The densities of two solid spheres A and B of | vedclass

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(B) The moment of inertia of a spherical shell of radius $r$ and thickness $dr$ is $dI = \frac{2}{3} (dm) r^2$. Since $dm = \rho(r) \cdot 4\pi r^2 dr$

Vedclass · Accepted Answer

$6$

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(B) The moment of inertia of a spherical shell of radius $r$ and thickness $dr$ is $dI = \frac{2}{3} (dm) r^2$. Since $dm = \rho(r) \cdot 4\pi r^2 dr$,we have $dI = \frac{8}{3} \pi \rho(r) r^4 dr$. For sphere $A$: $I_A = \int_0^R \frac{8}{3} \pi \left( k \frac{r}{R} \right) r^4 dr = \frac{8\pi k}{3R} \int_0^R r^5 dr = \frac{8\pi k}{3R} \left[ \frac{r^6}{6} \right]_0^R = \frac{8\pi k R^5}{18} = \frac{4\pi k R^5}{9}$. For sphere $B$: $I_B = \int_0^R \frac{8}{3} \pi \left( k \frac{r^5}{R^5} \right) r^4 dr = \frac{8\pi k}{3R^5} \int_0^R r^9 dr = \frac{8\pi k}{3R^5} \left[ \frac{r^{10}}{10} \right]_0^R = \frac{8\pi k R^5}{30} = \frac{4\pi k R^5}{15}$. Now,$\frac{I_B}{I_A} = \frac{4\pi k R^5 / 15}{4\pi k R^5 / 9} = \frac{9}{15} = \frac{3}{5} = \frac{6}{10}$. Comparing this with $\frac{n}{10}$,we get $n = 6$.

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