A square plate lies in the $x y$ plane with its centre at the origin and its edges parallel to the $x$ and $y$ axes. Its moments of inertia about the $x, y$ and $z$ axes are $I_{x}$ , $I_{y}$ and $I_{z}$ respectively, and about a diagonal it is $I_{D}$ . Then:

I_{x} = I_{y} = \frac{1}{2} I_{z}

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I_{x} = I_{y} = {2 I}_{2}

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I_{D} = I_{X}

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I_{D} = I_{Z}

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Solution

The correct option is C $I_{D} = I_{X}$

According to $⊥$ axis theorem,
$I_{x} + I_{y} = I_{z}$
Also, from symmetry we get $I_{x} = I_{y}$
$∴ 2 I_{x} = I_{z}$ or $I_{x} = \frac{I_{z}}{2}$
Again, from $p e r p$ axis theorem, $I_{D} + I_{D} = 2 I_{D} = I_{z}$
$\Rightarrow I_{x} = I_{D}$

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Two thin identical rods are arranged along X-axis and Y-axis with their ends meeting at the origin. If $I_{x}, I_{y} a n d I_{z}$ are the moments of inertia of the system about the X, Y and Z axes respectively, then which of the following are correct?