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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

  1. 4ml2
  2. 2ml2
  3. 3ml2
  4. ml2


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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