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A thin wire of lenght $L$ is is bent into a circular wire of uniform linear density $ρ$ . When circular wire is in a vertical plane find the moment of inertia of loop about an axis $B C$ , pasing through centre of the loop and which makes an angle $θ$ with the tangent at the topmost point of the loop

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$\frac{ρ L^{3}}{8 π^{2}}$

$\frac{ρ L^{3}}{2 \sqrt{2} π^{2}}$

$\frac{ρ L^{3}}{4 π^{2}}$

$\frac{ρ L^{3}}{3 π^{2}}$

answer is A.

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Detailed Solution

$M = ρ L$ ; $L = 2 πR$ or $R = L / 2 π$

$I'_{x x} = \frac{1}{2} M R^{2}$ or $I'_{y y} = \frac{1}{2} M R^{2}$

$I_{0} = I_{z} = I'_{x x} + I'_{y y} = M R^{2}$

Now, $I_{B C} + I_{B' C'} = I_{0} = I_{z} = M R^{2}$

So $I_{B C} = \frac{M R^{2}}{2} = \frac{1}{2} (ρ L) \frac{L^{2}}{4 π^{2}} = \frac{ρ L^{2}}{8 π^{2}}$

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