From a given sample of uniform wire, two circular loops `P` and `Q` are made, `P` of radius `r` and `Q` of radius `nr`. If the M.I. of `Q` about its axis is four times that of `P` about its axis (assuming the wire to be of diameter much smaller than either radius), the value of n is

Options

A. `(4)^(2/3)`
B. `(4)^(1/3)`
C. `(4)^(1/2)`
D. `(4)^(1/4)`

Explanation

To solve the problem, we need to find the value of \( n \) given the relationship between the moment of inertia of two circular loops made from a uniform wire.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two circular loops, \( P \) with radius \( r \) and \( Q \) with radius \( nr \).
- The moment of inertia (M.I.) of loop \( Q \) is four times that of loop \( P \).

2. **Mass of Each Loop**:
- The mass of loop \( P \) can be calculated using its circumference and linear mass density \( \lambda \):
\[ m_P = 2\pi r \lambda \]
- The mass of loop \( Q \) is:
\[ m_Q = 2\pi (nr) \lambda = 2\pi nr \lambda \]

3. **Moment of Inertia Formula**:
- The moment of inertia \( I \) of a circular loop about its axis is given by:
\[ I = m r^2 \]
- Therefore, for loop \( P \):
\[ I_P = m_P r^2 = (2\pi r \lambda) r^2 = 2\pi r^3 \lambda \]
- For loop \( Q \):
\[ I_Q = m_Q (nr)^2 = (2\pi nr \lambda) (nr)^2 = 2\pi nr \lambda n^2 r^2 = 2\pi n^3 r^3 \lambda \]

4. **Setting Up the Equation**:
- According to the problem, the moment of inertia of \( Q \) is four times that of \( P \):
\[ I_Q = 4 I_P \]
- Substituting the expressions we found:
\[ 2\pi n^3 r^3 \lambda = 4 (2\pi r^3 \lambda) \]

5. **Simplifying the Equation**:
- We can cancel \( 2\pi r^3 \lambda \) from both sides (assuming they are not zero):
\[ n^3 = 4 \]

6. **Finding \( n \)**:
- Taking the cube root of both sides:
\[ n = 4^{1/3} \]

### Final Answer:
The value of \( n \) is \( 4^{1/3} \).