A thin disc of mass M and radius R has mass p | vedclass

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(D) First,we find the total mass $M$ of the disc:$M = \int_0^R \sigma(r) \cdot 2\pi r \, dr = \int_0^R (kr^2) \cdot 2\pi r \, dr = 2\pi k \int_0^R r^3

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

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(D) First,we find the total mass $M$ of the disc:$M = \int_0^R \sigma(r) \cdot 2\pi r \, dr = \int_0^R (kr^2) \cdot 2\pi r \, dr = 2\pi k \int_0^R r^3 \, dr = 2\pi k \left[ \frac{r^4}{4} \right]_0^R = \frac{\pi k R^4}{2}$From this,we get $k = \frac{2M}{\pi R^4}$.Next,we calculate the moment of inertia $I$ about the central axis:$I = \int_0^R (dm) r^2 = \int_0^R (\sigma(r) \cdot 2\pi r \, dr) r^2 = \int_0^R (kr^2) \cdot 2\pi r^3 \, dr = 2\pi k \int_0^R r^5 \, dr$$I = 2\pi k \left[ \frac{r^6}{6} \right]_0^R = 2\pi k \frac{R^6}{6} = \frac{\pi k R^6}{3}$Substituting $k = \frac{2M}{\pi R^4}$ into the expression for $I$:$I = \frac{\pi}{3} \left( \frac{2M}{\pi R^4} \right) R^6 = \frac{2}{3} MR^2$

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

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