Q164 / 200 · MHT CET - Medical 2026 (21 April Shift 2) · Physics
Two particles A and B move in concentric circles of radii $r_1$ and $r_2$ respectively in such a way that position of particles A, B and the centre of the circle (O) always lie on a straight line, the ratio of the their angular velocities ($\omega_1 : \omega_2$) is
- A$r_1 : r_2$
- B$r_2 : r_1$
- C$1 : 2$
- Correct answerD$1 : 1$
Solution
Since particles A, B and the centre O always lie on a straight line, the angular displacement of both particles in any given time interval must be the same.
Let $\theta_1$ and $\theta_2$ be the angular displacements of particles A and B respectively. We have:
$\theta_1 = \theta_2$
Differentiating with respect to time $t$, we get:
$\dfrac{d\theta_1}{dt} = \dfrac{d\theta_2}{dt}$
$\omega_1 = \omega_2$
Therefore, the ratio of their angular velocities is $\omega_1 : \omega_2 = 1 : 1$.
Answer: $1 : 1$
Let $\theta_1$ and $\theta_2$ be the angular displacements of particles A and B respectively. We have:
$\theta_1 = \theta_2$
Differentiating with respect to time $t$, we get:
$\dfrac{d\theta_1}{dt} = \dfrac{d\theta_2}{dt}$
$\omega_1 = \omega_2$
Therefore, the ratio of their angular velocities is $\omega_1 : \omega_2 = 1 : 1$.
Answer: $1 : 1$
Hence, the correct answer is option: 'D'.