An annular disc has mass 10 kg and inner and outer radii are 5 m and 10 m respectively. Its moment of inertia about an axis passing through its centre and perpendicular to its plane is

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The moment of inertia of an annular disc of mass $M$ , inner radius $R_{1}$ , and outer radius $R_{2}$ about an axis passing through its centre and perpendicular to its plane is given by:

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

Given $M = 10 kg$ , $R_{1} = 5 m$ , and $R_{2} = 10 m$ .

Substituting the values:

$I = \frac{1}{2} \times 10 \times (5^{2} + 1 0^{2})$

$I = 5 \times (25 + 100)$

$I = 5 \times 125 = 625 kg m^{2}$

Answer: $625 kg m^{2}$

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