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There is a thin rod of uniform cross-section of mass M and length L. If this rodis is bent at 90^(@) from the mid-point, then the moment of inertia about the axis passing through mid-point and perpendicular to the plane which includes both parts of rod is .........

Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>^(2))/(<a href="https://interviewquestions.tuteehub.com/tag/24-295400" style="font-weight:bold;" target="_blank" title="Click to know more about 24">24</a>)`<br/>`(ML^(2))/(12)`<br/>`(ML^(2))/(6)`<br/>`(sqrt(2)ML^(2))/(24)` </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P1_C07_E05_060_S01.png" width="80%"/> <br/> After the <a href="https://interviewquestions.tuteehub.com/tag/rod-1190628" style="font-weight:bold;" target="_blank" title="Click to know more about ROD">ROD</a> bent at right angle <br/> mass of its energy part = `(M)/(2)` <br/> Length of every part = `(<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>)/(2)` <br/> Now total moment of inertia, <br/> I = sum of moment of inertia of both the part <br/> `therefore I=(1)/(3)((M)/(2))((L)/(2))^(2)+(1)/(3)((M)/(2))((L)/(2))^(2)` <br/> `=2((1)/(3)((M)/(2))((L)/(2))^(2))` <br/> `therefore I=(ML^(2))/(12)`</body></html>


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