Question

A uniform circular disc of radius $R$ lies in the $XY$ plane with the centre coinciding with the origin. The moment of inertia about an axis passing through a point on the $X$ axis at a distance $x=2R$ and perpendicular to the $XY$ plane is equal to its moment of inertia about an axis passing through a point on the $Y$ axis at a distance $y=d$ and parallel to the $X$ axis in the X-Y plane. The value of $d$ is:

• A. $\frac{4R}{3}$
• B. $\sqrt{17}\left(\frac{R}{2}\right)$
• C. $\sqrt{15}\left(\frac{R}{2}\right)$
• D. $\sqrt{13}\left(\frac{R}{2}\right)$

Answer 1

The correct option is B $\sqrt{17}\left(\frac{R}{2}\right)$

Moment of inertia of disc about an axis through origin and perpendicular to $X-Y$ plane is $\frac{m{R}^2}{2}$

Using Parallel Axis Theorem

$I_c=I_0+m{d}^2$ where $I_C$ is the moment about an axis parallel through axis at center of mass and passing through point C and $d$ is the distance of point C from center of mass

The moment of inertia at $x=2R$ and perpendicular to plane is $=\frac{1}{2}m{R}^2+m(2R{)}^2=\frac{9m{R}^2}{2}$

Using perpendicular axis theorem and parallel axis theorem, the moment of inertia value of any line through center of mass and in the plane is $\frac{m{R}^2}{4}$

Moment of inertia value at y=d and in the plane is $\frac{m{R}^2}{4}+m{d}^2$

By equating both,
we get $d=\sqrt{1}7\frac{R}{2}$