Two circular rings A and B of radii nR and R are made from the same wire. The Ml of A about an axis passing through the centre and perpendicular to the plane of A is 27 times that of the smaller loop B. What is the value of n if the length of A = n (length of B) ?

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To solve the problem, we need to find the value of \( n \) given the conditions about the two circular rings A and B.

### Step-by-step Solution:

1. **Understanding the Problem**:
- We have two circular rings A and B.
- The radius of ring A is \( nR \) and the radius of ring B is \( R \).
- Both rings are made from the same wire, which means they have the same mass per unit length.

2. **Relating Lengths and Masses**:
- The length of ring A, \( L_A \), is given by the circumference of the circle:
\[ L_A = 2\pi(nR) = 2\pi nR \]
- The length of ring B, \( L_B \), is:
\[ L_B = 2\pi R \]
- Since the rings are made from the same wire, the mass of ring A, \( M_A \), is proportional to its length:
\[ M_A = nM_B \]
- Here, \( M_B \) is the mass of ring B.

3. **Moment of Inertia Calculation**:
- The moment of inertia \( I \) of a ring about an axis passing through its center and perpendicular to its plane is given by:
\[ I = MR^2 \]
- For ring A:
\[ I_A = M_A (nR)^2 = nM_B (nR)^2 = nM_B n^2 R^2 = n^3 M_B R^2 \]
- For ring B:
\[ I_B = M_B R^2 \]

4. **Setting Up the Ratio**:
- According to the problem, the moment of inertia of ring A is 27 times that of ring B:
\[ I_A = 27 I_B \]
- Substituting the expressions for \( I_A \) and \( I_B \):
\[ n^3 M_B R^2 = 27 (M_B R^2) \]

5. **Simplifying the Equation**:
- We can cancel \( M_B R^2 \) from both sides (assuming \( M_B \neq 0 \)):
\[ n^3 = 27 \]

6. **Finding the Value of \( n \)**:
- Taking the cube root of both sides:
\[ n = \sqrt[3]{27} = 3 \]

### Final Answer:
The value of \( n \) is \( 3 \).