Question

A uniform circular disc of radius $R$ lies in the $XY$ plane with its centre coinciding with the origin of the coordinate system. Its moment of inertia about an axis, lying in the $XY$ plane, parallel to the $X$ axis and passing through a point on the $Y$ axis at a distance $y=2R$ is $I_1$. Its moment of inertia about an axis lying in a plane perpendicular to $XY$ plane passing through a point on the $X$-axis at a distance $x=d$ is $I_2$. If $I_1=I_2$, the value of $d$ is:

• A. $\frac{\sqrt{19}}{2}R$
• B. $\frac{\sqrt{17}}{2}R$
• C. $\frac{\sqrt{15}}{2}R$
• D. $\frac{\sqrt{13}}{2}R$

Answer 1

The correct option is C $\frac{\sqrt{15}}{2}R$

Moment of Inertia of disc in the plane of itself along one of its diameter is $\frac{M{R}^2}{4}$

Using parallel axis theorem

$I_1=\frac{M{R}^2}{4}+M(2R{)}^2=\frac{17M{R}^2}{4}$

Using parallel axis theorem

$I_2=\frac{M{R}^2}{2}+M{d}^2$.

equating $I_1=I_2$.

$d=\frac{\sqrt{15}R}{2}$