| ⇦ | 
| ⇨ |
Two springs with spring constants K₁=1500 N/m and K₂=3000 N/m are stretched by the same force. The ratio of potential energy stored in springs will be
Options
(a) 1:2
(b) 2:1
(c) 4:1
(d) 1:4
Correct Answer:
2:1
Explanation:
Force F=-Kxdisplacement (x),
Potential Energy = U =(1/2) Kx²
U = (1/2) K(F/K)² = (1/2)(F²/K)
U₁/U₂=(1/2)(F²/K₁)x(2K₂/F²)
K₂/K₁=3000/1500=2/1.
Related Questions: - The magnetic susceptibility of a material of a rod is 299. Permeability of vacuum
- In communication with help of antenna if height is doubled then the range
- When ₉₀Th²²⁸ transforms to ₈₃Bi²¹², the number of emitted α and β particles
- Resistances n, each r Ω, when connected in parallel give an equivalent resistance
- which of the following can not be emitted by radioactive substances during their decay?
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The magnetic susceptibility of a material of a rod is 299. Permeability of vacuum
- In communication with help of antenna if height is doubled then the range
- When ₉₀Th²²⁸ transforms to ₈₃Bi²¹², the number of emitted α and β particles
- Resistances n, each r Ω, when connected in parallel give an equivalent resistance
- which of the following can not be emitted by radioactive substances during their decay?
Topics: Laws of Motion (103)
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