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Two rotating bodies A and B of masses m ...

Two rotating bodies `A` and `B` of masses `m` and `2m` with moments of inertia `I_(A)` and `I_(B) (I_(B) gt I_(A))` have equal kinetic energy of rotation. If `L_(A)` and `L_(B)` be their angular momenta respectively, then

A

`L_(A) gt L_(B)`

B

`L_(A)=(L_(B))/(2)`

C

`L_(A)=2L_(B)`

D

`L_(B) gt L_(A)`

Text Solution

Verified by Experts

The correct Answer is:
A
`KE_(A)=KE_(B)`
`(1)/(2)I_(A)omega_(A)^(2)=(1)/(2)I_(B)omega_(B)^(2)implies` Since `I_(B) lt I_(A)` , So `omega_(B) gt omega_(A)`
`(1)/(2)L_(A)omega_(A)=(1)/(2)L_(B)omega_(B)impliesL_(B) lt L_(A)`.
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