question_answer

Two particles A, B are moving on two concentric circles of radii \[{{R}_{1}}\]and \[{{R}_{2}}\]with equal angular speed\[\omega \]. At t = 0, their positions and direction of motion are shown in the figure. The relative velocity \[{{\vec{v}}_{A}}-{{\vec{v}}_{B}}\]at \[t=\frac{\pi }{2\omega }\]is given by [JEE Main Online Paper Held On 12-Jan-2019 Evening]

A) \[-\omega ({{R}_{1}}+{{R}_{2}})\hat{i}\]                                

B) \[\omega ({{R}_{1}}-{{R}_{2}})\hat{i}\]

C) \[\omega ({{R}_{2}}-{{R}_{1}})\hat{i}\]                       

D)   \[-\omega ({{R}_{1}}+{{R}_{2}})\hat{i}\]

Correct Answer: C

Solution :

The angle transversed in time \[\frac{\pi }{2\omega }\]is\[\theta =\omega t=\frac{\omega \pi }{2\omega }=\frac{\pi }{2}\]i.e., at\[t=\frac{\pi }{2\omega },\]the position of two particles is shown in the figure \[\therefore \] The relative velocity\[{{\vec{v}}_{A}}-{{\vec{v}}_{B}}\]is \[=-{{R}_{1}}\omega \hat{i}-(-{{R}_{2}}\omega \hat{i})\] \[=\omega ({{R}_{2}}-{{R}_{1}})\hat{i}\]