Two loops P and Q are made from a uniform wire. The redii of P and Q are `r_(1)and r_(2)` respectively, and their moments of inertia are `I_(1)and I_(2)` respectively, If`I_(2)=4I_(1),"then" r_(2)/r_(1)"equals"`-

Options

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

Answer

To solve the problem, we need to find the ratio of the radii \( \frac{r_2}{r_1} \) given that the moment of inertia \( I_2 = 4 I_1 \).

### Step-by-step Solution:

1. **Understanding the Moment of Inertia for Circular Loops**:
The moment of inertia \( I \) for a circular loop is given by the formula:
\[ I = m \cdot r^2 \]
where \( m \) is the mass of the loop and \( r \) is the radius.

2. **Expressing Mass in Terms of Radius**:
The mass \( m \) of a loop made from a uniform wire can be expressed in terms of its circumference and linear mass density \( \lambda \):
\[ m = \text{Circumference} \cdot \lambda = 2\pi r \cdot \lambda \]
Therefore, for loop P (with radius \( r_1 \)):
\[ m_1 = 2\pi r_1 \cdot \lambda \]
and for loop Q (with radius \( r_2 \)):
\[ m_2 = 2\pi r_2 \cdot \lambda \]

3. **Calculating the Moments of Inertia**:
Now substituting the expressions for mass into the moment of inertia formula:
- For loop P:
\[ I_1 = m_1 \cdot r_1^2 = (2\pi r_1 \cdot \lambda) \cdot r_1^2 = 2\pi \lambda r_1^3 \]
- For loop Q:
\[ I_2 = m_2 \cdot r_2^2 = (2\pi r_2 \cdot \lambda) \cdot r_2^2 = 2\pi \lambda r_2^3 \]

4. **Using the Given Relation**:
We know from the problem statement that:
\[ I_2 = 4 I_1 \]
Substituting the expressions we derived:
\[ 2\pi \lambda r_2^3 = 4 \cdot (2\pi \lambda r_1^3) \]

5. **Simplifying the Equation**:
We can cancel \( 2\pi \lambda \) from both sides (assuming \( \lambda \neq 0 \)):
\[ r_2^3 = 4 r_1^3 \]

6. **Finding the Ratio of Radii**:
To find the ratio \( \frac{r_2}{r_1} \), we take the cube root of both sides:
\[ \frac{r_2}{r_1} = \sqrt[3]{4} \]

### Final Answer:
Thus, the ratio of the radii is:
\[ \frac{r_2}{r_1} = 4^{1/3} \]