Two circular loops A and B of radii R and 2R respectively are made of the same wire. Their moments of inertia about the axis passing through the centre and perpendicular to their plane are $I_{A}$ and $I_{B}$ respectively. The ratio $I_{A} / I_{B}$ is

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$\frac{1}{8}$

$\frac{1}{2}$

$\frac{1}{4}$

answer is D.

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Detailed Solution

Let $μ$ be the mass per unit length of the wire. The mass of loop A is $M_{A} = 2 π R μ$ and mass of loop B is $M_{B} = 4 π R μ$ .

Their moments of inertia respectively are
$I_{A} = M_{A} R_{A}^{2} = 2 π R μ \times R^{2} = 2 π μ R^{3} I_{B} = M_{B} R_{B}^{2} = 4 π R μ \times (2 R)^{2} = 16 π μ R^{3} \frac{I_{A}}{I_{B}} = \frac{1}{8}$

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