question_answer

A particle of mass m is moving along a trajectory given by \[x={{x}_{0}}+a\cos {{\omega }_{1}}t\] \[y={{y}_{0}}+b\sin {{\omega }_{2}}t\] The torque, acting on the particle about the origin, at t = 0 is :                     [JEE Main 10-4-2019 Morning]

A) \[m(-{{x}_{0}}b+{{y}_{0}}a)\omega _{1}^{2}\hat{k}\]

B) \[+m{{y}_{0}}a\omega _{1}^{2}\hat{k}\]

C) \[-m({{x}_{0}}b\omega _{2}^{2}-{{y}_{0}}a\omega _{1}^{2})\hat{k}\] D)  Zero Correct Answer: B Solution : [b]