Mass per unit area of a circular disc of radius 'a' depends on the distance r from its center as $σ (r) = A + Br$ . The moment of inertia of the disc about the axis, perpendicular to the plane and passing its centre is

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$2 {πa}^{4} (\frac{aA}{4} + \frac{B}{5})$

${πa}^{4} (\frac{A}{4} + \frac{aB}{5})$

$2 {πa}^{4} (\frac{A}{4} + \frac{aB}{5})$

$2 {πa}^{4} (\frac{A}{4} + \frac{B}{5})$

answer is A.

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Detailed Solution

$I = \int {dmr}^{2} = \int (σ 2 πrdr) r^{2}$

$I = 2 π \int_{o}^{a} (A + Br) r^{3} dr = 2 π (\frac{{Aa}^{4}}{4} + \frac{{Ba}^{5}}{5}) = 2 {πa}^{4} (\frac{A}{4} + \frac{Ba}{5})$

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