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) \[3R\]               

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

Correct Answer: B

Solution :

[b]: An axis passing through \[x=2R,y=0\]is in \[\otimes \] direction as shown in figure. Moment of inertia about this axis will be \[{{I}_{1}}=\frac{1}{2}m{{R}^{2}}+m{{(2R)}^{2}}=\frac{9}{2}m{{R}^{2}}\] Axis passing through y = d, z = 0 is shown by dotted line in figure. Moment of inertia about this axis will be \[{{I}_{2}}=\frac{1}{2}m{{R}^{2}}+m{{d}^{2}}\]                           ...(ii) By equations (i) and (ii), we get \[\frac{1}{4}m{{R}^{2}}+m{{d}^{2}}=\frac{9}{2}m{{R}^{2}}\]or\[d=\frac{\sqrt{17}}{2}R\]