A) \[\frac{4R}{3}\]
B) \[\sqrt{17}\left( \frac{R}{2} \right)\]
C) \[\sqrt{15}\left( \frac{R}{2} \right)\]
D) \[\sqrt{13}\left( \frac{R}{2} \right)\]
Correct Answer: B
Solution :
Applying perpendicular axis theorem moment of inertia of disc about the diameter is \[2{{I}_{D}}=\frac{M{{R}^{2}}}{2}\] \[{{I}_{D}}=\frac{M{{R}^{2}}}{4}\] From the parallel axis theorem, the moment of inertia at the distance = 2R from \[=\frac{M{{R}^{2}}}{2}+M{{(2R)}^{2}}=\frac{9}{2}M{{R}^{2}}\] ?(i) Now, moment of inertia about an axis parallel to \[x-\]axis is given by \[\frac{M{{R}^{2}}}{4}+M{{d}^{2}}\] ?(ii) where y = d From Eqs. (i) and (ii), we have \[\frac{M{{R}^{2}}}{4}+M{{d}^{2}}=\frac{9}{2}M{{R}^{2}}\] \[{{d}^{2}}=\frac{9{{R}^{2}}}{2}-\frac{{{R}^{2}}}{4}=\frac{17}{4}{{R}^{2}}\] \[d=\frac{\sqrt{17}}{2}R\]
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