| ⇦ | 
| ⇨ |
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: - Diameter of the objective of a telescope is 200 cm. What is the resolving power
- When you make ice cubes, the entropy of water
- An electric kettle has two heating elements. One brings it to boil in ten minutes
- Weber ampere per metre is equal to
- When temperature of an ideal gas is increased from 27⁰C to 227⁰C,
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Diameter of the objective of a telescope is 200 cm. What is the resolving power
- When you make ice cubes, the entropy of water
- An electric kettle has two heating elements. One brings it to boil in ten minutes
- Weber ampere per metre is equal to
- When temperature of an ideal gas is increased from 27⁰C to 227⁰C,
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