(1)

Moment of inertia of a solid sphere of mass M and radius R is

          $I=\frac{2}{5}M{R}^2$

Mass of smaller sphere, $M_S=\frac{M}{8}$

Mass of remaining part, $M_r=M-\frac{M}{8}=\frac{7M}{8}$

Using parallel-axes theorem, moment of inertia of smaller sphere about y-axis,

            $I_S=\frac{2}{5}{M}_S{R}^2+M_S{R}^2=(\frac{{\displaystyle 2}}{{\displaystyle 5}}+1)M_S{R}^2=\frac{7}{5}{M}_S{R}^2$

or  $I_S=\frac{7}{5}\times \frac{M}{8}\times R^2=\frac{7}{40}M{R}^2$

Moment of inertia of the larger part of the sphere about y-axis,

         $I_L=\frac{2}{5}M{\left(2R\right)}^2=\frac{8}{5}M{R}^2$

Now, moment of inertia of the rest part of the sphere is

         $I_r=I_L-I_S=\frac{8}{5}M{R}^2-\frac{7}{40}M{R}^2=\frac{57}{40}M{R}^2$

$\therefore$    $\frac{I_S}{I_r}=\frac{\frac{7}{40}M{R}^2}{\frac{57}{40}M{R}^2}=\frac{7}{57}$

(2)

The radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane is,

    $k=\frac{R}{\sqrt{2}}$                                                    ...(i)

Radius of gyration of disc about its diameter is $k\text{'}=\frac{R}{2}$                ...(ii)

Then, $\frac{k}{k\text{'}}=\frac{\frac{R}{\sqrt{2}}}{\frac{R}{2}}=\sqrt{2}:1$

(2)

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Here: M is the mass and R is the radius of the ring.

Moment of inertia of a ring, $I=M{R}^2$

If $90°$ sector is removed, the moment of inertia of the remaining part of the ring is

$I=M{R}^2-\frac{M{R}^2}{4} \Rightarrow KM{R}^2=\frac{3}{4}M{R}^2$

Here   $K=\frac{3}{4}$

(4)

[IMAGE 66]

Mass per unit area of disc $=\frac{M}{\pi R^2}$

Mass of removed portion of disc,

$M\text{'}=\frac{M}{\pi R^2}\times \pi {\left(\frac{{\displaystyle R}}{{\displaystyle 2}}\right)}^2=\frac{M}{4}$

Moment of inertia of removed portion about an axis passing through centre of disc O and perpendicular to the plane of disc,

       $I{\text{'}}_O=I_{o\text{'}}+M\text{'}{d}^2$

$=\frac{1}{2}\times \frac{M}{4}\times {\left(\frac{{\displaystyle R}}{{\displaystyle 2}}\right)}^2+\frac{M}{4}\times {\left(\frac{{\displaystyle R}}{{\displaystyle 2}}\right)}^2$

$=\frac{M{R}^2}{32}+\frac{M{R}^2}{16}=\frac{3M{R}^2}{32}$

The moment of inertia of complete disc about centre O before removing the portion of the disc

                 $I_O=\frac{1}{2}M{R}^2$

So, moment of inertia of the disc with removed portion is

        $I=I_O-I{\text{'}}_O=\frac{1}{2}M{R}^2-\frac{3M{R}^2}{32}=\frac{13M{R}^2}{32}$

(2)

Net moment of inertia of the system,

   $I=I_1+I_2+I_3$

The moment of inertia of a shell about its diameter,

    $I_1=\frac{2}{3}m{r}^2$

The moment of inertia of a shell about its tangent is given by

$I_2=I_3=I_1+m{r}^2=\frac{2}{3}m{r}^2+m{r}^2=\frac{5}{3}m{r}^2$

$\therefore$      $I=2\times \frac{5}{3}m{r}^2+\frac{2}{3}m{r}^2=\frac{12m{r}^2}{3}=4m{r}^2$