A) \[{{A}_{1}}{{\omega }_{1}}={{A}_{2}}{{\omega }_{2}}={{A}_{3}}{{\omega }_{3}}\]
B) \[{{A}_{1}}{{\omega }_{1}}^{2}={{A}_{2}}{{\omega }_{2}}^{2}={{A}_{3}}{{\omega }_{3}}^{2}\]
C) \[{{A}_{1}}^{2}{{\omega }_{1}}={{A}_{2}}^{2}{{\omega }_{2}}={{A}_{3}}^{2}{{\omega }_{3}}\]
D) \[{{A}_{1}}^{2}{{\omega }_{1}}^{2}={{A}_{2}}^{2}{{\omega }_{2}}^{2}={{A}^{2}}\]
Correct Answer: A
Solution :
Velocity is same. So by using \[v=a\omega \] Þ \[{{A}_{1}}{{\omega }_{1}}={{A}_{2}}{{\omega }_{2}}={{A}_{3}}{{\omega }_{3}}\]You need to login to perform this action.
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