question_answer

A uniform disc of radius R lies in x-y plane with its centre at origin. Its moment of inertia about the axis \[x\,=\,2R\] and \[y\,=\,0\] is equal to the moment of inertia about the axis \[y\,=\,\,d\] and \[z\,=\,\,0\], where d is equal to:

A) \[\frac{4}{3}R\]

B) \[\frac{\sqrt{17}}{2}R\]

C) \[\sqrt{13}R\]

D) \[\frac{\sqrt{15}}{2}R\]

Correct Answer: B

Solution :

An axis passing through \[x\,=\,2R\], \[y\,=\,0\] is in direction as shown in figure. Moment of inertia about this axis will be:

\[{{l}_{1}}=\frac{1}{2}m{{R}^{2}}+m{{\left( 2R \right)}^{2}}=\frac{9}{2}m{{R}^{2}}\]     ...(1)

Axis passing through \[y=d,z=0\]is shown as dotted line in figure. Moment of inertia about this axis will be:

\[{{l}_{2}}=\frac{1}{4}m{{R}^{2}}+m{{d}^{2}}\]   ...(2)

Eqs. (1) and (2), we get, \[\frac{1}{4}m{{R}^{2}}+m{{d}^{2}}=\frac{9}{2}m{{R}^{2}}\] or \[d=\frac{\sqrt{17}}{2}R\]