Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is:
A
√3ml2
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B
3ml2
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C
ml2
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D
2ml2
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Solution
The correct option is Cml2 IA=0 (on the Axis .) IB=ID=mr2 =m(l√2)2 =ml22 IC=m(√2l)2 IC=2ml2 I=IA+IB+IC+ID I=0+ml22+ml22+2ml2 I=3ml2