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Question

# Four point masses, each of value m are placed at the corners of square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is

A
4ml2
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B
2ml2
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C
3ml2
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D
ml2
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Solution

## The correct option is C 3ml2 From diagram, Distance of mass at A from axis r1=0 m r2=lsin45∘=l√2 simillarly, r3=lsin45∘=l√2 r4=√(AB)2+(BC)2 =√(l)2+(l)2 =l√2 Total moment of inertia of system about given axis of rotation = Sum of moment of inertia of individual particle about given axis of rotation. (I)=I1+I2+I3+I4 =mr21+mr22+mr23+mr24 =m(0)2+m(l√2)2+m(l√2)2+m(√2l)2 =ml22+ml22+2ml2 =3ml2

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