The ratio of `I_(AA)/I_(BB)` will be (both are identical thin rods of length L and mass M, joined together as shown in figure)

Options

A. `1:1`
B. `4:9`
C. `17:16`
D. `4:1``

Answer

C

Explanation

About `A A'=(ML^(2))/3+[(mL^(2))/12+mL^(2)]`
By parallel axis theoram.
`I A pi^(2) =(17 ml^(2))/12`
About `B B'=(mL^(2))/3+mL^(2)`
`I_("BB'")=17/4xx3/4=17/16` So, 17:16