac{t}{4}\right) = \pi (16R^2) \left(\frac{t}{4}\right) = 4\pi R^2 t \]

The mass \( M_y \) of disc \( Y \) is:
\[ M_y = \rho V_y = \rho (4\pi R^2 t) \]

### Step 5: Calculate the Moment of Inertia for Disc \( Y \)
Now substituting \( M_y \) into the moment of inertia formula:
\[ I_y = \frac{1}{2} M_y R_y^2 = \frac{1}{2} (\rho (4\pi R^2 t)) (4R)^2 = \frac{1}{2} (\rho (4\pi R^2 t)) (16R^2) = \frac{32}{2} \rho \pi R^4 t = 16 \rho \pi R^4 t \]

### Step 6: Find the Relation Between \( I_x \) and \( I_y \)
Now we have:
\[ I_x = \frac{1}{2} \rho \pi R^4 t \]
\[ I_y = 16 \rho \pi R^4 t \]

To find the ratio \( \frac{I_x}{I_y} \):
\[ \frac{I_x}{I_y} = \frac{\frac{1}{2} \rho \pi R^4 t}{16 \rho \pi R^4 t} = \frac{1}{2} \cdot \frac{1}{16} = \frac{1}{32} \]

Thus, we can express \( I_y \) in terms of \( I_x \):
\[ I_y = 32 I_x \]

### Final Result
The relation between the moments of inertia is:
\[ I_y = 32 I_x \]

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