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(D) Case $1$: When both move in opposite directions.Let the angular positions be $	heta_1 = \omega t$ and $	heta_2 = 5\omega t$.They meet when the s

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All of the above

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(D) Case $1$: When both move in opposite directions.Let the angular positions be $	heta_1 = \omega t$ and $	heta_2 = 5\omega t$.They meet when the sum of their angular displacements equals $2\pi$ radians: $	heta_2 + 	heta_1 = 2\pi$.$5\omega t + \omega t = 2\pi \implies 6\omega t = 2\pi \implies \omega t = \frac{\pi}{3} = 60^{\circ}$.Thus,they cross at points subtending $60^{\circ}$ at the center.The time interval between crossings is $T = \frac{2\pi}{\omega_{rel}} = \frac{2\pi}{5\omega + \omega} = \frac{2\pi}{6\omega} = \frac{\pi}{3\omega}$.Case $2$: When both move in the same direction.They meet when the difference of their angular displacements equals $2\pi$ radians: $	heta_2 - 	heta_1 = 2\pi$.$5\omega t - \omega t = 2\pi \implies 4\omega t = 2\pi \implies \omega t = \frac{\pi}{2} = 90^{\circ}$.Thus,they cross at points subtending $90^{\circ}$ at the center.Since all statements are correct,the answer is $D$.

Two particles move on a circular path (one just inside and the other just outside) with angular velocities $\omega$ and $5\omega$ starting from the same point. Then

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