InterviewSolution
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InterviewSolution
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State an expression for the moment of inertia of a solid uniform disc rotating about an axis passing through its centre, perpendicular to its plane. Hence derive an expression for the moment of inertia and radius of gyration. (i) about a tangent in the plane of the disc, and (ii) about a tangent perpendicular to the plane of the disc. In a set, 21 tuning forks are arranged in a series of decresing frequencies. Each tuning fork produces 4 beats per second with the preceding fork. If the first fork is an octave of the last fork , find the frequencies of the first and tenth forks. |
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Answer» Solution :Moment of inertia for a solid uniform disc rotating about an axis passing through its centre, perpendicular to its plane, `I_(C )=(MR^(2))/(2)` `(i)` `M.I.` about DIAMETER `=(MR^(2))/(4)` `I_(t)=I_(d)+MR^(2)` [Using parallel axis theorem] `=(MR^(2))/(4)+MR^(2)=(5MR^(2))/(4)` RADIUS of gyration , `k=sqrt((I)/(M))=sqrt((5MR^(2))/(4M))` `k=(sqrt(5)R)/(2)` `(ii) I_(t)'=I_(c )+MR^(2)` [Using parallel axis theorem] `I_(t)'=(MR^(2))/(2)+MR^(2)=(3MR^(2))/(2)` `Mk'^(2)=(3MR^(2))/(2)` `k'=sqrt((3)/(2))*R` Numerical : Let, `n_(1)=`frequency of first tuning fork `n_(21)=`frequency of last tuning fork Difference in frequencies `(-d)` `=-4` beats/second `P=` number of tuning forks `n_(P)=n_(1)+(P-1)d` `n_(21)=n_(1)+(21-1)(-4)` `n_(21)=n_(1)-80`........`(i)` `n_(1)=2(n_(21))` (Given) `:.(n_(1))/(2)=n_(21)` `:. (n_(1))/(2)=n_(1)-80` [From equation `(i)`] `n_(1)=160Hz` `n_(10)=` frequency of `10th` tuning fork `n_(10)=n_(1)+(10-1)d` `n_(10)=160+9(-4)` `n_(10)=124Hz` |
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