Moment Of Inertia Of Hollow Sphere MCQ - Practice Questions with Answers

Let I = Moment of inertia of a SOLID SPHERE about an axis through its centre

 And $I_x$ = Moment of inertia of a SOLID SPHERE about x-axis through its centre

And $I_y$= Moment of inertia of a SOLID SPHERE about y-axis through its centre

And $I_z$= Moment of inertia of a SOLID SPHERE about z-axis through its centre

As hollow sphere is symmetric about any axis passing through its centre 

So it will be symmetric about the x, y, and z axis passing through its centre 

So we can say that  $I_x=I_y=I_z=I$

So take an elemental point P of  mass dm at distance R from the centre whose coordinates are (x,y,z)

And R is given as  $R=\sqrt{x^2+y^2+z^2}$

Now let's see point P on the coordinate axis as shown in figure

Let $r_x=$ Perpendicular distance of P from x-axis

And $r_z=$ Perpendicular distance of P from z-axis
From figure we can say that

$$
\begin{aligned}
& r_x=\sqrt{y^2+z^2} \\
& r_y=\sqrt{x^2+z^2} \\
& r_z=\sqrt{y^2+x^2}
\end{aligned}
$$
 

Now,

$$
\begin{aligned}
& d I_x=d m\left(r_x\right)^2 \\
& d I_y=d m\left(r_y\right)^2 \\
& d I_z=d m\left(r_z\right)^2
\end{aligned}
$$

As, $I_x+I_y+I_z=3 I$
$\mathrm{Sa}_1$

$$
\begin{aligned}
& d I_x+d I_y+d I_z=3 d I \\
& \int 2 d m\left(x^2+y^2+z^2\right)=3 \int d I \\
& 3 I=2 \int 2 d m\left(x^2+y^2+z^2\right)=2 \int d m\left(R^2\right)=2 M R^2 \\
& I=\frac{2}{3} M R^2
\end{aligned}
$$