| ⇦ | 
| ⇨ |
The instantaneous angular position of a point on a rotating wheel is given by the equation θ(t) = 2t³ – 6t². The torque on the wheel becomes zero at
Options
(a) t = 1 s
(b) t = 0.5 s
(c) t = 0.25 s
(d) t = 2 s
Correct Answer:
t = 1 s
Explanation:
When angular acceleration is zero then torque on the wheel becomes zero.
θ(t) = 2t – 6t
dθ / dt = 6t – 12t
= dθ / dt = 12t – 12 = 0
t = 1 s
Related Questions: - The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV
- In R-L-C series circuit, the potential differences across each element is 20 V.
- 34 gram of water at 30⁰C is poured into a calorimeter at 15⁰C.
- The magnifying power of a telescope is 9. When it is adjusted for parallel rays
- A parallel plate condenser with a dielectric of dielectric constant K between the plates
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
- The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV
- In R-L-C series circuit, the potential differences across each element is 20 V.
- 34 gram of water at 30⁰C is poured into a calorimeter at 15⁰C.
- The magnifying power of a telescope is 9. When it is adjusted for parallel rays
- A parallel plate condenser with a dielectric of dielectric constant K between the plates
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