If `I_(1)` is the moment of inertia of a uniform rod about an axis perpendicular to its length and passing through its one end. Now ring formed by bending the rod, if the moment of inertia about the diameter of ring `I_(1)` then `(I_(1))/(I_(2))` = ……….

Options

A. `(pi^(2))/(3)`
B. `(2pi^(2))/(3)`
C. `(4pi^(2))/(3)`
D. `(8pi^(2))/(3)`

Answer

D

Explanation

Suppose the length of rod is l. Its moment of inertia about axis perpendicular to the rod and passing from one end of rod is :
`I_(1)=I_(0)+Md^(2)` [where M = mass, `d=(l)/(2)`]
`=(Ml^(2))/(12)+M((l)/(2))^(2)`
`=(Ml^(2))/(12)+(Ml^(2))/(4)`
`=(Ml^(2)+3Ml^(2))/(12)`
`=(4Ml^(2))/(12)`
`therefore I_(1)=(Ml^(2))/(3)" "...(1)`
Radius of ring formed by rod of lengt l is r then `l=2pir`
`therefore r=(l)/(2pi)" "....(2)`
`therefore` Moment of inertia of ring about an axis passing through the centre and perpendicular the plane of ring.
`I_(2)=(Mr^(2))/(2)`
`=(Ml^(2))/(2(4pi^(2)))` [`because` from result (2)]
`therefore I_(2)=(Ml^(2))/(8pi^(2))`
`therefore (I_(1))/(I_(2))=(Ml^(2))/(3)xx(8pi^(2))/(Ml^(2))`
`=(8pi^(2))/(3)`