| ⇦ | 
| ⇨ |
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 moment of inertia of a thin uniform rod of mass M and length L about an axis
- At what distance from the centre of earth, the value of acceleration due to gravity
- A transformer rated at 10 kW is used to connect a 5 kV transmission line to a 240 V
- Two radioactive materials X₁ and X₂ have decay constants 5λ and λ respectively
- The output of OR gate is 1(one)
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The moment of inertia of a thin uniform rod of mass M and length L about an axis
- At what distance from the centre of earth, the value of acceleration due to gravity
- A transformer rated at 10 kW is used to connect a 5 kV transmission line to a 240 V
- Two radioactive materials X₁ and X₂ have decay constants 5λ and λ respectively
- The output of OR gate is 1(one)
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