Find the moment of inertia of a uniform disc of radius $R_{1}$ having an empty symmetric annular region of radius $R_{2}$ in between, about an axis passing through the geometrical centre and perpendicular to the disc.

I = \frac{M (R_{1}^{2} + R_{2}^{2})}{2}

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I = \frac{M (R_{1}^{2} - R_{2}^{2})}{2}

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I = \frac{M (R_{1}^{2} + R_{2}^{2})}{3}

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I = \frac{M (R_{1}^{2} - R_{2}^{2})}{3}

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Solution

The correct option is A $I = \frac{M (R_{1}^{2} + R_{2}^{2})}{2}$
$ρ = \frac{M}{π (R_{1}^{2} - R_{2}^{2})} \Rightarrow I = ρ \times (\frac{π R_{1}^{4} - π R_{2}^{4}}{2})$
$I = \frac{M (R_{1}^{2} + R_{2}^{2})}{2}$

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Find the moment of inertia of a uniform disc of radius R1 having an empty symmetric annular region of radius R2 in between, about an axis passing through the geometrical centre and perpendicular to the disc.

The correct option is A I=M(R21+R22)2ρ=Mπ(R21−R22)⇒I=ρ×(πR41−πR422)I=M(R21+R22)2

Find the moment of inertia of a uniform disc of radius $R_{1}$ having an empty symmetric annular region of radius $R_{2}$ in between, about an axis passing through the geometrical centre and perpendicular to the disc.

The correct option is A $I = \frac{M (R_{1}^{2} + R_{2}^{2})}{2}$
$ρ = \frac{M}{π (R_{1}^{2} - R_{2}^{2})} \Rightarrow I = ρ \times (\frac{π R_{1}^{4} - π R_{2}^{4}}{2})$
$I = \frac{M (R_{1}^{2} + R_{2}^{2})}{2}$