A uniform disc of radius R lies in $xy$ plane with its centre at origin. Its moment of inertia about the axis $x=2\text{R}$  and $y=0$  is equal to the moment of inertia about the axis $y=d$  and$z=0$ . What is $d$  equal to?

The axis $x=2\text{R,}y=0$  is in Z direction as shown in. using theorem of $\parallel$ axes,

${\text{I}}_1=\frac{1}{2}m{\text{R}}^2+m{\left(2\text{R}\right)}^2=\frac{9}{2}m{\text{R}}^2$

The axis $y=d,z=0$  is shown as dotted line in. Again, using the theorem of parallel axes,

${\text{I}}_2=\frac{1}{4}m{\text{R}}^2+m{d}^2$

As ${\text{I}}_2={\text{I}}_1$

$\therefore$  $\frac{1}{4}m{\text{R}}^2+m{d}^2=\frac{9}{2}m{\text{R}}^2$

$m{d}^2=\left(\frac{9}{2}-\frac{1}{4}\right)m{\text{R}}^2=\frac{17}{4}m{\text{R}}^2$

$d=\frac{\sqrt{17}}{2}\text{R}$ .