Consider a sphere of mass M and radius R cent | vedclass

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(D) Given, density of sphere, $\rho = A r^\alpha$ (where $r$ is the radial distance and $A$ and $\alpha$ are constants).Consider an elemental spherica

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-$12$

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(D) Given, density of sphere, $ho = A r^\alpha$ (where $r$ is the radial distance and $A$ and $\alpha$ are constants).Consider an elemental spherical shell of radius $r$ and thickness $dr$.Mass of elemental spherical shell, $dm = 	ext{Volume} 	imes 	ext{Density} = (4 \pi r^2) dr \cdot A r^\alpha = 4 \pi A r^{2+\alpha} dr$.Mass of entire solid sphere, $M = 4 \pi A \int_0^R r^{2+\alpha} dr = 4 \pi A \left[ \frac{r^{3+\alpha}}{3+\alpha} ight]_0^R = \frac{4 \pi A}{3+\alpha} R^{3+\alpha}$.Moment of inertia of the elemental spherical shell is $dI = \frac{2}{3} (dm) r^2 = \frac{2}{3} (4 \pi A r^{2+\alpha} dr) r^2 = \frac{8}{3} \pi A r^{4+\alpha} dr$.Moment of inertia of the entire solid sphere, $I = \int_0^R dI = \frac{8}{3} \pi A \int_0^R r^{4+\alpha} dr = \frac{8}{3} \pi A \left[ \frac{r^{5+\alpha}}{5+\alpha} ight]_0^R = \frac{8 \pi A}{3(5+\alpha)} R^{5+\alpha}$.Substituting $M$ into the expression for $I$, we get $I = \left( \frac{4 \pi A R^{3+\alpha}}{3+\alpha} ight) \cdot \frac{2}{3} \cdot \frac{3+\alpha}{5+\alpha} R^2 = M R^2 \left( \frac{2(3+\alpha)}{3(5+\alpha)} ight)$.Given $I = \frac{6}{7} M R^2$, we equate: $\frac{2(3+\alpha)}{3(5+\alpha)} = \frac{6}{7}$.$14(3+\alpha) = 18(5+\alpha) \Rightarrow 42 + 14\alpha = 90 + 18\alpha \Rightarrow -4\alpha = 48 \Rightarrow \alpha = -12$.

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