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

Options

A. `L_(A)=(L_(B))/(2)`
B. `L_(A)=2L_(B)`
C. `L_(B) gt L_(A)`
D. `L_(A) gt L_(B)`

Answer

C

Explanation

`K.E_(A)=K.E_(B)`
`(1)/(2)I_(A)omega_(A)^(2)=(1)/(2)I_(B)omega_(B)^(2)`
`(omega_(A))/(omega_(B))=sqrt((I_(B))/(I_(A)))` . . . (i) `{I_(B) gt I_(A)implies(I_(A))/(I_(B)) lt 1}`
`L_(A)=I_(A)omega_(A)" "L_(B)=I_(B)omega_(B)`
`(L_(A))/(L_(B))=(I_(A))/(I_(B))xx(omega_(A))/(omega_(B))=(I_(A))/(I_(B))xxsqrt((I_(A))/(I_(B)))`
`=sqrt((I_(A))/(I_(B))) lt 1`
`[L_(A) lt L_(B)]`