A uniform circular disc of radius R lies in the XY - plane with its centre at the origin of 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 I₁. 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 I₂. If I₁ = I₂ , the value of ‘d’ is

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$\frac{\sqrt{19}}{2} R$

$\frac{\sqrt{17}}{2} R$

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

$\frac{\sqrt{13}}{2} R$

answer is C.

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Detailed Solution

The disc's moment of inertia in its own plane along one of its diameters is $\frac{{MR}^{2}}{4}$ .

Using the parallel axis theorem

$I_{1} = \frac{{MR}^{2}}{4} + M {(2 R)}^{2} = \frac{17 {MR}^{2}}{4}$

Using parallel axis theorem

$I_{2} = \frac{{MR}^{2}}{2} + {Md}^{2}$

$If I_{1} = I_{2},$

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

Hence the correct answer is $\frac{\sqrt{15} R}{2} .$

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