A) \[\frac{2}{3}M{{R}^{2}}\]
B) \[\frac{3}{2}M{{R}^{2}}\]
C) \[\frac{4}{5}M{{R}^{2}}\]
D) \[\frac{5}{4}M{{R}^{2}}\]
Correct Answer: D
Solution :
Moment of inertia of a disc about its diameter is \[{{I}_{d}}=\frac{1}{4}M{{R}^{2}}\] Now, according to perpendicular axis theorem moment of inertia of disc about a tangent passing through rim and in the plane of discs \[I={{I}_{d}}+M{{R}^{2}}\] \[=\frac{1}{4}M{{R}^{2}}+M{{R}^{2}}\] \[=\frac{5}{4}M{{R}^{2}}\]You need to login to perform this action.
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