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The moment of inertia of a thin square p...

The moment of inertia of a thin square plate ABCD, fig, of uniform thickness about an axis passing through the centre O and perpendicular to the plane of the plate is
where `l_1, l_2, l_3 and l_4` are respectively the moments of intertial about axis 1,2,3 and 4 which are in the plane of the plate.

A

`l_1 + l_2`

B

`l_3 + l_4`

C

`l_1 + l_3`

D

`l_1 + l_2 + l_3 + l_4`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C

(a,b,c) To find the moment of inertia of ABCD about an axis
passing through the centre O and perpendiuclar to the plane
of the plate, we use perpendicular axis theorem. If we
consider ABCD to be in the X-Y plane then we know that
`I_(zz') = I_(xx') + I_(yy')`
`:. I_(zz') = I_1+ I_2`
Also, `I_(zz)` = `I_3` +`I_4`
Adding (i) and (ii),
`2I_(zz`) = I_1 + I_2 + I_3 +I_4`
But `I_2 = I_2 and I_3 = I_4`
(By symmetry)
`:. 2I_(zz') = I_1 + I_1 + I_3 + I_3`
`=2l_1 + 2l_3`
`rArr I_(zz') = I_1 + I_3.`
.
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