Two rings of radius R and nR made of same material have the ratio of moment of inertia about an axis passing through center is `1:8`. The value of n is

Options

A. (a)2
B. (b)`2sqrt2`
C. (c)4
D. (d)`1/2`

Answer

To solve the problem, we need to find the value of \( n \) given that the ratio of the moments of inertia of two rings is \( 1:8 \). The first ring has a radius \( R \) and the second ring has a radius \( nR \).

### Step-by-Step Solution:

1. **Understanding the Moment of Inertia of a Ring**:
The moment of inertia \( I \) of a ring about an axis passing through its center is given by the formula:
\[ I = mR^2 \]
where \( m \) is the mass of the ring and \( R \) is its radius.

2. **Defining the Mass of Each Ring**:
The mass of each ring can be expressed in terms of its linear density \( \lambda \) and its circumference:
- For the first ring (radius \( R \)):
\[ L_1 = 2\pi R \quad \text{(circumference)} \]
\[ m_1 = \lambda L_1 = \lambda (2\pi R) \]
- For the second ring (radius \( nR \)):
\[ L_2 = 2\pi (nR) = 2\pi nR \]
\[ m_2 = \lambda L_2 = \lambda (2\pi nR) \]

3. **Calculating the Moments of Inertia**:
- Moment of inertia for the first ring:
\[ I_1 = m_1 R^2 = (\lambda (2\pi R)) R^2 = 2\pi \lambda R^3 \]
- Moment of inertia for the second ring:
\[ I_2 = m_2 (nR)^2 = (\lambda (2\pi nR)) (nR)^2 = 2\pi \lambda n R^3 n^2 = 2\pi \lambda n^3 R^3 \]

4. **Setting Up the Ratio**:
According to the problem, the ratio of the moments of inertia is given as:
\[ \frac{I_1}{I_2} = \frac{1}{8} \]
Substituting the expressions for \( I_1 \) and \( I_2 \):
\[ \frac{2\pi \lambda R^3}{2\pi \lambda n^3 R^3} = \frac{1}{8} \]
The \( 2\pi \lambda R^3 \) terms cancel out:
\[ \frac{1}{n^3} = \frac{1}{8} \]

5. **Solving for \( n^3 \)**:
This implies:
\[ n^3 = 8 \]

6. **Finding \( n \)**:
Taking the cube root of both sides:
\[ n = \sqrt[3]{8} = 2 \]

### Conclusion:
The value of \( n \) is \( 2 \).