A thin disc of mass M and radius R has mass per unit area $σ (r) = {kr}^{2}$ where r is the distance from its centre. Its moment of inertia about an axis going through its center of mass and perpendicular to its plane is

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$\frac{{MR}^{2}}{2}$

$\frac{{MR}^{2}}{3}$

$\frac{2 {MR}^{2}}{3}$

$\frac{{MR}^{2}}{6}$

answer is B.

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Detailed Solution

$M = \int_{o}^{R} ({kr}^{2}) 2 πrdr = \frac{{πkR}^{4}}{2}$

$\Rightarrow k = \frac{2 M}{{πR}^{4}}$

$I = \int dI = \int r^{2} dm = \int {kr}^{2} (2 πrdr) r^{2}$

$= \frac{{πkR}^{6}}{3} = \frac{π (\frac{2 M}{{πR}^{4}}) \times R^{6}}{3} = \frac{2}{3} {MR}^{2}$

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