| ⇦ | 
| ⇨ |
If a body is rolling on the surface, its total energy will be
Options
(a)(1/2) mv² – (Gm₁m₂/R)
(b) (1/2) mv² + (1/2) Iω²
(c) (1/2) mv² – (1/2) mR²ω²
(d) None
Correct Answer:
(1/2) mv² + (1/2) Iω²
Explanation:
Total energy = P.E. + K.E. of (linear motion + Rotation)
= 0 + (1/2) mv² + (1/2) Iω²
Total energy = (1/2) mv² + (1/2) Iω².
Related Questions: - The dimensional formula for Boltzmann’s constant is
- A nucleus ᴢXᴬ emits an α- particle with velocity v. The recoil speed of the daughter
- Acceleration of a particle, executing SHM, at its mean position is
- What is the wavelength of light for the least energetic photon emitted
- In a potentiometer experiment, the balancing with a cell is at length 240 cm.
Topics: Work Energy and Power
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The dimensional formula for Boltzmann’s constant is
- A nucleus ᴢXᴬ emits an α- particle with velocity v. The recoil speed of the daughter
- Acceleration of a particle, executing SHM, at its mean position is
- What is the wavelength of light for the least energetic photon emitted
- In a potentiometer experiment, the balancing with a cell is at length 240 cm.
Topics: Work Energy and Power (94)
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