d mass $$M$$, the moment of inertia about its own axis can be given by:

$$ \mathrm{I} = M R^2 $$

Given two loops P and Q with radii $$R_1$$ and $$R_2$$ respectively, and their moments of inertia $$\mathrm{I_P}$$ and $$\mathrm{I_Q}$$, we have the following relationship:

$$ \frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I_Q}} = \frac{1}{8} $$

Using the formula for the moment of inertia, we can write:

$$ \mathrm{I_P} = M_1 R_1^2 $$

$$ \mathrm{I_Q} = M_2 R_2^2 $$

Since the loops are made of the same uniform wire, the mass (which is proportional to the length of the wire used) will be different for different radii. The mass of a loop is proportional to the circumference so:

$$ M_1 \propto 2 \pi R_1 $$

$$ M_2 \propto 2 \pi R_2 $$

We can express the masses in terms of a common unit length mass $$\lambda$$:

$$ M_1 = \lambda (2 \pi R_1) $$

$$ M_2 = \lambda (2 \pi R_2) $$

Substitute these into the moments of inertia:

$$ \mathrm{I_P} = \lambda (2 \pi R_1) R_1^2 = 2 \pi \lambda R_1^3 $$

$$ \mathrm{I_Q} = \lambda (2 \pi R_2) R_2^2 = 2 \pi \lambda R_2^3 $$

Using the given ratio:

$$ \frac{ \mathrm{I_P} }{ \mathrm{I_Q} } = \frac{ 2 \pi \lambda R_1^3 }{ 2 \pi \lambda R_2^3 } = \frac{R_1^3}{R_2^3} $$

Given that:

$$ \frac{R_1^3}{R_2^3} = \frac{1}{8} $$

We take the cube root of both sides:

$$ \frac{R_1}{R_2} = \sqrt[3]{\frac{1}{8}} $$

Simplify the cube root:

$$ \frac{R_1}{R_2} = \frac{1}{2} $$

Therefore:

$$ \frac{R_2}{R_1} = 2 $$

So the correct answer is Option C: 2.

Rotational Motion

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