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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, thenA. `L_(A)=(L_(B))/(2)`B. `L_(A)=2L_(B)`C. `L_(B) gt L_(A)`D. `L_(A) gt L_(B)` |
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Answer» Correct Answer - C `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)]` |
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