

InterviewSolution
Saved Bookmarks
1. |
The moment of inertia of thin square plate `ABCD` of uniform thickness about an axis passing through the center `O` and perpendicular to the plane of the plate is (`i`) `I_(1)+I_(2)` (`ii`) `I_(2)+I_(4)` (`iii`) `I_(1)+I_(3)` (`iv`) `I_(1)+I_(2)+I_(3)+I_(4)` where `I_(1)`, `I_(2)`, `I_(3)` and `I_(4)` are repectively moments of inertia about axes `1`, `2`, `3` and `4` which are in the plane of the planeA. (`i`),(`ii`)B. (`i`), (`ii`), (`iii`)C. (`ii`), (`iii`)D. (`i`), (`iii`) |
Answer» As explained in the previous problem `I_(1)=I_(2)=I_(3)=I_(4)` `I_(0)=I_(1)+I_(2)=I_(2)+I_(4)` (`bot^(ar)` axes theorem) `=I_(1)+I_(3)` |
|