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Question

The moments of inertia of a thin square plate ABCD of uniform thickness about an axis posing through the centre 0 and perpendicular to the plate are

(where /2, 1 and 14 are, respectively, the moments of inertia about axes I, 2, 3 and 4, where axes are in the plane of the plate)

  1. I subscript 1 plus I subscript 2
  2. I subscript 3 plus I subscript 4
  3. I subscript 1 plus I subscript 3
  4. all the above 

The correct answer is: all the above


    According to the theorem of perpendicular axis: Moment of inertia of the plate about an axis passing through centre 0 and perpendicular to the plate is I0 = I1+I2 =I3+I4 (Because diagonals 1 and 2 as well as 3 and 4 are mutually perpendicular) Now, according to symmetry,
    table attributes columnalign left columnspacing 1em rowspacing 4 pt end attributes row cell I subscript 1 equals I subscript 2 equals I left parenthesis text say end text right parenthesis end cell row cell a n d space space I subscript 3 equals I subscript 4 equals I to the power of straight prime left parenthesis text say end text right parenthesis end cell end table
    For a square plate,
    I subscript 0 equals fraction numerator M left parenthesis L squared plus L squared right parenthesis over denominator 12 end fraction equals fraction numerator M L squared over denominator 6 end fraction
    Hence I subscript 0 equals I subscript 1 plus I subscript 2 equals I subscript 3 plus I subscript 4 equals I subscript 1 plus I subscript 3
    i.e option 1,2&3 are correct.

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