Two circular loops A and B of radii R and 2R respectively are made of the similar wire. Their moments of inertia about the axis passing through the centre of perpendicular to their plane are `I_(A)" and "I_(B)` respectively. The ratio`(I_(A))/(I_(B))` is :

Options

A. `(1)/(4)`
B. `(1)/(8)`
C. `(1)/(2)`
D. 1

Answer

To find the ratio of the moments of inertia \( \frac{I_A}{I_B} \) for two circular loops A and B with radii R and 2R respectively, we can follow these steps:

### Step 1: Understand the Moment of Inertia for a Ring
The moment of inertia \( I \) of a ring about an axis perpendicular to its plane and passing through its center is given by the formula:
\[ I = m r^2 \]
where \( m \) is the mass of the ring and \( r \) is its radius.

### Step 2: Define the Masses of the Rings
Since both loops are made of similar wire, they have the same linear mass density. The mass of loop A (radius R) can be denoted as \( m_A \) and the mass of loop B (radius 2R) as \( m_B \).

Using the relationship of mass and circumference:
- For loop A:
\[ m_A = \sigma \cdot (2\pi R) \]
- For loop B:
\[ m_B = \sigma \cdot (2\pi (2R)) = \sigma \cdot (4\pi R) \]

### Step 3: Relate the Masses
Since both loops are made of similar wire, we can express the mass of loop B in terms of loop A:
\[ m_B = 2 \cdot m_A \]

### Step 4: Calculate the Moments of Inertia
Now we can calculate the moments of inertia for both loops:
- For loop A:
\[ I_A = m_A R^2 \]
- For loop B:
\[ I_B = m_B (2R)^2 = m_B \cdot 4R^2 \]

### Step 5: Substitute the Mass of Loop B
Substituting \( m_B = 2m_A \) into the equation for \( I_B \):
\[ I_B = (2m_A) \cdot 4R^2 = 8m_A R^2 \]

### Step 6: Find the Ratio of Moments of Inertia
Now we can find the ratio \( \frac{I_A}{I_B} \):
\[ \frac{I_A}{I_B} = \frac{m_A R^2}{8m_A R^2} \]
The \( m_A R^2 \) terms cancel out:
\[ \frac{I_A}{I_B} = \frac{1}{8} \]

### Final Answer
Thus, the ratio \( \frac{I_A}{I_B} \) is:
\[ \frac{I_A}{I_B} = \frac{1}{4} \]