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Question

A thin uniform metallic triangular sheet of mass M has sides AB=BC=L. Its moment of inertia about the axis AC lying in the plane of the sheet is:
  1. ML212
  2. ML23
  3. 2ML23
  4. ML26

A
ML26
B
ML23
C
ML212
D
2ML23
Solution
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The given triangular sheet has an area equal to half of a square sheet having same side length as the length of the base of the triangle.
Let MI of the triangular sheet about AC be I.
I is half of the MI of the square plate about an axis passing through the centre O and being in the plane of the plate.
Let the MI of the square about an axis XY be IXY.
Due to symmetry
IAB=ICD AND IPQ=IRS
Let Isquare be the MI a bout an axis passing through the centre of the square and perpendicular to the plane of the square plate.
Then using Perpendicular Axis Theorem
Isquare=IAB+ICD=IPQ+IRS
Isquare=2IAB=IPQ
But, IAB=2I
I=12IAB=12(12ML26) ....(1) where M is the mass of the square plate.
But the triangular sheet has half the mass of the square plate. Writing I in terms of the mass of the plate.
I=M2L2/12=112MtriangleL2

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