Moment Of Inertia Of Hollow Sphere - Practice Questions & MCQ

Let I = Moment of inertia of a hollow SPHERE about an axis through its centre
And $I_x=$ Moment of inertia of a hollow SPHERE about x - axis through its centre
And $I_{y=\text { Moment of inertia of a hollow SPHERE about } \mathrm{y} \text { - axis through its centre }}$
And $I_z=$ Moment of inertia of a hollow 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 $x, y, z$ 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 $d m$ at distance $R$ from centre whose coordinates are ( $x, y, z)$
And R is given as $R=\sqrt{x^2+y^2+z^2}$

Now lets see point P on coordinate axis as shown in figure

Let $r_x=$ Perpendicular distance of P from x -axis
And $r_{y=\text { Perpendicular distance of } \mathrm{P} \text { from } \mathrm{y} \text {-axis }}$
And $r_z=$ Perpendicular distance of P from z -axis
And 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, $I_x+I_y+I_z=3 I$
So,

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