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. [JEE Main 12-Jan-2019 Evening]

The relative velocity \[{{\vec{v}}_{A}}-{{\vec{v}}_{B}}\]at \[t=\frac{\pi }{2\omega }\]is given by

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

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

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

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

Correct Answer: C

Solution :

[c] 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

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

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