Question

A uniform circular disc of radius R lies in the X- Y plane with its centre coinciding with the origin of the co-ordinate system. Its moment of inertia about an axis, lying in the X-Y 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 X-Y 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

Answer: (C)

$I_1=\frac{M{R}^2}{4}+M{(2R)}^2$ $=\frac{M{R}^2}{4}+4M{R}^2$ $=\frac{17M{R}^2}{4}$ $I_2=\frac{M{R}^2}{2}+M{d}^2$ Given, $I_1=I_2$ $\therefore$ $\frac{17M{R}^2}{4}=\frac{M{R}^2}{2}+M{d}^2$ $\Rightarrow$ $M{d}^2=\frac{17}{4}M{R}^2-\frac{M{R}^2}{2}$ $\Rightarrow$ $d^2=\frac{15}{4}{R}^2$ $\therefore$ $d=\sqrt{\frac{15}{2}}R$