Two loops P and Q are made from a uniform wire. The radii of P and Q are `R_(1)` and `R_(2)`, respectively, and their moments of inertia about their axis of rotation are `I_(1)` and `I_(2)`, respectively. If `(I_(1))/(I_(2))=4`, then `(R_(2))/(R_(1))` is

Options

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

Answer

To solve the problem, we start by using the relationship between the moment of inertia and the radius of the loops.

1. **Understanding Moment of Inertia**:
The moment of inertia \( I \) for a loop (ring) can be expressed as:
\[ I = m r^2 \]
where \( m \) is the mass of the loop and \( r \) is the radius.

2. **Mass of the Loops**:
Since both loops are made from a uniform wire, we can express the mass in terms of the linear density \( \mu \) (mass per unit length) and the circumference of the loops:
\[ m_1 = \mu \cdot L_1 = \mu \cdot (2 \pi R_1) \]
\[ m_2 = \mu \cdot L_2 = \mu \cdot (2 \pi R_2) \]

3. **Substituting Mass into Moment of Inertia**:
Now substituting the expressions for mass into the moment of inertia:
\[ I_1 = m_1 R_1^2 = (\mu \cdot 2 \pi R_1) R_1^2 = 2 \pi \mu R_1^3 \]
\[ I_2 = m_2 R_2^2 = (\mu \cdot 2 \pi R_2) R_2^2 = 2 \pi \mu R_2^3 \]

4. **Using the Given Ratio**:
We are given that:
\[ \frac{I_1}{I_2} = 4 \]
Substituting the expressions for \( I_1 \) and \( I_2 \):
\[ \frac{2 \pi \mu R_1^3}{2 \pi \mu R_2^3} = 4 \]
The \( 2 \pi \mu \) cancels out:
\[ \frac{R_1^3}{R_2^3} = 4 \]

5. **Finding the Ratio of Radii**:
Taking the cube root of both sides:
\[ \frac{R_1}{R_2} = 4^{1/3} \]
Therefore, we can express \( \frac{R_2}{R_1} \):
\[ \frac{R_2}{R_1} = \frac{1}{4^{1/3}} = 4^{-1/3} \]

Thus, the final answer is:
\[ \frac{R_2}{R_1} = 4^{-1/3} \]