A) \[{{I}_{xy}}={{I}_{x'y'}}\]
B) \[{{I}_{xy}}=\frac{1}{2}{{I}_{x'y'}}\]
C) \[{{I}_{x'y'}}=\frac{3}{4}{{I}_{xy}}\]
D) \[{{I}_{xy}}=\frac{3}{4}{{I}_{x'y'}}\]
Correct Answer: D
Solution :
[d] \[{{I}_{xy}}\], moment of inertia of a ring about its tangent in the plane of ring \[{{I}_{x'y}}=\frac{3}{2}M{{R}^{2}}\] Moment of inertia about a tangent perpendicular to the plane of ring \[{{I}_{xy}}=2M{{R}^{2}}\] \[\therefore \,\,\,{{I}_{xy}}=\frac{3}{4}(2M{{R}^{2}})\] or \[{{I}_{xy}}=\frac{3}{4}{{I}_{{{x}^{1}}{{y}^{1}}}}\]
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