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 inerita `I_X` and `I_Y` is

Options

A. `I_Y = 32I_X`
B. `I_Y = 16 l_X`
C. `I_Y = i_X`
D. `I_Y = 64 I_X`

Explanation

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

### Step 1: Define the Moments of Inertia
The moment of inertia \( I \) for a disc about an axis perpendicular to its plane through its center 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 \( t_X = t \)
- Volume \( V_X = \text{Area} \times \text{Thickness} = \pi R^2 t \)
- Mass \( M_X = \rho V_X = \rho \pi R^2 t \)

### Step 3: Calculate the Moment of Inertia for Disc X
Using the mass calculated:
\[ I_X = \frac{1}{2} M_X R_X^2 = \frac{1}{2} (\rho \pi R^2 t) R^2 = \frac{1}{2} \rho \pi R^4 t \]

### Step 4: Calculate the Mass of Disc Y
For disc Y:
- Radius \( R_Y = 4R \)
- Thickness \( t_Y = \frac{t}{4} \)
- Volume \( V_Y = \text{Area} \times \text{Thickness} = \pi (4R)^2 \left(\frac{t}{4}\right) = \pi (16R^2) \left(\frac{t}{4}\right) = 4\pi R^2 t \)
- Mass \( M_Y = \rho V_Y = \rho (4\pi R^2 t) \)

### Step 5: Calculate the Moment of Inertia for Disc Y
Using the mass calculated:
\[ I_Y = \frac{1}{2} M_Y R_Y^2 = \frac{1}{2} (\rho (4\pi R^2 t)) (4R)^2 = \frac{1}{2} (\rho (4\pi R^2 t)) (16R^2) = \frac{32}{2} \rho \pi R^4 t = 16 \rho \pi R^4 t \]

### Step 6: Relate \( I_X \) 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 relation:
\[ I_Y = 16 \cdot I_X \]
Thus, we can conclude:
\[ I_Y = 32 I_X \]

### Final Relation
The relation between the moments of inertia is:
\[ I_Y = 32 I_X \]