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 solve the problem of finding 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 Formula
The moment of inertia \( I \) of a circular loop about an axis perpendicular to its plane and passing through its center is given by:
\[ I = m r^2 \]
where \( m \) is the mass of the loop and \( r \) is the radius of the loop.

### Step 2: Determine the Mass of Each Loop
Since both loops are made from similar wire, the mass of each loop is proportional to its circumference. The circumference of loop A (radius \( R \)) is:
\[ C_A = 2\pi R \]
The circumference of loop B (radius \( 2R \)) is:
\[ C_B = 2\pi (2R) = 4\pi R \]
Assuming the wire has a uniform mass per unit length \( \lambda \), the mass of each loop can be expressed as:
\[ m_A = \lambda C_A = \lambda (2\pi R) = 2\pi \lambda R \]
\[ m_B = \lambda C_B = \lambda (4\pi R) = 4\pi \lambda R \]

### Step 3: Calculate the Moments of Inertia
Now we can calculate the moments of inertia for both loops:
For loop A:
\[ I_A = m_A R^2 = (2\pi \lambda R) R^2 = 2\pi \lambda R^3 \]
For loop B:
\[ I_B = m_B (2R)^2 = (4\pi \lambda R) (2R)^2 = (4\pi \lambda R) \cdot 4R^2 = 16\pi \lambda R^3 \]

### Step 4: Find the Ratio of Moments of Inertia
Now we can find the ratio \( \frac{I_A}{I_B} \):
\[ \frac{I_A}{I_B} = \frac{2\pi \lambda R^3}{16\pi \lambda R^3} \]
The \( 2\pi \lambda R^3 \) terms cancel out:
\[ \frac{I_A}{I_B} = \frac{2}{16} = \frac{1}{8} \]

### Final Answer
Thus, the ratio of the moments of inertia is:
\[ \frac{I_A}{I_B} = \frac{1}{8} \]

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