Two discs of same mass and same thickness have densities as `17 g//cm^(3)` and `51 g//cm^(3)`. The ratio of their moment of inertia about their central axes is

Options

A. `(1)/(3)`
B. `(2)/(3)`
C. `(3)/(1)`
D. `(3)/(2)`

Answer

To find the ratio of the moment of inertia of two discs with different densities but the same mass and thickness, we can follow these steps:

### Step 1: Understand the Moment of Inertia Formula
The moment of inertia \( I \) of a disc about its central axis is given by the formula:
\[ I = \frac{1}{2} m r^2 \]
where \( m \) is the mass of the disc and \( r \) is the radius.

### Step 2: Relate Mass and Density
The mass \( m \) of the disc can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ m = \rho V \]
For a disc, the volume \( V \) is given by:
\[ V = \text{Area} \times \text{Thickness} = \pi r^2 t \]
where \( t \) is the thickness of the disc. Therefore, the mass can be rewritten as:
\[ m = \rho (\pi r^2 t) \]

### Step 3: Substitute Mass into the Moment of Inertia Formula
Substituting the expression for mass into the moment of inertia formula:
\[ I = \frac{1}{2} (\rho \pi r^2 t) r^2 = \frac{1}{2} \rho \pi t r^4 \]
This shows that the moment of inertia \( I \) is directly proportional to the density \( \rho \) and the fourth power of the radius \( r \).

### Step 4: Analyze the Given Information
We have two discs with the following properties:
- Disc 1: Density \( \rho_1 = 17 \, \text{g/cm}^3 \)
- Disc 2: Density \( \rho_2 = 51 \, \text{g/cm}^3 \)

Since both discs have the same mass and thickness, we can express the ratio of their moments of inertia as:
\[ \frac{I_1}{I_2} = \frac{\frac{1}{2} \rho_1 \pi t r_1^4}{\frac{1}{2} \rho_2 \pi t r_2^4} \]
The constants \( \frac{1}{2} \), \( \pi \), and \( t \) cancel out, leading to:
\[ \frac{I_1}{I_2} = \frac{\rho_1 r_1^4}{\rho_2 r_2^4} \]

### Step 5: Use the Relationship of Densities
Since the discs have the same mass and thickness, we can assume that the radius \( r \) is related to the density. The mass being constant implies:
\[ \rho_1 \pi r_1^2 t = \rho_2 \pi r_2^2 t \]
This implies:
\[ \rho_1 r_1^2 = \rho_2 r_2^2 \]
From this, we can derive the relationship between the radii:
\[ \frac{r_1^2}{r_2^2} = \frac{\rho_2}{\rho_1} \]

### Step 6: Substitute Back into the Moment of Inertia Ratio
Now substituting this back into our ratio of moments of inertia:
\[ \frac{I_1}{I_2} = \frac{\rho_1 r_1^4}{\rho_2 r_2^4} = \frac{\rho_1}{\rho_2} \left(\frac{r_1^2}{r_2^2}\right)^2 = \frac{\rho_1}{\rho_2} \left(\frac{\rho_2}{\rho_1}\right)^2 = \frac{\rho_1}{\rho_2} \cdot \frac{\rho_2^2}{\rho_1^2} = \frac{\rho_2}{\rho_1} \]

### Step 7: Calculate the Ratio
Substituting the values:
\[ \frac{I_1}{I_2} = \frac{51}{17} = 3 \]
Thus, the ratio of the moments of inertia is:
\[ \frac{I_1}{I_2} = 3:1 \]

### Final Answer
The ratio of their moment of inertia about their central axes is \( 3:1 \).

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