Ratio of Moments of Inertia for Sphere and Disc

To find the ratio of the moments of inertia $I_2/I_1$, we analyze the two newly formed bodies step-by-step:

1. Analysis of the Smaller Sphere:

Let the original solid sphere have mass $M$, radius $R$, and uniform density $\rho$.

• The mass of the smaller part is given as: $m_1 = \frac{M}{8}$

• Assuming the density $\rho$ remains constant, the volume $V_1$ of the smaller sphere of radius $r$ is: $V_1 = \frac{4}{3}\pi r^3 = \frac{m_1}{\rho} = \frac{M}{8\rho} = \frac{1}{8}\left(\frac{4}{3}\pi R^3\right)$ $\implies r^3 = \frac{R^3}{8} \implies r = \frac{R}{2}$

• The moment of inertia $I_1$ of this smaller solid sphere about an axis through its centre is: $I_1 = \frac{2}{5} m_1 r^2$ Substitute $m_1 = \frac{M}{8}$ and $r = \frac{R}{2}$: $I_1 = \frac{2}{5} \left(\frac{M}{8}\right) \left(\frac{R}{2}\right)^2 = \frac{2}{5} \times \frac{M}{8} \times \frac{R^2}{4} = \frac{M R^2}{80}$

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2. Analysis of the Circular Disc:

• The mass of the remaining larger part is: $m_2 = M - m_1 = M - \frac{M}{8} = \frac{7M}{8}$

• The radius of the disc is given as $R_d = 2R$.

• The moment of inertia $I_2$ of a thin circular disc about any of its diameters is given by: $I_2 = \frac{1}{4} m_2 R_d^2$ Substitute $m_2 = \frac{7M}{8}$ and $R_d = 2R$: $I_2 = \frac{1}{4} \left(\frac{7M}{8}\right) (2R)^2 = \frac{1}{4} \times \frac{7M}{8} \times 4R^2 = \frac{7}{8} M R^2$

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3. Ratio of Moments of Inertia:

Now, taking the ratio of $I_2$ to $I_1$: $\frac{I_2}{I_1} = \frac{\frac{7}{8} M R^2}{\frac{1}{80} M R^2} = \frac{7}{8} \times 80 = 70$

Thus, the correct option is B (value is 70).