Ratio of Moments of Inertia for Spheres with Variable Density

To find the moment of inertia of a spherically symmetric mass distribution about an axis passing through its centre, we consider a thin spherical shell of radius $r$ and thickness $dr$.

The volume of this spherical shell is: $dV = 4\pi r^2 dr$

Its mass is: $dm = \rho(r) \, dV = 4\pi r^2 \rho(r) \, dr$

The moment of inertia of a thin spherical shell of radius $r$ and mass $dm$ about any diameter is given by: $dI = \frac{2}{3} dm \, r^2 = \frac{2}{3} \left( 4\pi r^2 \rho(r) \, dr \right) r^2 = \frac{8\pi}{3} r^4 \rho(r) \, dr$

Integrating from $r = 0$ to $r = R$, the total moment of inertia of a sphere is: $I = \frac{8\pi}{3} \int_{0}^{R} r^4 \rho(r) \, dr$

1. Moment of Inertia of Sphere $A$:

For sphere $A$, the density distribution is $\rho_A(r) = k \left( \frac{r}{R} \right)$. $I_A = \frac{8\pi}{3} \int_{0}^{R} r^4 \left[ k \left( \frac{r}{R} \right) \right] dr = \frac{8\pi k}{3R} \int_{0}^{R} r^5 \, dr$ $I_A = \frac{8\pi k}{3R} \left[ \frac{r^6}{6} \right]_{0}^{R} = \frac{8\pi k}{3R} \cdot \frac{R^6}{6} = \frac{4\pi k R^5}{9}$

2. Moment of Inertia of Sphere $B$:

For sphere $B$, the density distribution is $\rho_B(r) = k \left( \frac{r}{R} \right)^5$. $I_B = \frac{8\pi}{3} \int_{0}^{R} r^4 \left[ k \left( \frac{r}{R} \right)^5 \right] dr = \frac{8\pi k}{3R^5} \int_{0}^{R} r^9 \, dr$ $I_B = \frac{8\pi k}{3R^5} \left[ \frac{r^{10}}{10} \right]_{0}^{R} = \frac{8\pi k}{3R^5} \cdot \frac{R^{10}}{10} = \frac{4\pi k R^5}{15}$

3. Ratio of Moments of Inertia:

Taking the ratio $\frac{I_B}{I_A}$: $\frac{I_B}{I_A} = \frac{\frac{4\pi k R^5}{15}}{\frac{4\pi k R^5}{9}} = \frac{9}{15} = \frac{6}{10}$

Comparing this with the given expression $\frac{I_B}{I_A} = \frac{n}{10}$, we obtain: $n = 6$