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For the given uniform square lamina ABCD, whose centre is O ,

Option 1)
$I_{AC}=\sqrt{2}I_{EF}$
Option 2)
$\sqrt{2}I_{AC}=I_{EF}$

Option 3)
$I_{AD}=3I_{EF}$
Option 4)
$I_{AC}=I_{EF}$

Answers (1)

best_answer

As we learnt in

Perpendicular Axis theorem -

$I_{z}= I_{x}+I_{y}$

(for a body in XY plane )

- wherein

$I_{z}$ = moment of inertia about z axis

$I_{x}$ . $I_{y}$ : the moment of inertia about x & y axis in the plane of the body respectively.

$I_{EF}=M\frac{\left ( a^{2}+b^{2} \right )}{12}$

a=b

$I_{EF}=M\frac{\left ( a^{2}+a^{2} \right )}{12}=\frac{Ma^{2}}{6}$

$I_{Z}=\frac{Ma^{2}}{6}+\frac{Ma^{2}}{6}=\frac{Ma^{2}}{6}$

By a theorem of the perpendicular axis.

$I_{AC}+I_{BD}=I_{Z}=I_{AC}=\frac{I_{2}}{2}=\frac{Ma^{2}}{6}$

Similarly

$I_{EF}=\frac{I_{Z}}{2}=\frac{Ma^{2}}{6}$

$\therefore\ \; I_{AC}=I_{EF}$

Option 1)

$I_{AC}= \sqrt{2}I_{EF}$

This is an incorrect option.

Option 2)

$\sqrt{2}I_{AC}= I_{EF}$

This is an incorrect option.

Option 3)

$I_{AD}= 3I_{EF}$

This is an incorrect option.

Option 4)

$I_{AC}= I_{EF}$

This is the correct option.

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