| ⇦ |
| ⇨ |
A thin circular ring of mass M and radius R rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity ?. Four small spheres each of mass m (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be
Options
(a) M?/4m
(b) [M+4m/M] ?
(c) [M/M-4m] ?
(d) [M/M+4m] ?
Correct Answer:
[M/M+4m] ?
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - In an orbital motion, the angular momentum vector is
- What will be the ratio of the distance moved by a freely falling body from rest in 4th
- A vibrating string of certain length l under a tension T resonates with a mode
- A train of mass 2x 10⁵ kg has a constant speed of 20 m/s up a hill inclined
- Weber/m² is equal to
Topics: Motion of system of Particles and Rigid Body
(73)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In an orbital motion, the angular momentum vector is
- What will be the ratio of the distance moved by a freely falling body from rest in 4th
- A vibrating string of certain length l under a tension T resonates with a mode
- A train of mass 2x 10⁵ kg has a constant speed of 20 m/s up a hill inclined
- Weber/m² is equal to
Topics: Motion of system of Particles and Rigid Body (73)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply