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Question

Physics Question on System of Particles & Rotational Motion

For the given uniform square lamina ABCDABCD, whose centre is OO,

A

2IAC=IEF\sqrt 2I_{AC}=I_{EF}

B

IAD=3IEFI_{AD}=3I_{EF}

C

IAD=4IEFI_{AD}=4I_{EF}

D

IAC=2IEFI_{AC}=\sqrt 2I_{EF}

Answer

IAD=4IEFI_{AD}=4I_{EF}

Explanation

Solution

Let the each side of square lamina is d. So, \hspace5mm I_{EF}=I_{GH} \hspace10mm (due \, to \, symmetry) and \hspace5mm I_{AC}=I_{BD} \hspace10mm (due \, to \, symmetry) Now, according to theorem of perpendicular axis \hspace10mm I_{AC}+I_{BD}=I_0 or \hspace10mm 2I_{AC}=I_0 \hspace5mm ...(i) and \hspace10mm I_{EF}+I_{GH}=I_0 or \hspace10mm 2I_{EF}=I_0 \hspace5mm ...(ii) From Eqs. (i) and (ii), we get IAC=IEFI_{AC}=I_{EF} \therefore \hspace5mm I_{AD}=I_{EF}+ \frac {md^2}{4} \hspace5mm =\frac {md^2}{12}+ \frac {md^2}{4} \hspace5mm \bigg (as \, I_{EF}= \frac {md^2}{12}\bigg ) So, \hspace5mm I_{AD}= \frac {md^2}{3}=4I_{EF}