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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=l2
simillarly,
r3=lsin45=l2

r4=(AB)2+(BC)2 =(l)2+(l)2 =l2

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(l2)2+m(l2)2+m(2l)2

=ml22+ml22+2ml2

=3ml2

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