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)

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} :moment of inertia about x & y  axis in the plane of 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 theorem of 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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